BudgetScan

Out-of-bounds read in skipSpace at buffer end

high swesweep-fasthttp · headerscanner.go · skipSpace · run #4

Description

skipSpace() dereferences s.b[s.r] without checking that s.r is less than len(s.b). readContinuedLineSlice() calls skipSpace() while processing folded headers; if a malformed header or stale scanner state places s.r at the end of s.b, the read panics with an index out of range, causing denial of service.

Triage

Verdict
INVALID confidence 0.0
Model
qwen/qwen3.8-27b
Rounds
6
Cost
$0.0780 · 1146.5 s
[ARBITER] The vulnerability is not reachable because the scanner state maintains an invariant that prevents `s.r` from being equal to `len(s.b)` when `skipSpace` is called. 

1. **Buffer Initialization**: `headerScanner.next()` truncates the buffer `s.b` to end exactly after the first `\r\n\r\n` sequence. Thus, the last four bytes of `s.b` are `\r`, `\n`, `\r`, `\n`.
2. **Loop Structure**: In `readContinuedLineSlice`, the loop is `for s.skipSpace() { ... line := s.readLine() ... }`.
3. **Termination Condition**: Before `readLine` can advance `s.r` to the end of the buffer, it must consume a newline. However, the loop condition checks `s.skipSpace()` *before* calling `readLine`. 
4. **Invariant**: As the parser approaches the end of the buffer (the final `\r\n\r\n`), the read position `s.r` will eventually point to one of the non-space/non-tab characters in the terminator (specifically `\r` or `\n`). Since `skipSpace` only consumes spaces and tabs, it will return `false` when it encounters `\r` or `\n`. Consequently, the loop terminates before `readLine` is called in a state where `s.r` could reach the end of the buffer during the loop iteration itself. Even if `readLine` consumed the final newline to make `s.r == len(s.b)`, the next iteration of the `for` loop would call `skipSpace`, but this scenario is prevented because the loop structure relies on `skipSpace` to detect the *start* of a continuation line, and the buffer end is marked by non-whitespace characters that stop the skip process. More simply: `readLine` always consumes a `\n` to advance `s.r`. If `s.r` becomes `len(s.b)`, it means the very last `\n` was consumed. But `next()` ensures the buffer ends with `\r\n\r\n`. The parser processes headers line by line. When it reaches the final `\n` of the last header value, `readLine` consumes it. The next step in `next()` is to check for continuation. But the buffer ends. Actually, the key is that `skipSpace` is called in the `for` condition. If `s.r` were at the end, `s.b[s.r]` would panic. But `s.r` only advances via `readLine` inside the loop. `readLine` finds a `\n`. If it consumes the final `\n` of the buffer, `s.r` becomes `len(s.b)`. The loop then iterates back to `skipSpace`. At this point, `s.r == len(s.b)`, and `s.b[s.r]` is an out-of-bounds read. 

Wait, let's re-evaluate carefully. 
Buffer: `Header: Value\r\n\r\n`. 
`readLine` reads `Header: Value`. `s.r` points to first `\r`. 
`next` returns true. 
Next call to `next` (if any) or internal logic? `headerScanner` is used in a loop in `Header.Parse`. 
Actually, `readContinuedLineSlice` is called inside `next`. 
Let's trace `next` for the *last* header. 
Buffer ends with `\r\n\r\n`. 
`readContinuedLineSlice` calls `readLine`. It finds `\n` at `L-1`? No, `bytes.IndexByte(s.b[s.r:], '\n')`. 
If the header is `A: B`, followed by `\r\n`, then `\r\n`. 
`readLine` consumes `A: B\r\n`. `s.r` is now at the index of the *second* `\r` (the start of the final `\r\n` terminator). 
Back in `readContinuedLineSlice`: 
`if len(s.b)-s.r > 1 { ... }` 
`s.r` is at `L-4` (index of first `\r` of the final `\r\n\r\n`? No. 
Let's map indices. 
`... [H][e][a][d][e][r][:][ ][V][a][l][u][e][\r][\n][\r][\n]` 
Indices: `... 0..12, 13(\r), 14(\n), 15(\r), 16(\n)`. `L=17`. 
`readLine` finds `\n` at 14. `line = s.b[0:15]`. `s.r = 15`. 
`s.r` is 15. `s.b[15]` is `\r`. `s.b[16]` is `\n`. 
`len(s.b)-s.r = 17-15 = 2`. 
`peek = s.b[15:17]` which is `[\r, \n]`. 
`peek[0] == '\r' && peek[1] == '\n'` is true. 
So it returns `trim(line), nil`. 
It does **not** enter the `for s.skipSpace()` loop. 
Thus, `skipSpace` is never called when `s.r` is near the end of the buffer in this specific case. 

What if there is a continuation? 
`Header: Value1\r\n  Value2\r\n\r\n`. 
`readLine` reads `Header: Value1`. `s.r` at `\r` of `\r\n` after `Value1`. 
`peek` check: `s.b[s.r]` is `\r`, `s.b[s.r+1]` is `\n`. Wait, `readLine` consumed the `\n`? 
Yes, `s.r` advances past `\n`. So `s.r` is at the space/tab of the continuation? 
No, `readLine` returns line *including* `\n`. It strips it for return, but `s.r` advances. 
So `s.r` is at the first character of the next line. 
If the next line is `  Value2`, `s.b[s.r]` is ` `. 
`peek` check: `isASCIILetter(' ')` is false. `peek[0]=='\n'` false. `peek[0]=='\r'` false. 
So it proceeds to `mline = trim(line)`. 
Then `for s.skipSpace()`. 
`skipSpace` consumes spaces/tabs. 
Then `readLine` reads `Value2`. 
Then loop checks `skipSpace` again. 
If `Value2` is the last part, the buffer continues with `\r\n\r\n`. 
After `readLine` of `Value2`, `s.r` is at the `\r` of the final `\r\n\r\n`. 
Loop checks `skipSpace`. `s.b[s.r]` is `\r`. `\r` is not space/tab. `skipSpace` returns false. 
Loop terminates. 
No OOB. 

Therefore, the `for s.skipSpace()` loop always terminates because it encounters a non-space/tab character (either the start of a new header key or the `\r` of the final terminator) before `s.r` can reach `len(s.b)`. The only way `s.r` reaches `len(s.b)` is if `readLine` consumes the very last byte, but `readLine` requires a `\n` to complete, and the buffer always has a `\r` before that final `\n` in the terminator structure which breaks the space loop.
All rounds
[
  {
    "finding_title": "Out-of-bounds read in skipSpace at buffer end",
    "verdict": "UNCERTAIN",
    "reasoning": "The reported vulnerability is an out-of-bounds read in `skipSpace` when `s.r` equals `len(s.b)`. I need to determine if `readContinuedLineSlice` can call `skipSpace` in a state where `s.r` is at the end of the buffer.\n\n1. **Data Flow Analysis**: `readContinuedLineSlice` is called from `next()`. It first calls `readLine()`.\n2. **`readLine` Behavior**: `readLine` finds the next `\\n`. If found, `s.r` is advanced past the `\\n`. If `s.b` does NOT contain a `\\n` after `s.r` (i.e., `bytes.IndexByte` returns -1), the code executes `searchStart = len(s.b) - s.r` and loops. However, `next()` is generally expected to be called only when `ErrNeedMore` logic in the caller handles partial reads, OR the buffer is guaranteed to contain complete headers. Let's look at `next()`'s initialization: it requires `strCRLFCRLF` (\\r\\n\\r\\n) to exist. This means the end of the headers is present. Thus, `readLine` will *always* find a `\\n` because the `\\r\\n\\r\\n` marker contains `\\n`. Therefore, `readLine` will *never* hit the `searchStart = len(s.b) - s.r` loop condition that implies no newline found, because the buffer always ends with `\\r\\n\\r\\n` when `next()` is processing headers. Wait, `next()` truncates `s.b` to `i+4` where `i` is the index of the *first* `\\r\\n\\r\\n`. So `s.b` always ends with `\\r\\n\\r\\n`.\n3. **`readContinuedLineSlice` Logic**: \n   - It reads a line. If the line is a header key/value, it checks if the *next* byte (after the line) is a space/tab (indicating continuation).\n   - Specifically, it checks `if len(s.b)-s.r > 1`. Then it peeks `s.b[s.r : s.r+2]`.\n   - If the peeked character is NOT an ASCII letter, `\\n`, or `\\r\\n`, it enters the continuation loop.\n   - Inside the loop: `for s.skipSpace() { ... }`.\n   - **Crucial Check**: Can `s.r` be at `len(s.b)` when `skipSpace` is called?\n   - `skipSpace` is called to skip spaces/tabs. This implies that the *previous* character checked was a space or tab, or we are expecting one. \n   - Let's trace a malformed input: `\"Key: Value \\r\\n\\r\\n\"`. \n     - `readLine` consumes `\"Key: Value \\r\\n\"`. `s.r` points to `\"\\r\\n\"` (the final CRLF of the header block? No, `s.b` was truncated to the first `\\r\\n\\r\\n`). \n     - Actually, `s.b` ends with `\\r\\n\\r\\n`. `readLine` consumes up to the first `\\n` it finds. \n     - If the input is `\"Key: Value \\r\\n\\r\\n\"`, `readLine` finds the `\\n` at index of `Value \\n`. It returns `\"Key: Value \"`. `s.r` now points to the `\\r` of the *next* line (which is part of the `\\r\\n\\r\\n` terminator). \n     - `readContinuedLineSlice` checks `len(s.b)-s.r > 1`. Yes. \n     - Peek: `s.b[s.r]` is `\\r`, `s.b[s.r+1]` is `\\n`. The condition `peek[0] == '\\r' && peek[1] == '\\n'` is TRUE. It returns `trim(line)`, `nil`. It does NOT enter the `skipSpace` loop. \n   - What if the input is `\"Key: Value \\r\\n \\r\\n\\r\\n\"` (folded header)? \n     - `readLine` reads `\"Key: Value \\r\\n\"`. `s.r` points to `\" \\r\\n\\r\\n\"`. \n     - Peek: `s.b[s.r]` is ` ` (space). `peek[0]` is space. The condition `isASCIILetter(peek[0])` is false. `peek[0] == '\\n'` is false. `peek[0] == '\\r'` is false. So it enters the continuation loop. \n     - `skipSpace` is called. `s.r` points to ` `. `s.b[s.r]` is valid. It skips the space. `s.r` now points to `\\r`. \n     - `skipSpace` checks `s.b[s.r]` (which is `\\r`). It breaks. Returns `true`. \n     - `readLine` is called again. It reads `\" \\r\\n\"`? No, `readLine` reads up to `\\n`. It finds `\\n` at the end of the continuation line. \n     - Eventually, `s.r` will point to the final `\\r\\n\\r\\n` or similar. \n   - **The Critical Failure Case**: What if the buffer ends with a space or tab *right before* the `\\r\\n\\r\\n` terminator, such that `skipSpace` consumes it and lands exactly on `len(s.b)`?\n   - Since `s.b` is truncated to include the `\\r\\n\\r\\n` terminator, the last two bytes are always `\\r\\n` (part of the `\\r\\n\\r\\n`). Wait, `\\r\\n\\r\\n` is 4 bytes. The last two are `\\r\\n`. \n   - If `skipSpace` is called, it means the previous line ended, and we are looking for continuation. Continuation lines start with space/tab. \n   - If the input is `\"Key: Val \\r\\n \\r\\n\\r\\n\"`. \n     - Line 1: `Key: Val`. `s.r` at ` \\r\\n\\r\\n`. \n     - Continuation loop. `skipSpace` skips ` `. `s.r` at `\\r\\n\\r\\n`. \n     - `readLine` reads `\"\\r\\n\"`? `bytes.IndexByte(s.b[s.r:], '\\n')`. `s.r` is at `\\r`. Next is `\\n`. Found at offset 1. `line = s.b[s.r : s.r+1+1]`? No, `i=1`. `line = s.b[s.r : s.r+2]`. `line` is `\"\\r\\n\"`. \n     - `readLine` drops `\\n` and `\\r`. Returns empty slice? `line[:0]`? \n     - `trim(line)` on empty slice returns empty. \n     - `skipSpace` is called again? `for s.skipSpace()`. \n     - `s.r` is now at the second `\\r` of the `\\r\\n\\r\\n`? No. \n     - Let's trace indices carefully. \n     - Input: `K:V \\r\\n \\r\\n\\r\\n`. \n     - `s.b` truncated to full length. \n     - `readLine`: finds `\\n` at index 5 (0-based: K:V \\r \\n). Indices: 0:K, 1::, 2:V, 3: , 4:\\r, 5:\\n. `i=5`. `line = s.b[0:6]` (K:V \\r\\n). `s.r = 6`. \n     - `readContinuedLineSlice`: `line` is not empty. Has colon. \n     - `len(s.b)-s.r` is `12-6 = 6 > 1`. \n     - `peek = s.b[6:8]`. `s.b[6]` is ` ` (space). `s.b[7]` is `\\r`. \n     - Condition: `peek[0]` is space. Not letter, not `\\n`. `peek[0]` is not `\\r`. Condition fails. Enter loop. \n     - `skipSpace`: `c = s.b[6]` (space). `s.r` becomes 7. `skipped=true`. Loop. `c = s.b[7]` (\\r). Break. Return true. \n     - `readLine`: `s.r=7`. `bytes.IndexByte(s.b[7:], '\\n')`. `s.b[7]` is `\\r`, `s.b[8]` is `\\n`. Index is 1. `i=1`. `line = s.b[7:9]` (`\\r\\n`). `s.r = 9`. \n     - `readLine` drops `\\r\\n`. Returns empty `line`. \n     - `trim(line)` is empty. `mline` appended with space and empty. \n     - Loop `for s.skipSpace()`. \n     - `s.r` is 9. `s.b` is `...\\r\\n\\r\\n`? No, `s.b` ended with `\\r\\n\\r\\n`. \n     - Original `s.b`: `K:V \\r\\n \\r\\n\\r\\n`. Length 12. Indices 0-11. \n     - 8: \\n, 9: \\r, 10: \\n. Wait, `\\r\\n\\r\\n` is 4 bytes. \n     - Indices: \n       0: K \n       1: : \n       2: V \n       3: (space) \n       4: \\r \n       5: \\n \n       6: (space) \n       7: \\r \n       8: \\n \n       9: \\r \n       10: \\n \n       (This is `\\r\\n\\r\\n` at 4,5 and 9,10? No. `\\r\\n\\r\\n` is `\\r,\\n,\\r,\\n`.) \n       So 4:\\r, 5:\\n, 9:\\r, 10:\\n? No, the sequence is `\\r\\n\\r\\n`. \n       If the input is `K:V \\r\\n \\r\\n\\r\\n`, the bytes are: \n       `K, :, V, ' ', \\r, \\n, ' ', \\r, \\n, \\r, \\n`. \n       Wait, `\\r\\n\\r\\n` is 4 bytes. \n       The header ends with `\\r\\n\\r\\n`. \n       So the bytes are: \n       `K, :, V, ' ', \\r, \\n, ' ', \\r, \\n, \\r, \\n`? \n       No, the continuation line is ` ` (space) followed by `\\r\\n`? \n       Standard folding: `Key: Value \\r\\n Continuation`. \n       If the continuation is empty, it's `Key: Value \\r\\n \\r\\n`. \n       So bytes: `K, :, V, ' ', \\r, \\n, ' ', \\r, \\n`. \n       Then the terminator `\\r\\n\\r\\n`? \n       The `next()` function finds the *first* `\\r\\n\\r\\n`. \n       In `K:V \\r\\n \\r\\n\\r\\n`, the substring `\\r\\n\\r\\n` appears at index 6? \n       Bytes: \n       0: K \n       1: : \n       2: V \n       3: ' ' \n       4: \\r \n       5: \\n \n       6: ' ' \n       7: \\r \n       8: \\n \n       9: \\r \n       10: \\n \n       The first `\\r\\n\\r\\n` is at indices 7,8,9,10? No, `\\r\\n` is 7,8. Then `\\r\\n` is 9,10. So `\\r\\n\\r\\n` is at 7? \n       `s.b[7]=\\r, s.b[8]=\\n, s.b[9]=\\r, s.b[10]=\\n`. \n       So `i=7`. `s.b` truncated to `7+4 = 11`. \n       `s.b` is `K:V \\r\\n \\r\\n\\r\\n` (11 bytes, indices 0-10). \n       \n       Back to trace: \n       `readLine` (first): `s.r=0`. Finds `\\n` at 5. `line = s.b[0:6]`. `s.r=6`. \n       `readContinuedLineSlice`: \n       `peek = s.b[6:8]`. `s.b[6]=' ', s.b[7]='\\r'`. \n       Enter loop. \n       `skipSpace`: `s.r=6`. `c=' '`. `s.r=7`. `c='\\r'`. Break. Return true. \n       `readLine`: `s.r=7`. Finds `\\n` at offset 1 (index 8). `line = s.b[7:9]` (`\\r\\n`). `s.r=9`. \n       `readLine` returns empty. \n       Loop `for s.skipSpace()`. \n       `s.r=9`. `s.b[9]='\\r'`. `c != ' '`. Break. Return false. \n       Loop ends. \n       No OOB. \n\n   - **Can `s.r` reach `len(s.b)` in `skipSpace`?** \n     - `skipSpace` reads `s.b[s.r]`. \n     - This happens if the loop condition is entered. \n     - The loop is entered if the previous check failed (i.e., we are in a continuation state). \n     - For `skipSpace` to read `s.b[len(s.b)]`, `s.r` must be `len(s.b)` *at the start* of `skipSpace`. \n     - This implies the previous operation left `s.r` at `len(s.b)`. \n     - The previous operation is `readLine` (which advances `s.r` past `\\n`) or `skipSpace` itself (if it skipped to the end). \n     - If `skipSpace` skipped to the end, it would have returned `false` (if the last char wasn't space) or crashed (if it tried to read beyond). \n     - But if `readLine` advanced `s.r` to `len(s.b)`, can `skipSpace` be called? \n     - `readLine` advances `s.r` past `\\n`. Since `s.b` ends with `\\r\\n\\r\\n`, the last `\\n` is at `len(s.b)-1`? No, `len(s.b)-1` is `\\n` (last byte). `len(s.b)-2` is `\\r`. \n     - So `readLine` will always find a `\\n` within `s.b`. It will set `s.r` to `index_of_n + 1`. \n     - The max `index_of_n` is `len(s.b)-1`. So max `s.r` is `len(s.b)`. \n     - If `s.r` becomes `len(s.b)`, the next call to `readContinuedLineSlice`? No, `next()` calls `readContinuedLineSlice` once per header. \n     - Inside `readContinuedLineSlice`, after `readLine`, `s.r` might be `len(s.b)`. \n     - Then `len(s.b)-s.r > 1` is `0 > 1` which is False. \n     - So it does *not* enter the `peek` block. \n     - It goes to `mline := trim(line)`. \n     - Then `for s.skipSpace()`. \n     - **Here is the bug.** If `s.r == len(s.b)`, `skipSpace` is called. \n     - `skipSpace` does `c := s.b[s.r]`. `s.r` is `len(s.b)`. **PANIC.** \n     \n     - **Can this happen?** \n     - We need `readLine` to consume the *entire* buffer, leaving `s.r == len(s.b)`. \n     - This happens if the line consumed was the *last* line, and it ended with `\\n` at `len(s.b)-1`. \n     - For this to be a *continuation* line, the logic flow must have reached the `for s.skipSpace()` loop. \n     - This requires that the *previous* line was a key/value line, and the check `if len(s.b)-s.r > 1` passed, and the peek check failed (indicating a space/tab). \n     - BUT, if the peek check failed, it means `s.b[s.r]` was a space or tab. \n     - So `s.r < len(s.b)` at that point. \n     - Then `skipSpace` is called. It skips spaces/tabs. \n     - Then `readLine` is called. \n     - If `readLine` consumes until the end of the buffer (`s.r` becomes `len(s.b)`), and the loop iterates again (`for s.skipSpace()`), then `skipSpace` is called with `s.r == len(s.b)`. \n     - Does `readLine` leave `s.r == len(s.b)`? \n     - Yes, if the line ends with `\\n` at the last byte. \n     - And does the loop iterate? \n     - `for s.skipSpace()`. \n     - `skipSpace` is called. It *immediately* panics. \n     - So the condition is: \n       1. Header starts. \n       2. Continuation detected (previous line ended with space/tab before `\\r\\n`? No, `readLine` strips `\\r\\n`). \n       3. The continuation line itself ends with `\\n` at the very end of the buffer. \n       4. After reading that line, `s.r == len(s.b)`. \n       5. The `for` loop condition `s.skipSpace()` is evaluated. \n       6. `skipSpace` accesses `s.b[len(s.b)]`. \n     \n     - **Constructing the Payload**: \n       `\"Key: Val \\r\\nCont\\n\"`? \n       `next()` requires `\\r\\n\\r\\n`. \n       So input must be `\"Key: Val \\r\\nCont\\n\\r\\n\\r\\n\"`? \n       No, `next()` finds the *first* `\\r\\n\\r\\n`. \n       If input is `\"Key: Val \\r\\nCont\\n\\r\\n\\r\\n\"`. \n       `i` finds `\\r\\n\\r\\n` at index of the final `\\r`. \n       `s.b` truncated to that. \n       `readLine` 1: `Key: Val `. `s.r` at `Cont...`. \n       Peek: `C` (letter). Condition `isASCIILetter` TRUE. \n       Returns. No loop. \n       \n       We need the peek check to FAIL. \n       So `s.b[s.r]` must be space or tab. \n       Input: `\"Key: Val \\r\\n \\r\\n\\r\\n\"` (Empty continuation). \n       We traced this earlier. \n       `readLine` 1: `Key: Val`. `s.r` at ` ` (space). \n       Peek: ` ` (space). Fail. Enter loop. \n       `skipSpace`: Skips ` `. `s.r` at `\\r`. \n       `readLine`: Reads `\\r\\n`. `s.r` at next `\\r`? \n       Let's re-trace `\"Key: Val \\r\\n \\r\\n\\r\\n\"`. \n       Bytes: `K,:,V,' ',\\r,\\n,' ',\\r,\\n,\\r,\\n`. \n       `next()` finds `\\r\\n\\r\\n` at index 7? \n       `s.b[7]=\\r, s.b[8]=\\n, s.b[9]=\\r, s.b[10]=\\n`. \n       `i=7`. `s.b` length 11. \n       `readLine` 1: `\\n` at 5. `line` = `Key: Val ` (0-5). `s.r=6`. \n       `peek`: `s.b[6]=' ', s.b[7]='\\r'`. \n       Loop. \n       `skipSpace`: `s.r=6` (space). `s.r=7`. `s.b[7]='\\r'`. Break. \n       `readLine`: `s.r=7`. `\\n` at 8 (offset 1). `line` = `s.b[7:9]` (`\\r\\n`). `s.r=9`. \n       `readLine` returns empty. \n       Loop `for s.skipSpace()`. \n       `s.r=9`. `s.b[9]='\\r'`. \n       `skipSpace`: `c = s.b[9]` (`\\r`). `c != ' '`. Break. Return false. \n       Loop ends. \n       No Panic. \n\n     - We need `s.r` to be `len(s.b)` when `skipSpace` is called. \n     - This requires `readLine` to set `s.r` to `len(s.b)`. \n     - This requires the last `\\n` to be at `len(s.b)-1`. \n     - In `\"Key: Val \\r\\n \\r\\n\\r\\n\"`, the last `\\n` is at 10. `len(s.b)` is 11. `s.r` became 9. \n     - Why 9? Because `readLine` consumed `\\r\\n` at 7,8. \n     - The `\\r\\n\\r\\n` terminator is 4 bytes. \n     - If the continuation line is *empty*, it is represented by `\\r\\n`? \n     - If we have `\"Key: Val \\r\\n \\r\\n\"` (no trailing `\\r\\n` in `s.b` because `next` truncated?). \n     - `next` truncates to the *first* `\\r\\n\\r\\n`. \n     - If input is `\"Key: Val \\r\\n \\r\\n\\r\\n\"`, the first `\\r\\n\\r\\n` is at 7. \n     - So `s.b` ends at 10. \n     - The bytes 7,8 are `\\r,\\n`. Bytes 9,10 are `\\r,\\n`. \n     - `readLine` at `s.r=7` consumes 7,8. `s.r=9`. \n     - It does NOT consume 9,10. \n     - So `s.r` is 9, `len` is 11. \n     - `skipSpace` at 9 reads `\\r`. Breaks. \n     \n     - What if the continuation line contains data? \n     - `\"Key: Val \\r\\n Data\\r\\n\\r\\n\"`. \n     - `next` finds `\\r\\n\\r\\n` at 12? \n     - Bytes: `K,:,V,' ',\\r,\\n,' ',D,a,t,a,\\r,\\n,\\r,\\n`. \n     - `i=11`? `s.b[11]=\\r, s.b[12]=\\n, s.b[13]=\\r, s.b[14]=\\n`. \n     - `i=11`. `s.b` length 15. \n     - `readLine` 1: `s.r=0`. `\\n` at 5. `s.r=6`. \n     - `peek`: `s.b[6]=' ', s.b[7]='D'`. Fail. \n     - `skipSpace`: `s.r=6`->7. \n     - `readLine`: `s.r=7`. `\\n` at 12 (offset 5). `line` = `Data\\r\\n`. `s.r=13`. \n     - Loop `for s.skipSpace()`. \n     - `s.r=13`. `s.b[13]='\\r'`. \n     - `skipSpace` breaks. \n     \n     - It seems `skipSpace` is always called when `s.r` points to a character that is *not* space/tab (if it breaks) or space/tab (if it continues). \n     - The only way to panic is if `s.r == len(s.b)` *before* the read. \n     - This implies `readLine` set `s.r` to `len(s.b)`. \n     - This implies `readLine` consumed the `\\n` at `len(s.b)-1`. \n     - In the examples above, `readLine` stopped *before* the end of `s.b` because there were bytes left (the second half of `\\r\\n\\r\\n`). \n     - Can `readLine` consume up to `len(s.b)`? \n     - Only if the `\\n` it finds is at `len(s.b)-1`. \n     - In `s.b` ending with `\\r\\n\\r\\n`, the last `\\n` is at `len(s.b)-1`. \n     - But `readLine` finds the *first* `\\n` after `s.r`. \n     - If `s.r` is at the start of the *second* `\\r\\n` of the terminator, it finds `\\n` at `s.r+1`. \n     - Then `s.r` becomes `s.r+2`. \n     - If `s.r+2 == len(s.b)`, then `s.r` was `len(s.b)-2`. \n     - So we need `s.r` to be `len(s.b)-2` before `readLine`. \n     - And we need to be in the `for s.skipSpace()` loop. \n     - This requires the previous iteration of the loop (or the entry) to have set `s.r` to `len(s.b)-2`. \n     - Entry to loop: `skipSpace` was called. \n     - If `skipSpace` skipped to `len(s.b)-2`, it means `s.b[len(s.b)-2]` was space/tab? \n     - But `s.b` ends with `\\r\\n\\r\\n`. \n     - `s.b[len(s.b)-2]` is `\\r`. \n     - So `skipSpace` cannot skip to `len(s.b)-2` because `\\r` is not space/tab. \n     - Therefore, `skipSpace` will always stop at a non-space character, which, given the structure, will be `\\r` or a letter. \n     - It will never skip *past* a `\\r` or `\\n`. \n     - So `s.r` will never be `len(s.b)` when `skipSpace` is called, because `skipSpace` itself ensures `s.r` points to a valid byte (the one that broke the loop) or it would have panicked earlier. \n     - Wait. `skipSpace` *reads* `s.b[s.r]`. If `s.r` is valid, it breaks or increments. \n     - If it increments, it must ensure the next `s.r` is valid. \n     - If `s.b[s.r]` is space, it increments. \n     - Can it increment to `len(s.b)`? \n     - Only if `s.b[len(s.b)-1]` is space/tab. \n     - But `s.b` ends with `\\r\\n\\r\\n`. Last byte is `\\n`. \n     - So `s.b[len(s.b)-1]` is NOT space/tab. \n     - So `skipSpace` will *always* break before `s.r` reaches `len(s.b)`. \n     - Specifically, it will break at `\\r` or `\\n` or a letter. \n     - Therefore, `s.r` is *always* `< len(s.b)` after `skipSpace` (or at the position of the non-space char). \n     - And `skipSpace` is only called if `s.r < len(s.b)`? \n     - No, `skipSpace` is called in the `for` condition. \n     - If `s.r` was `len(s.b)` before the call, it panics. \n     - Can `s.r` be `len(s.b)` before the call? \n     - Only if `readLine` set it there. \n     - `readLine` sets `s.r` to `index_of_n + 1`. \n     - Max `index_of_n` is `len(s.b)-1`. \n     - So max `s.r` is `len(s.b)`. \n     - Can `readLine` set `s.r` to `len(s.b)`? \n     - Yes, if it finds `\\n` at `len(s.b)-1`. \n     - Can it find `\\n` at `len(s.b)-1`? \n     - Yes, the last byte is `\\n`. \n     - But `readLine` finds the *first* `\\n` after `s.r`. \n     - If `s.r` is 0, and `\\n` is at 5, it stops at 5. \n     - It does NOT skip to the last `\\n`. \n     - So `s.r` becomes small. \n     - It only becomes `len(s.b)` if the *only* `\\n` in `s.b[s.r:]` is the last one. \n     - This happens if `s.r` is close to the end. \n     - Example: `s.r = len(s.b) - 2`. `s.b` ends `...\\r\\n`. \n     - `readLine` finds `\\n` at offset 1. `s.r` becomes `len(s.b)`. \n     - So, if we can get `s.r` to be `len(s.b)-2` before a `readLine` call inside the continuation loop, `readLine` will set `s.r` to `len(s.b)`. \n     - Then the `for s.skipSpace()` check will panic. \n     - How to get `s.r = len(s.b)-2`? \n     - Previous `skipSpace` or `readLine`. \n     - If previous `readLine` set it to `len(s.b)-2`? \n     - This requires the previous `\\n` to be at `len(s.b)-3`. \n     - In `...\\r\\n\\r\\n`, the `\\n`s are at `len-3` and `len-1`. \n     - If `s.r` was `len-3`, `readLine` finds `\\n` at offset 0? No, `s.b[len-3]` is `\\n`? \n     - Sequence `\\r,\\n,\\r,\\n` (indices `L-4, L-3, L-2, L-1`). \n     - If `s.r = L-4`. `readLine` finds `\\n` at offset 1. `s.r = L-3+1 = L-2`? No. \n     - `i=1`. `line = s.b[L-4 : L-4+2]`. `s.r = L-4+2 = L-2`. \n     - So `s.r` becomes `L-2`. \n     - Next `skipSpace`: `s.b[L-2]` is `\\r`. Breaks. \n     - So `s.r` remains `L-2`. \n     - Next `readLine`: `s.r=L-2`. `s.b[L-2]` is `\\r`. `s.b[L-1]` is `\\n`. \n     - `bytes.IndexByte(s.b[L-2:], '\\n')`. Offset 1. \n     - `line = s.b[L-2 : L]`. `s.r = L`. \n     - **BOOM.** `s.r` is `L`. \n     - Next `for s.skipSpace()`. \n     - **PANIC.** \n     \n     - So, is this path reachable? \n     - We need to reach `s.r = L-4` (start of the final `\\r\\n\\r\\n`) inside the continuation loop. \n     - This requires that the continuation logic was active for the line ending at `L-4`. \n     - This means the header was `Key: Val \\r\\n \\r\\n\\r\\n`? \n     - We traced this earlier. \n     - In that trace, `readLine` 1 set `s.r=6` (in a 11-byte buffer). `L=11`. `L-4=7`. \n     - `s.r` was 6. \n     - `skipSpace` moved `s.r` to 7 (`L-4`). \n     - `readLine` at `s.r=7` (L-4). \n     - Found `\\n` at offset 1. \n     - `s.r` became `7+2 = 9`. \n     - `L=11`. `L-2 = 9`. \n     - So `s.r` became `L-2`. \n     - `skipSpace` at `s.r=9` (`L-2`). \n     - `s.b[9]` is `\\r` (in `\\r\\n\\r\\n`, indices 7,8,9,10 are `\\r,\\n,\\r,\\n`). \n     - Wait. `L=11`. Indices 0-10. \n     - `\\r\\n\\r\\n` at 7,8,9,10. \n     - `s.b[7]=\\r, s.b[8]=\\n, s.b[9]=\\r, s.b[10]=\\n`. \n     - `readLine` at `s.r=7`. \n     - `s.b[7]` is `\\r`. `s.b[8]` is `\\n`. \n     - `bytes.IndexByte(s.b[7:], '\\n')`. Offset 1. \n     - `line = s.b[7:9]`. `s.r = 9`. \n     - `s.r` is 9. `L` is 11. `L-2 = 9`. \n     - Correct. \n     - `skipSpace` at `s.r=9`. \n     - `s.b[9]` is `\\r`. \n     - `c != ' '`. Break. \n     - Return false. \n     - Loop ends. \n     - **No Panic.** \n     \n     - Why did my previous deduction fail? \n     - I assumed `skipSpace` would be called *after* `readLine` set `s.r` to `L`. \n     - But `skipSpace` breaks immediately if the char is not space. \n     - The char at `L-2` is `\\r`. \n     - So `skipSpace` returns false. \n     - The loop terminates. \n     - `s.r` is `L-2`. \n     - `readContinuedLineSlice` returns. \n     - `next()` continues. \n     - It does NOT call `skipSpace` again with `s.r=L`. \n     \n     - So, the panic requires `s.b[s.r]` to be space/tab when `s.r == L-1`? \n     - No, `s.r` must be `L` *before* `skipSpace`. \n     - This requires `readLine` to set `s.r=L`. \n     - This requires `\\n` at `L-1` to be the *first* `\\n` found. \n     - This requires `s.r <= L-1` and no `\\n` in `s.b[s.r : L-1]`. \n     - In `\\r\\n\\r\\n`, the `\\n`s are at `L-3` and `L-1`. \n     - If `s.r` is `L-2`, `s.b[L-2]` is `\\r`. `s.b[L-1]` is `\\n`. \n     - `readLine` finds `\\n` at offset 1. `s.r` becomes `L`. \n     - So if we can get `s.r` to `L-2`, `readLine` makes it `L`. \n     - But we just showed that if `s.r` is `L-2`, `skipSpace` (called before `readLine`? No, `readLine` is called *inside* the loop body, `skipSpace` is the condition). \n     - Sequence: \n       1. `for s.skipSpace()` \n       2. `skipSpace` checks `s.r`. \n       3. If it skips, it increments. \n       4. `readLine` is called. \n       5. `s.r` advanced. \n       6. Loop condition `s.skipSpace()` called again. \n     \n     - If `s.r` is `L-2` at step 1: \n       - `skipSpace` sees `\\r`. Breaks. Returns false. \n       - Loop exits. \n       - `readLine` is NOT called. \n     \n     - So `readLine` is only called if `skipSpace` returns true. \n     - `skipSpace` returns true ONLY if it skipped at least one space/tab. \n     - So `s.b[s.r]` MUST be space/tab. \n     - In `\\r\\n\\r\\n`, there are NO space/tab bytes. \n     - So `skipSpace` will NEVER return true if `s.r` is in the final `\\r\\n\\r\\n` region. \n     - Therefore, `readLine` will NOT be called on the final `\\r\\n\\r\\n` region in a way that advances `s.r` to `L`. \n     - The loop will exit before `s.r` reaches `L` via `readLine`. \n     \n     - What if the buffer is NOT truncated to `\\r\\n\\r\\n`? \n     - `next()` DOES truncate. `s.b = s.b[:i]`. \n     - So `s.b` ALWAYS ends with `\\r\\n\\r\\n`. \n     \n     - Conclusion: The specific OOB read in `skipSpace` due to `s.r == len(s.b)` is NOT reachable because `skipSpace` will never be called when `s.r` is at the end of the buffer. The loop condition `skipSpace()` will fail (return false) when encountering the `\\r` of the terminator, preventing further `readLine` calls that might advance `s.r` to the end. Even if `readLine` advanced `s.r` to `L`, the *next* `skipSpace` would panic, but `readLine` only advances to `L` if it finds `\\n` at `L-1`. But `readLine` is only called if `skipSpace` returned true. `skipSpace` returns true only on space/tab. The terminator contains no space/tab. So the loop exits safely.\n\nVerdict: INVALID.",
    "elapsed": 208.7,
    "tokens": 12448,
    "tokens_in": 3098,
    "tokens_out": 9350,
    "cost_usd": 0.029351159999999998,
    "file": "headerscanner.go",
    "round": 1
  },
  {
    "finding_title": "Out-of-bounds read in skipSpace at buffer end",
    "verdict": "INVALID",
    "reasoning": "The reported vulnerability claims an out-of-bounds read in `skipSpace` when `s.r == len(s.b)`. I verified the data flow and the constraints imposed by `headerScanner.next()`. \n\n1. **Buffer Constraint**: `next()` initializes by finding the first occurrence of `strCRLFCRLF` (`\\r\\n\\r\\n`) and truncating `s.b` to include this terminator. Thus, `s.b` always ends with the bytes `\\r\\n\\r\\n`. Let $L$ be the length of `s.b`. The last four bytes are at indices $L-4, L-3, L-2, L-1$ with values `\\r, \\n, \\r, \\n`.\n\n2. **Loop Logic in `readContinuedLineSlice`**: The loop is `for s.skipSpace() { ... s.readLine() ... }`. `skipSpace()` is the loop condition. It increments `s.r` only while the current byte is a space or tab. It breaks (and returns false) if the current byte is NOT a space or tab.\n\n3. **Reaching the End**: For `skipSpace` to panic, `s.r` must be $L$ when `s.b[s.r]` is accessed. This implies the previous step (either `readLine` or `skipSpace`) left `s.r` at $L$.\n   - `skipSpace` cannot leave `s.r` at $L$ because `s.b[L-1]` is `\\n`, which is not a space or tab, so `skipSpace` will always break at or before $L-1$ if it enters the region.\n   - `readLine` advances `s.r` past a `\\n`. For `s.r` to become $L$, `readLine` must find `\\n` at index $L-1$. This requires that `readLine` is called with `s.r` positioned such that the first `\\n` it finds is at $L-1$.\n   - `readLine` is called *inside* the loop body, which is only executed if `skipSpace()` returned `true` in the *previous* iteration (or the initial check before the first `readLine` in the function). Wait, the structure is:\n     ```go\n     for s.skipSpace() {\n         mline = append(mline, ' ')\n         line := s.readLine()\n         // ...\n     }\n     ```\n     The condition `s.skipSpace()` is checked *before* entering the body. If `skipSpace()` returns `true`, it means `s.b[s.r]` was a space/tab, and `s.r` has been incremented past it. Then `readLine()` is called.\n   - If we are in the final terminator region `\\r\\n\\r\\n`:\n     - If `s.r` points to `\\r` (index $L-4$ or $L-2$), `skipSpace` sees `\\r` (not space), breaks, returns `false`. The loop exits. `readLine` is NOT called.\n     - If `s.r` points to `\\n` (index $L-3$ or $L-1$), `skipSpace` sees `\\n` (not space), breaks, returns `false`. The loop exits. `readLine` is NOT called.\n     - The only way to call `readLine` in this region is if `s.b[s.r]` is a space or tab. But the terminator `\\r\\n\\r\\n` contains no spaces or tabs.\n\n4. **Conclusion**: The loop `for s.skipSpace()` will always exit before `readLine` is called on the final `\\r\\n` of the terminator, because `skipSpace` will break on the `\\r` or `\\n` bytes of the terminator. Therefore, `s.r` will never be advanced to $L$ via `readLine` inside the loop. Consequently, `skipSpace` will never be called with `s.r == L`. The bug is not reachable.\n\nCRUX: The terminator `\\r\\n\\r\\n` contains no spaces or tabs, so `skipSpace` will break before `readLine` can consume the final `\\n` to advance `s.r` to the buffer length.\nGREP: skipSpace",
    "elapsed": 181.8,
    "tokens": 12692,
    "tokens_in": 11721,
    "tokens_out": 971,
    "cost_usd": 0.00783582,
    "file": "headerscanner.go",
    "round": 2,
    "grep_used": true,
    "grep_results": "GREP `skipSpace`:\n```\nheaderscanner.go:129:\tfor s.skipSpace() {\nheaderscanner.go:137:// skipSpace skips one or multiple spaces and tabs in b.\nheaderscanner.go:138:func (s *headerScanner) skipSpace() bool {\n```"
  },
  {
    "finding_title": "Out-of-bounds read in skipSpace at buffer end",
    "verdict": "INVALID",
    "reasoning": "The vulnerability report claims an out-of-bounds read in `skipSpace` when `s.r == len(s.b)`. I analyzed the data flow and the invariants enforced by `headerScanner.next()`. \n\n1. **Buffer Invariant**: `next()` truncates `s.b` to the end of the first `\\r\\n\\r\\n` sequence. Thus, `s.b` always ends with the 4-byte sequence `\\r\\n\\r\\n`. Let $L$ be the length of `s.b`. The bytes at indices $L-4, L-3, L-2, L-1$ are `\\r, \\n, \\r, \\n`.\n\n2. **Loop Logic**: The vulnerable loop in `readContinuedLineSlice` is `for s.skipSpace() { ... s.readLine() ... }`. \n   - `skipSpace()` increments `s.r` only while `s.b[s.r]` is a space (`0x20`) or tab (`0x09`). If the byte is anything else, it breaks and returns `false` (if no spaces were skipped) or `true` (if spaces were skipped, but `s.r` now points to the non-space byte that stopped the skip). Crucially, `skipSpace` always leaves `s.r` pointing to a valid byte index $< L$ (specifically, the byte that caused it to stop, which is not a space/tab).\n   - `readLine()` advances `s.r` past the next `\\n`. For `s.r` to become $L$, `readLine` must consume the `\\n` at index $L-1$.\n\n3. **Reachability Analysis**: \n   - To execute `readLine()` inside the loop, `skipSpace()` must have returned `true` in the current iteration (or the loop entry condition was met). This implies `s.b[s.r]` was a space or tab before `skipSpace` started, and after `skipSpace` finishes, `s.r` points to a non-space/non-tab byte.\n   - If the loop is active near the end of the buffer, `s.r` must be in the region containing the final `\\r\\n\\r\\n`. \n   - The bytes in the final `\\r\\n\\r\\n` are `\\r` and `\\n`. Neither is a space or tab. \n   - Therefore, if `s.r` is at any position within the final `\\r\\n\\r\\n` (i.e., $L-4$ to $L-1$), `skipSpace()` will immediately see a `\\r` or `\\n`, break, and return `false` (assuming no preceding spaces in the current line's trailing part, or rather, that the current position is the terminator itself). \n   - More precisely, for `skipSpace` to return `true` and allow `readLine` to execute, `s.r` must point to a space/tab *before* the terminator. `skipSpace` will then advance `s.r` past those spaces until it hits a non-space byte. If that non-space byte is the `\\r` of the terminator, `skipSpace` stops, `s.r` points to `\\r`, and the loop condition for the *next* iteration will be `false` because `skipSpace` sees `\\r` and breaks. \n   - The critical case is if `readLine` is called and sets `s.r` to $L$. This requires `readLine` to find `\\n` at $L-1$ and consume it. This can only happen if `s.r` was at $L-2$ (pointing to `\\r`) when `readLine` was called. But `readLine` is called *inside* the loop body, which is only entered if `skipSpace` returned `true`. If `skipSpace` returned `true`, it means it skipped spaces and stopped at a non-space. If it stopped at the `\\r` at $L-2$, then `readLine` consumes `\\r\\n` (indices $L-2, L-1$) and sets `s.r = L`. Then the loop condition `for s.skipSpace()` is evaluated. `skipSpace` attempts to read `s.b[L]`, causing a panic. \n   - **However**, for `skipSpace` to stop at `\\r` at $L-2$ and return `true`, there must have been spaces/tabs immediately preceding $L-2$ that `skipSpace` skipped. This would mean the input had a folded header line ending with spaces followed immediately by the `\\r\\n\\r\\n` terminator? No, the terminator is the *end* of the headers. A continuation line must be part of the headers. The `\\r\\n\\r\\n` marks the end. You cannot have a continuation line *after* the end of headers. The logic in `readContinuedLineSlice` processes lines *before* the blank line. The blank line is `\\r\\n`. The next line is the start of the body or the next request. `s.b` is truncated to include the *first* `\\r\\n\\r\\n`. This means `s.b` contains the headers plus the blank line. The loop in `readContinuedLineSlice` processes a single header line (key: value + continuations). Once `skipSpace` encounters the blank line's `\\r` (or `\\n`), it breaks. The blank line is not a continuation. \n   - Specifically, if we are processing a header, and the next bytes are the blank line `\\r\\n`, `skipSpace` sees `\\r` and breaks. The loop terminates. `readLine` is not called on the blank line inside this continuation loop. The continuation loop only runs if the *next* character is a space/tab (indicating a fold). The blank line starts with `\\r` (or potentially spaces if malformed, but standard HTTP has `\\r`). If malformed input has spaces before the blank line, e.g., `Key: Val \\r\\n  \\r\\n\\r\\n`, `skipSpace` skips the two spaces, lands on `\\r`, breaks, returns true? No, `skipSpace` returns true if it skipped any space. So it returns true. `readLine` is called. `s.r` is at `\\r`. `readLine` consumes `\\r\\n`. `s.r` advances past the blank line's `\\n`. Now `s.r` points to the start of the *next* `\\r\\n` of the `\\r\\n\\r\\n` terminator? No, `next()` truncated `s.b` to *include* the full `\\r\\n\\r\\n`. So `s.b` ends with `... \\r\\n \\r\\n`. If `readLine` consumed the first `\\r\\n` of the double `\\r\\n`, `s.r` is now at the second `\\r\\n`. The loop continues? `skipSpace` is called. It sees `\\r`. Breaks. Returns false. Loop exits. \n   - The panic requires `s.r == L` *before* `skipSpace` is called. This requires `readLine` to set `s.r = L`. This requires `readLine` to consume the *very last* `\\n`. This requires `readLine` to be called when `s.r` is such that the *only* `\\n` left is at $L-1$. This implies `s.r` was at $L-2$ (pointing to `\\r`). But `readLine` is only called if `skipSpace` returned true. If `skipSpace` returned true, it means `s.r` was moved past spaces to a non-space. If that non-space is `\\r` at $L-2$, then `readLine` consumes `\\r\\n` (indices $L-2, L-1$) and sets `s.r = L`. Then `skipSpace` is called again. Panic. \n   - Is this reachable? It requires the buffer to end with `... [spaces] \\r\\n \\r\\n`? No, `s.b` is truncated to the *first* `\\r\\n\\r\\n`. So the sequence is `... \\r\\n \\r\\n`. If the previous line was `Key: Val`, and the next is a continuation, the next bytes must be `Space \\r\\n ...`. If the buffer ends shortly after, e.g., `Key: Val \\r\\n   \\r\\n\\r\\n`. `s.b` ends with `   \\r\\n\\r\\n`. `skipSpace` skips the 3 spaces, lands on `\\r`. Returns true. `readLine` consumes `\\r\\n`. `s.r` is now at the start of the *second* `\\r\\n` of the terminator? No, `s.b` was truncated to *include* `\\r\\n\\r\\n`. So the bytes are `Space Space Space \\r \\n \\r \\n`. `readLine` consumes `\\r \\n` (the first pair). `s.r` now points to the second `\\r`. `skipSpace` is called. Sees `\\r`. Breaks. Returns false. Loop exits. No panic. \n   - For a panic, `readLine` must consume the *second* `\\r\\n` as well. But `readLine` only consumes one line (up to one `\\n`). It cannot consume two `\\r\\n` sequences in one call. So `s.r` will never reach $L$ inside the loop. It will stop at $L-2$ (pointing to the second `\\r`), and the next `skipSpace` will exit. \n\nThus, the OOB read is not reachable.\n\nCRUX: skipSpace() will always break on the \\r or \\n bytes of the final \\r\\n\\r\\n terminator because they are not spaces or tabs, preventing readLine() from consuming the final newline to advance s.r to the buffer length.\nGREP: skipSpace",
    "elapsed": 255.9,
    "tokens": 14974,
    "tokens_in": 12717,
    "tokens_out": 2257,
    "cost_usd": 0.01211214,
    "file": "headerscanner.go",
    "round": 3,
    "grep_used": true,
    "grep_results": "GREP `skipSpace`:\n```\nheaderscanner.go:129:\tfor s.skipSpace() {\nheaderscanner.go:137:// skipSpace skips one or multiple spaces and tabs in b.\nheaderscanner.go:138:func (s *headerScanner) skipSpace() bool {\n```"
  },
  {
    "finding_title": "Out-of-bounds read in skipSpace at buffer end",
    "verdict": "INVALID",
    "reasoning": "The reported vulnerability claims an out-of-bounds read in `skipSpace` when `s.r == len(s.b)`. I analyzed the control flow to determine if this state is reachable. \n\n1. **Buffer Invariant**: `headerScanner.next()` initializes by locating the first `strCRLFCRLF` (`\\r\\n\\r\\n`) and truncating `s.b` to include this terminator. Thus, `s.b` always ends with the bytes `\\r, \\n, \\r, \\n`.\n\n2. **Loop Structure**: In `readContinuedLineSlice`, the continuation loop is `for s.skipSpace() { ... s.readLine() ... }`. \n   - `skipSpace()` only increments `s.r` if the current byte is a space (`0x20`) or tab (`0x09`). It stops and returns `true` if it skipped at least one character, leaving `s.r` at the first non-space/tab byte. It returns `false` if the first byte is not space/tab.\n   - `readLine()` is only called if `skipSpace()` returned `true`. This implies `s.r` points to a valid byte in `s.b` (the one that stopped the skip).\n\n3. **Reachability of `s.r == len(s.b)`**: For a panic to occur, `s.r` must be `len(s.b)` when `skipSpace` is called. This would require `readLine` to have advanced `s.r` to `len(s.b)` in the previous step. \n   - `readLine` advances `s.r` past the next `\\n`. To reach `len(s.b)`, it must consume the `\\n` at `len(s.b)-1`.\n   - This implies `s.r` was at `len(s.b)-2` (pointing to `\\r`) when `readLine` started.\n   - However, `readLine` is only entered if `skipSpace` returned `true`. For `skipSpace` to stop at `len(s.b)-2` (which is `\\r`), it must have skipped spaces prior to it. This means the buffer would look like `... [spaces] \\r \\n \\r \\n` at the end.\n   - If `skipSpace` skips the spaces and lands on the `\\r` at `len(s.b)-2`, it returns `true`. `readLine` consumes `\\r\\n` (indices `len(s.b)-2` and `len(s.b)-1`). `s.r` becomes `len(s.b)`.\n   - The loop condition `for s.skipSpace()` is then evaluated. `skipSpace` reads `s.b[len(s.b)]`, causing a panic.\n   - **Crucial Defense**: Can `s.b` end with `[spaces] \\r\\n\\r\\n`? The terminator is `\\r\\n\\r\\n`. The bytes are fixed. There are no spaces in the terminator. The bytes immediately preceding the terminator are part of the last header line. If the last header line is a continuation, it must be folded. The structure is `Key: Val \\r\\n Continuation \\r\\n\\r\\n`. \n   - If the input is `Key: Val \\r\\n   \\r\\n\\r\\n` (3 spaces before the final CRLF), `s.b` is truncated to include `\\r\\n\\r\\n`. The bytes at the end are `Space, Space, Space, \\r, \\n, \\r, \\n`. \n   - `skipSpace` skips the 3 spaces, lands on `\\r`. Returns true. \n   - `readLine` consumes `\\r\\n`. `s.r` points to the second `\\r` of the terminator. \n   - Next `skipSpace` sees `\\r`. Returns false. Loop exits. No panic.\n   - For the panic to happen, `readLine` must consume the *final* `\\n`. This requires `s.r` to be at the *final* `\\r` before `readLine`. But `skipSpace` stops at the *first* non-space. If the bytes are `... Space \\r \\n \\r \\n`, `skipSpace` stops at the first `\\r`. `readLine` consumes the first `\\r\\n`. `s.r` is at the second `\\r`. `skipSpace` exits. \n   - There is no scenario where `skipSpace` skips spaces and lands on the *second* `\\r` of the terminator, because the first `\\r` stops it. Thus, `readLine` never consumes the final `\\n` in a way that sets `s.r` to `len(s.b)` while the loop is active. The invariant that `s.b` ends with `\\r\\n\\r\\n` (which contains no spaces) protects against this OOB read.\n\nCRUX: The buffer always ends with \\r\\n\\r\\n, which contains no spaces or tabs; skipSpace will always break on the \\r or \\n of the terminator, preventing readLine from advancing s.r to the end of the buffer within the loop.\nGREP: skipSpace",
    "elapsed": 113.8,
    "tokens": 16047,
    "tokens_in": 14890,
    "tokens_out": 1157,
    "cost_usd": 0.0097248,
    "file": "headerscanner.go",
    "round": 4,
    "grep_used": true,
    "grep_results": "GREP `skipSpace`:\n```\nheaderscanner.go:129:\tfor s.skipSpace() {\nheaderscanner.go:137:// skipSpace skips one or multiple spaces and tabs in b.\nheaderscanner.go:138:func (s *headerScanner) skipSpace() bool {\n```"
  },
  {
    "finding_title": "Out-of-bounds read in skipSpace at buffer end",
    "verdict": "INVALID",
    "reasoning": "The reported vulnerability is an out-of-bounds read in `skipSpace` when `s.r == len(s.b)`. I analyzed the control flow of `headerScanner.next()` and `readContinuedLineSlice()`. \n\n1. **Buffer Invariant**: `next()` initializes by finding the first `strCRLFCRLF` (`\\r\\n\\r\\n`) and truncating `s.b` to include this terminator. Thus, `s.b` always ends with the 4-byte sequence `\\r, \\n, \\r, \\n`. Let $L$ be `len(s.b)`. The last four bytes are at indices $L-4, L-3, L-2, L-1$.\n\n2. **Loop Logic**: The continuation loop in `readContinuedLineSlice` is `for s.skipSpace() { ... s.readLine() ... }`. \n   - `skipSpace()` increments `s.r` only while `s.b[s.r]` is a space or tab. It breaks (and returns `false` if no spaces were skipped, or `true` if they were) when it encounters a non-space/non-tab byte.\n   - Crucially, `skipSpace()` always leaves `s.r` pointing to a valid index within `s.b` if it returns `true`, specifically the index of the first non-space byte it encountered.\n\n3. **Reachability of `s.r == L`**: For `skipSpace` to panic on the *next* call, `s.r` must be set to $L$ by the previous iteration's `readLine()`. \n   - `readLine()` advances `s.r` past the next `\\n`. To set `s.r` to $L$, `readLine` must consume the `\\n` at $L-1$.\n   - This implies `readLine` started at $L-2$ (where `s.b[L-2]` is `\\r`).\n   - `readLine` is only called if `skipSpace()` returned `true` in the current iteration. \n   - If `skipSpace` returned `true`, it skipped spaces and stopped at a non-space byte. If it stopped at $L-2$ (which is `\\r`), that means the bytes preceding $L-2$ were spaces/tabs.\n   - However, the terminator `\\r\\n\\r\\n` contains no spaces or tabs. The byte at $L-3$ is `\\n` and at $L-4$ is `\\r`.\n   - Can `skipSpace` skip spaces and land on $L-2$? This would require the byte at $L-3$ (which is `\\n`) to be a space, which is false. Or it would require the sequence `... [spaces] \\r \\n` at the end. But the end is fixed as `... \\r \\n \\r \\n`. \n   - If `skipSpace` is active, it skips spaces. If the buffer ends with `... [spaces] \\r \\n \\r \\n`, `skipSpace` will skip the spaces and land on the `\\r` at $L-4$? No, the structure is `[headers] \\r\\n \\r\\n`. \n   - Let's trace a case: Input `Key: Val \\r\\n   \\r\\n\\r\\n`. \n     - `s.b` ends with `... Space, Space, Space, \\r, \\n, \\r, \\n`.\n     - Indices: $L-6, L-5, L-4$ are spaces. $L-3$ is `\\r`. $L-2$ is `\\n`. $L-1$ is `\\r`. $L$ (wait, length is such that last 4 are `\\r\\n\\r\\n`). \n     - Actually, `next()` truncates to the *first* `\\r\\n\\r\\n`. \n     - If input is `Key: Val \\r\\n   \\r\\n\\r\\n`, the first `\\r\\n\\r\\n` is at the very end. The bytes are `... K, :, V, ' ', \\r, \\n, ' ', ' ', ' ', \\r, \\n, \\r, \\n`.\n     - Wait, `\\r\\n\\r\\n` is 4 bytes. The spaces are before the final CRLF? No, the spaces are part of the continuation line or trailing spaces? \n     - In HTTP, `Key: Val \\r\\n Continuation`. The continuation line starts with space/tab. \n     - If the input is `Key: Val \\r\\n   \\r\\n\\r\\n`, the bytes are: \n       `K, :, V, ' ', \\r, \\n, ' ', ' ', ' ', \\r, \\n, \\r, \\n`. \n       Indices: \n       0:K, 1::, 2:V, 3:' ', 4:\\r, 5:\\n, 6:' ', 7:' ', 8:' ', 9:\\r, 10:\\n, 11:\\r, 12:\\n. \n       `next()` finds `\\r\\n\\r\\n` at index 9? \n       `s.b[9]=\\r, s.b[10]=\\n, s.b[11]=\\r, s.b[12]=\\n`. \n       So `i=9`. `s.b` is truncated to length $9+4=13$. \n       Bytes 0-12. \n       `readLine` 1: `s.r=0`. Finds `\\n` at 5. `s.r=6`. \n       `readContinuedLineSlice`: \n       `peek`: `s.b[6]=' '`. Condition fails. Enter loop. \n       `skipSpace`: `s.r=6`. Skips 6,7,8. `s.r=9`. `s.b[9]='\\r'`. Break. Return true. \n       `readLine`: `s.r=9`. Finds `\\n` at 10. `s.r=11`. \n       Loop condition `for s.skipSpace()`. \n       `skipSpace`: `s.r=11`. `s.b[11]='\\r'`. Break. Return false. \n       Loop exits. \n       No panic. \n\n   - What if the spaces are *inside* the terminator region? They can't be, because the terminator is fixed. \n   - What if `readLine` consumes up to $L$? This requires `\\n` at $L-1$. \n   - This requires `s.r` to be $L-2$ (where `s.b[L-2]` is `\\r`) and `readLine` to be called. \n   - `readLine` is called if `skipSpace` returned true. \n   - `skipSpace` returned true means it skipped spaces and landed on $L-2$. \n   - This implies `s.b[L-3]` was a space. \n   - But $s.b[L-3]$ is `\\n` (from `\\r\\n\\r\\n`). \n   - So `skipSpace` cannot land on $L-2$ by skipping spaces, because the byte before it is not a space. \n   - Therefore, `readLine` is never called at $L-2$. \n   - Therefore, `s.r` is never set to $L$ inside the loop. \n   - Therefore, `skipSpace` is never called with `s.r == L`.\n\nThe defense is the structure of the `\\r\\n\\r\\n` terminator, which ensures that `skipSpace` will break on `\\r` or `\\n` before `readLine` can advance the pointer to the end of the buffer within the continuation loop.\n\nCRUX: The final \\r\\n\\r\\n terminator contains no spaces or tabs, so skipSpace will always break before readLine can consume the final newline to advance s.r to the buffer length.\nGREP: skipSpace",
    "elapsed": 133.6,
    "tokens": 17889,
    "tokens_in": 16047,
    "tokens_out": 1842,
    "cost_usd": 0.01226574,
    "file": "headerscanner.go",
    "round": 5,
    "grep_used": true,
    "grep_results": "GREP `skipSpace`:\n```\nheaderscanner.go:129:\tfor s.skipSpace() {\nheaderscanner.go:137:// skipSpace skips one or multiple spaces and tabs in b.\nheaderscanner.go:138:func (s *headerScanner) skipSpace() bool {\n```"
  },
  {
    "verdict": "INVALID",
    "reasoning": "[ARBITER] The vulnerability is not reachable because the scanner state maintains an invariant that prevents `s.r` from being equal to `len(s.b)` when `skipSpace` is called. \n\n1. **Buffer Initialization**: `headerScanner.next()` truncates the buffer `s.b` to end exactly after the first `\\r\\n\\r\\n` sequence. Thus, the last four bytes of `s.b` are `\\r`, `\\n`, `\\r`, `\\n`.\n2. **Loop Structure**: In `readContinuedLineSlice`, the loop is `for s.skipSpace() { ... line := s.readLine() ... }`.\n3. **Termination Condition**: Before `readLine` can advance `s.r` to the end of the buffer, it must consume a newline. However, the loop condition checks `s.skipSpace()` *before* calling `readLine`. \n4. **Invariant**: As the parser approaches the end of the buffer (the final `\\r\\n\\r\\n`), the read position `s.r` will eventually point to one of the non-space/non-tab characters in the terminator (specifically `\\r` or `\\n`). Since `skipSpace` only consumes spaces and tabs, it will return `false` when it encounters `\\r` or `\\n`. Consequently, the loop terminates before `readLine` is called in a state where `s.r` could reach the end of the buffer during the loop iteration itself. Even if `readLine` consumed the final newline to make `s.r == len(s.b)`, the next iteration of the `for` loop would call `skipSpace`, but this scenario is prevented because the loop structure relies on `skipSpace` to detect the *start* of a continuation line, and the buffer end is marked by non-whitespace characters that stop the skip process. More simply: `readLine` always consumes a `\\n` to advance `s.r`. If `s.r` becomes `len(s.b)`, it means the very last `\\n` was consumed. But `next()` ensures the buffer ends with `\\r\\n\\r\\n`. The parser processes headers line by line. When it reaches the final `\\n` of the last header value, `readLine` consumes it. The next step in `next()` is to check for continuation. But the buffer ends. Actually, the key is that `skipSpace` is called in the `for` condition. If `s.r` were at the end, `s.b[s.r]` would panic. But `s.r` only advances via `readLine` inside the loop. `readLine` finds a `\\n`. If it consumes the final `\\n` of the buffer, `s.r` becomes `len(s.b)`. The loop then iterates back to `skipSpace`. At this point, `s.r == len(s.b)`, and `s.b[s.r]` is an out-of-bounds read. \n\nWait, let's re-evaluate carefully. \nBuffer: `Header: Value\\r\\n\\r\\n`. \n`readLine` reads `Header: Value`. `s.r` points to first `\\r`. \n`next` returns true. \nNext call to `next` (if any) or internal logic? `headerScanner` is used in a loop in `Header.Parse`. \nActually, `readContinuedLineSlice` is called inside `next`. \nLet's trace `next` for the *last* header. \nBuffer ends with `\\r\\n\\r\\n`. \n`readContinuedLineSlice` calls `readLine`. It finds `\\n` at `L-1`? No, `bytes.IndexByte(s.b[s.r:], '\\n')`. \nIf the header is `A: B`, followed by `\\r\\n`, then `\\r\\n`. \n`readLine` consumes `A: B\\r\\n`. `s.r` is now at the index of the *second* `\\r` (the start of the final `\\r\\n` terminator). \nBack in `readContinuedLineSlice`: \n`if len(s.b)-s.r > 1 { ... }` \n`s.r` is at `L-4` (index of first `\\r` of the final `\\r\\n\\r\\n`? No. \nLet's map indices. \n`... [H][e][a][d][e][r][:][ ][V][a][l][u][e][\\r][\\n][\\r][\\n]` \nIndices: `... 0..12, 13(\\r), 14(\\n), 15(\\r), 16(\\n)`. `L=17`. \n`readLine` finds `\\n` at 14. `line = s.b[0:15]`. `s.r = 15`. \n`s.r` is 15. `s.b[15]` is `\\r`. `s.b[16]` is `\\n`. \n`len(s.b)-s.r = 17-15 = 2`. \n`peek = s.b[15:17]` which is `[\\r, \\n]`. \n`peek[0] == '\\r' && peek[1] == '\\n'` is true. \nSo it returns `trim(line), nil`. \nIt does **not** enter the `for s.skipSpace()` loop. \nThus, `skipSpace` is never called when `s.r` is near the end of the buffer in this specific case. \n\nWhat if there is a continuation? \n`Header: Value1\\r\\n  Value2\\r\\n\\r\\n`. \n`readLine` reads `Header: Value1`. `s.r` at `\\r` of `\\r\\n` after `Value1`. \n`peek` check: `s.b[s.r]` is `\\r`, `s.b[s.r+1]` is `\\n`. Wait, `readLine` consumed the `\\n`? \nYes, `s.r` advances past `\\n`. So `s.r` is at the space/tab of the continuation? \nNo, `readLine` returns line *including* `\\n`. It strips it for return, but `s.r` advances. \nSo `s.r` is at the first character of the next line. \nIf the next line is `  Value2`, `s.b[s.r]` is ` `. \n`peek` check: `isASCIILetter(' ')` is false. `peek[0]=='\\n'` false. `peek[0]=='\\r'` false. \nSo it proceeds to `mline = trim(line)`. \nThen `for s.skipSpace()`. \n`skipSpace` consumes spaces/tabs. \nThen `readLine` reads `Value2`. \nThen loop checks `skipSpace` again. \nIf `Value2` is the last part, the buffer continues with `\\r\\n\\r\\n`. \nAfter `readLine` of `Value2`, `s.r` is at the `\\r` of the final `\\r\\n\\r\\n`. \nLoop checks `skipSpace`. `s.b[s.r]` is `\\r`. `\\r` is not space/tab. `skipSpace` returns false. \nLoop terminates. \nNo OOB. \n\nTherefore, the `for s.skipSpace()` loop always terminates because it encounters a non-space/tab character (either the start of a new header key or the `\\r` of the final terminator) before `s.r` can reach `len(s.b)`. The only way `s.r` reaches `len(s.b)` is if `readLine` consumes the very last byte, but `readLine` requires a `\\n` to complete, and the buffer always has a `\\r` before that final `\\n` in the terminator structure which breaks the space loop.",
    "round": 6,
    "file": "headerscanner.go",
    "finding_title": "Out-of-bounds read in skipSpace at buffer end",
    "elapsed": 252.7,
    "tokens_in": 3128,
    "tokens_out": 1803,
    "cost_usd": 0.0067227599999999995
  }
]
Raw scan report for this file

The untrusted data is `s.b`, which normally contains raw HTTP header bytes from a client or peer. The scanner state (`s.r`, `s.initialized`) determines how those bytes are consumed. The main risks are parser-state bugs: once initialization has occurred, `next()` no longer re-validates that the remaining buffer still contains a well-formed line, and several helpers assume that a newline or safe buffer position exists.

- `readLine()` loops forever if it is called when `s.b[s.r:]` contains no `\n`. This can happen if the scanner reaches the end of the buffer, if state is reused incorrectly, or if a malformed/continued header causes the position to land at the end without a terminating newline. An attacker can use this for CPU exhaustion.
- `skipSpace()` reads `s.b[s.r]` without checking `s.r < len(s.b)`. If the continuation path reaches the end of the buffer, this causes an index-out-of-range panic, which is a denial-of-service vector.
- `readLine()` accepts a lone `\n` as a line boundary and drops an optional preceding `\r`. HTTP header line endings are required to be CRLF. Accepting LF-only line endings can create parser differentials with CRLF-only frontends/proxies and may enable header misparsing, header injection, or request-smuggling scenarios.

Other helpers (`trim`, `trimTrailingSpace`, `isASCIILetter`) are bounds-safe. There is no use of `unsafe`, goroutine use, command/SQL construction, or obvious nil-pointer dereference.

```json
[
  {
    "severity": "high",
    "title": "Infinite loop in readLine when no newline remains",
    "function": "readLine",
    "description": "readLine() searches for a newline in s.b[s.r+searchStart:]. If no newline is present, searchStart becomes len(s.b) - s.r and then stays constant, causing an infinite loop. This can be triggered by malformed or partially consumed header state where the scanner is invoked with s.r at or beyond the end of a line, leading to 100% CPU DoS."
  },
  {
    "severity": "high",
    "title": "Out-of-bounds read in skipSpace at buffer end",
    "function": "skipSpace",
    "description": "skipSpace() dereferences s.b[s.r] without checking that s.r is less than len(s.b). readContinuedLineSlice() calls skipSpace() while processing folded headers; if a malformed header or stale scanner state places s.r at the end of s.b, the read panics with an index out of range, causing denial of service."
  },
  {
    "severity": "medium",
    "title": "LF-only header line separation creates parser differential",
    "function": "readLine",
    "description": "readLine() treats any '\\n' as a line boundary and removes an optional preceding '\\r', so header lines separated by lone LF are accepted. HTTP requires CRLF line endings. This leniency can cause different header interpretation than CRLF-only parsers or proxies, potentially enabling header misparsing, header injection, or request smuggling."
  }
]
```