ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncasecmp>:
  endbr64
  push   %r13
  lea    -0x1(%rdi,%rdx,1),%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  jne    L19
  jmp    L50
L14:
  cmp    %r13,%rbx
  je     L50
  cmpb   $0x0,(%rbx)
  je     L50
  cmpb   $0x0,0x0(%rbp)
  je     L50
  add    $0x1,%rbx
  add    $0x1,%rbp
L19:
  movzbl (%rbx),%edi
  call   <tolower@plt>
  movzbl 0x0(%rbp),%edi
  mov    %eax,%r12d
  call   <tolower@plt>
  cmp    %eax,%r12d
  je     L14
  movzbl (%rbx),%edi
  call   <tolower@plt>
  movzbl 0x0(%rbp),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int strncasecmp(const char *s1, const char *s2, size_t n) {
  if (n == 0)
    return 0;
  while ((n-- != 0) && (tolower(*(unsigned char *)s1) == tolower(*(unsigned char *)s2))) {
    if (n == 0 || *s1 == '\0' || *s2 == '\0')
      return 0;
    s1++;
    s2++;
  }
  return tolower(*(unsigned char *)s1) - tolower(*(unsigned char *)s2);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strncasecmp>:
  endbr64
  push   %r13
  lea    -0x1(%rdi,%rdx,1),%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  jne    L19
  jmp    L26
L0:
  cmp    %r13,%rbx
  je     L26
  cmpb   $0x0,(%rbx)
  je     L26
  cmpb   $0x0,0x0(%rbp)
  je     L26
  add    $0x1,%rbx
  add    $0x1,%rbp
L19:
  movzbl (%rbx),%edi
  call   <tolower@plt>
  movzbl 0x0(%rbp),%edi
  mov    %eax,%r12d
  call   <tolower@plt>
  cmp    %eax,%r12d
  je     L0
  movzbl (%rbx),%edi
  call   <tolower@plt>
  movzbl 0x0(%rbp),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int strncasecmp(const char *s1, const char *s2, size_t n) {
  if (n == 0)
    return 0;
  while ((n-- != 0) && (tolower(*(unsigned char *)s1) == tolower(*(unsigned char *)s2))) {
    if (n == 0 || *s1 == '\0' || *s2 == '\0')
      return 0;
    s1++;
    s2++;
  }
  return tolower(*(unsigned char *)s1) - tolower(*(unsigned char *)s2);
}
```
user: What is the c source code of the assembly code below?

```asm
<EndSolve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    %r9d,-0x1c(%rbp)
  cmpl   $0x7,-0x18(%rbp)
  jle    L39
  mov    -0x1c(%rbp),%r9d
  mov    -0x18(%rbp),%r8d
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %rax,%rdi
  call   <FastestFirstEndSolve@plt>
  add    $0x10,%rsp
  jmp    L18
L39:
  cmpl   $0x4,-0x18(%rbp)
  jg     L4
  mov    -0x1c(%rbp),%r9d
  mov    -0x18(%rbp),%r8d
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %rax,%rdi
  call   <NoParEndSolve@plt>
  add    $0x10,%rsp
  jmp    L18
L4:
  mov    -0x1c(%rbp),%r9d
  mov    -0x18(%rbp),%r8d
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %rax,%rdi
  call   <ParEndSolve@plt>
  add    $0x10,%rsp
L18:
  leave
  ret
```

assistant: ```c
int EndSolve(unsigned char *board, int alpha, int beta, int color, int empties, int discdiff, int prevmove) {
  if (empties > 7)
    return FastestFirstEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  else {
    if (empties <= (2 > 4 ? 2 : 4))
      return NoParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
    else
      return ParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<EndSolve>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x7,%r8d
  jg     L6
  cmp    $0x4,%r8d
  jg     L17
  sub    $0x8,%rsp
  mov    0x18(%rsp),%eax
  push   %rax
  call   <NoParEndSolve@plt>
  add    $0x10,%rsp
L14:
  add    $0x8,%rsp
  ret
L6:
  sub    $0x8,%rsp
  mov    0x18(%rsp),%eax
  push   %rax
  call   <FastestFirstEndSolve@plt>
  add    $0x10,%rsp
  jmp    L14
L17:
  sub    $0x8,%rsp
  mov    0x18(%rsp),%eax
  push   %rax
  call   <ParEndSolve@plt>
  add    $0x10,%rsp
  jmp    L14
```

assistant: ```c
int EndSolve(unsigned char *board, int alpha, int beta, int color, int empties, int discdiff, int prevmove) {
  if (empties > 7)
    return FastestFirstEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  else {
    if (empties <= (2 > 4 ? 2 : 4))
      return NoParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
    else
      return ParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<EndSolve>:
  endbr64
  cmp    $0x7,%r8d
  jg     L11
  cmp    $0x4,%r8d
  jg     L5
  jmp    1090 <NoParEndSolve@plt>
  nopl   (%rax)
L5:
  jmp    1070 <ParEndSolve@plt>
  nopl   (%rax)
L11:
  jmp    1080 <FastestFirstEndSolve@plt>
```

assistant: ```c
int EndSolve(unsigned char *board, int alpha, int beta, int color, int empties, int discdiff, int prevmove) {
  if (empties > 7)
    return FastestFirstEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  else {
    if (empties <= (2 > 4 ? 2 : 4))
      return NoParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
    else
      return ParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<EndSolve>:
  endbr64
  cmp    $0x7,%r8d
  jg     L19
  cmp    $0x4,%r8d
  jg     L2
  jmp    1090 <NoParEndSolve@plt>
  nopl   (%rax)
L2:
  jmp    1070 <ParEndSolve@plt>
  nopl   (%rax)
L19:
  jmp    1080 <FastestFirstEndSolve@plt>
```

assistant: ```c
int EndSolve(unsigned char *board, int alpha, int beta, int color, int empties, int discdiff, int prevmove) {
  if (empties > 7)
    return FastestFirstEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  else {
    if (empties <= (2 > 4 ? 2 : 4))
      return NoParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
    else
      return ParEndSolve(board, alpha, beta, color, empties, discdiff, prevmove);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<L_4_inv>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rax
  mov    (%rdi),%ebx
  mov    0x10(%rdi),%esi
  mov    %esi,(%rdi)
  mov    0x4(%rdi),%r11d
  mov    0x14(%rdi),%edi
  mov    %edi,0x4(%rax)
  mov    0x8(%rax),%r10d
  mov    0x18(%rax),%r8d
  mov    %r8d,0x8(%rax)
  mov    0xc(%rax),%r9d
  mov    0x1c(%rax),%ecx
  mov    %ecx,0xc(%rax)
  mov    %esi,%edx
  xor    %edi,%edx
  xor    %r8d,%edx
  xor    %ecx,%edx
  mov    %edx,%ebp
  shl    $0x8,%ebp
  sar    $0x8,%edx
  or     %ebp,%edx
  xor    %ebx,%r8d
  xor    %edx,%r8d
  mov    %r8d,0x10(%rax)
  xor    %r11d,%edi
  xor    %edx,%edi
  mov    %edi,0x14(%rax)
  xor    %r10d,%esi
  xor    %edx,%esi
  mov    %esi,0x18(%rax)
  xor    %r9d,%ecx
  xor    %edx,%ecx
  mov    %ecx,0x1c(%rax)
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void L_4_inv(uint16_t *x) {
  uint16_t tmp;
  tmp = x[0];
  x[0] = x[4];
  x[4] = tmp;
  tmp = x[1];
  x[1] = x[5];
  x[5] = tmp;
  tmp = x[2];
  x[2] = x[6];
  x[6] = tmp;
  tmp = x[3];
  x[3] = x[7];
  x[7] = tmp;
  tmp = x[0] ^ x[1] ^ x[2] ^ x[3];
  tmp = (((tmp) << 8) | ((tmp) >> (16 - (8))));
  x[4] ^= x[2] ^ tmp;
  x[5] ^= x[1] ^ tmp;
  x[6] ^= x[0] ^ tmp;
  x[7] ^= x[3] ^ tmp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<L_4_inv>:
  endbr64
  push   %rbp
  mov    %rdi,%rax
  push   %rbx
  mov    0x10(%rdi),%esi
  mov    (%rdi),%ebx
  mov    0x4(%rdi),%r11d
  mov    %esi,(%rdi)
  mov    0x14(%rdi),%edi
  mov    %esi,%edx
  mov    0x18(%rax),%r8d
  mov    0x1c(%rax),%ecx
  xor    %edi,%edx
  mov    0x8(%rax),%r10d
  mov    0xc(%rax),%r9d
  mov    %edi,0x4(%rax)
  xor    %r8d,%edx
  mov    %r8d,0x8(%rax)
  xor    %r11d,%edi
  xor    %ebx,%r8d
  xor    %ecx,%edx
  mov    %ecx,0xc(%rax)
  xor    %r10d,%esi
  xor    %r9d,%ecx
  mov    %edx,%ebp
  sar    $0x8,%edx
  pop    %rbx
  shl    $0x8,%ebp
  or     %ebp,%edx
  pop    %rbp
  xor    %edx,%r8d
  xor    %edx,%edi
  xor    %edx,%esi
  xor    %edx,%ecx
  mov    %r8d,0x10(%rax)
  mov    %edi,0x14(%rax)
  mov    %esi,0x18(%rax)
  mov    %ecx,0x1c(%rax)
  ret
```

assistant: ```c
void L_4_inv(uint16_t *x) {
  uint16_t tmp;
  tmp = x[0];
  x[0] = x[4];
  x[4] = tmp;
  tmp = x[1];
  x[1] = x[5];
  x[5] = tmp;
  tmp = x[2];
  x[2] = x[6];
  x[6] = tmp;
  tmp = x[3];
  x[3] = x[7];
  x[7] = tmp;
  tmp = x[0] ^ x[1] ^ x[2] ^ x[3];
  tmp = (((tmp) << 8) | ((tmp) >> (16 - (8))));
  x[4] ^= x[2] ^ tmp;
  x[5] ^= x[1] ^ tmp;
  x[6] ^= x[0] ^ tmp;
  x[7] ^= x[3] ^ tmp;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<L_4_inv>:
  endbr64
  movdqu 0x10(%rdi),%xmm1
  movdqu (%rdi),%xmm3
  movdqa %xmm1,%xmm0
  pshufd $0xc6,%xmm1,%xmm2
  movups %xmm1,(%rdi)
  punpckhdq %xmm1,%xmm0
  pxor   %xmm3,%xmm2
  movd   %xmm0,%eax
  pshufd $0xff,%xmm1,%xmm0
  movd   %xmm0,%edx
  pshufd $0x55,%xmm1,%xmm0
  xor    %edx,%eax
  movd   %xmm0,%edx
  xor    %edx,%eax
  movd   %xmm1,%edx
  xor    %edx,%eax
  mov    %eax,%edx
  sar    $0x8,%eax
  shl    $0x8,%edx
  or     %eax,%edx
  movd   %edx,%xmm4
  pshufd $0x0,%xmm4,%xmm0
  pxor   %xmm2,%xmm0
  movups %xmm0,0x10(%rdi)
  ret
```

assistant: ```c
void L_4_inv(uint16_t *x) {
  uint16_t tmp;
  tmp = x[0];
  x[0] = x[4];
  x[4] = tmp;
  tmp = x[1];
  x[1] = x[5];
  x[5] = tmp;
  tmp = x[2];
  x[2] = x[6];
  x[6] = tmp;
  tmp = x[3];
  x[3] = x[7];
  x[7] = tmp;
  tmp = x[0] ^ x[1] ^ x[2] ^ x[3];
  tmp = (((tmp) << 8) | ((tmp) >> (16 - (8))));
  x[4] ^= x[2] ^ tmp;
  x[5] ^= x[1] ^ tmp;
  x[6] ^= x[0] ^ tmp;
  x[7] ^= x[3] ^ tmp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jle    L15
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <f@plt>
  mov    $0x0,%eax
  call   <__VERIFIER_error@plt>
  jmp    L6
L15:
  nop
L6:
  leave
  ret
```

assistant: ```c
void f2(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f2>:
  endbr64
  cmp    $0x2,%edi
  jg     L6
  ret
L6:
  sub    $0x8,%rsp
  sub    $0x1,%edi
  call   <f@plt>
  mov    $0x0,%eax
  call   <__VERIFIER_error@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void f2(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<f2>:
  endbr64
  cmp    $0x2,%edi
  jg     L3
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  sub    $0x8,%rsp
  sub    $0x1,%edi
  call   <f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <__VERIFIER_error@plt>
```

assistant: ```c
void f2(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: What is the c source code of the assembly code below?

```asm
<f2>:
  endbr64
  cmp    $0x2,%edi
  jg     L8
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  sub    $0x8,%rsp
  sub    $0x1,%edi
  call   <f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <__VERIFIER_error@plt>
```

assistant: ```c
void f2(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_inputstream>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x10000(%rsp),%r11
L26:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L26
  sub    $0x10,%rsp
  mov    %rdi,%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x10008(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r13d
  lea    D2(%rip),%r12
  jmp    L87
L38:
  mov    %rsp,%rdi
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  call   <memmove@plt>
  mov    %rbp,%r13
L72:
  cmp    $0x10,%rbp
  jbe    L20
L87:
  mov    $0x10000,%edx
  sub    %r13,%rdx
  lea    (%rsp,%r13,1),%rdi
  mov    %r14,%rcx
  mov    $0x1,%esi
  call   <fread@plt>
  test   %eax,%eax
  jle    L34
  cltq
  lea    (%rax,%r13,1),%rbp
  cmp    $0xb,%rbp
  jbe    L100
  mov    %rsp,%rbx
L99:
  mov    (%rbx),%esi
  mov    0x4(%rbx),%edx
  mov    0x8(%rbx),%ecx
  mov    0xc(%rbx),%r8d
  movslq %r8d,%rax
  lea    0xc(%rax),%rdi
  cmp    %rbp,%rdi
  ja     L38
  lea    0x10(%rbx),%r9
  lea    0x11(%rbx,%rax,1),%rbx
  lea    0x1(%r8),%eax
  cltq
  sub    %rax,%rbp
  lea    -0x10(%rbp),%rbp
  movsbl %cl,%ecx
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0xb,%rbp
  ja     L99
  jmp    L72
L100:
  mov    %rsp,%rbx
  jmp    L72
L20:
  mov    %rsp,%rdi
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  call   <memmove@plt>
  mov    %rbp,%r13
  jmp    L87
L34:
  mov    0x10008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L63
  add    $0x10010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L63:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
data_type: string

tool: D2: %i %i %c %u %s


assistant: ```c
void parse_inputstream(FILE *input_file) {
  char buf[64 * 1024];
  ssize_t buf_size = sizeof(buf);
  ssize_t buf_offset = 0;
  int read;
  while ((read = fread(buf + buf_offset, 1, buf_size - buf_offset, input_file)) > 0) {
    size_t buf_alive = buf_offset + read;
    const char *bufp = buf;
    while (buf_alive >= 3 * sizeof(uint64_t)) {
      int64_t timestamp_secs = ((int64_t *)bufp)[0];
      uint64_t size = ((uint64_t *)bufp)[1];
      char event_type = (char)((uint64_t *)bufp)[2];
      uint64_t len_of_path = ((uint64_t *)bufp)[3];
      if (3 * sizeof(uint64_t) + len_of_path > buf_alive) {
        buf_offset = buf_alive;
        memmove(buf, bufp, buf_alive);
        break;
      }
      const char *path = bufp + 4 * sizeof(uint64_t);
      bufp += len_of_path + 1 + 4 * sizeof(uint64_t);
      buf_alive -= len_of_path + 1 + 4 * sizeof(uint64_t);
      printf("%i %i %c %u %s\n", timestamp_secs, size, event_type, len_of_path, path);
    }
    if (4 * sizeof(uint64_t) >= buf_alive) {
      buf_offset = buf_alive;
      memmove(buf, bufp, buf_alive);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<parse_inputstream>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x10000(%rsp),%r11
L50:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L50
  sub    $0x10,%rsp
  xor    %ebx,%ebx
  lea    D8(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x10008(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r14
  mov    %rsp,%r13
  nopl   0x0(%rax,%rax,1)
L51:
  mov    $0x10000,%edx
  lea    0x0(%r13,%rbx,1),%rdi
  mov    %r14,%rcx
  mov    $0x1,%esi
  sub    %rbx,%rdx
  call   <fread@plt>
  test   %eax,%eax
  jle    L15
  cltq
  mov    %r13,%rbp
  add    %rax,%rbx
  cmp    $0xb,%rbx
  ja     L30
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L78:
  lea    0x10(%rbp),%r9
  lea    0x11(%rbp,%rax,1),%rbp
  movsbl %cl,%ecx
  mov    %r12,%rdi
  lea    0x1(%rax),%eax
  cltq
  sub    %rax,%rbx
  sub    $0x10,%rbx
  call   <printf@plt>
  cmp    $0xb,%rbx
  jbe    L4
L30:
  movslq 0xc(%rbp),%rax
  mov    0x0(%rbp),%esi
  mov    0x4(%rbp),%edx
  mov    0x8(%rbp),%ecx
  lea    0xc(%rax),%rdi
  mov    %rax,%r8
  cmp    %rbx,%rdi
  jbe    L78
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <memmove@plt>
  cmp    $0x10,%rbx
  ja     L51
  nop
L4:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <memmove@plt>
  jmp    L51
L15:
  mov    0x10008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D8
type_infered: string

tool: D8: %i %i %c %u %s


assistant: ```c
void parse_inputstream(FILE *input_file) {
  char buf[64 * 1024];
  ssize_t buf_size = sizeof(buf);
  ssize_t buf_offset = 0;
  int read;
  while ((read = fread(buf + buf_offset, 1, buf_size - buf_offset, input_file)) > 0) {
    size_t buf_alive = buf_offset + read;
    const char *bufp = buf;
    while (buf_alive >= 3 * sizeof(uint64_t)) {
      int64_t timestamp_secs = ((int64_t *)bufp)[0];
      uint64_t size = ((uint64_t *)bufp)[1];
      char event_type = (char)((uint64_t *)bufp)[2];
      uint64_t len_of_path = ((uint64_t *)bufp)[3];
      if (3 * sizeof(uint64_t) + len_of_path > buf_alive) {
        buf_offset = buf_alive;
        memmove(buf, bufp, buf_alive);
        break;
      }
      const char *path = bufp + 4 * sizeof(uint64_t);
      bufp += len_of_path + 1 + 4 * sizeof(uint64_t);
      buf_alive -= len_of_path + 1 + 4 * sizeof(uint64_t);
      printf("%i %i %c %u %s\n", timestamp_secs, size, event_type, len_of_path, path);
    }
    if (4 * sizeof(uint64_t) >= buf_alive) {
      buf_offset = buf_alive;
      memmove(buf, bufp, buf_alive);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_inputstream>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x10000(%rsp),%r11
L83:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L83
  sub    $0x10,%rsp
  xor    %ebx,%ebx
  lea    D9(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x10008(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r14
  mov    %rsp,%r13
  nopl   0x0(%rax,%rax,1)
L40:
  mov    $0x10000,%edx
  lea    0x0(%r13,%rbx,1),%rdi
  mov    %r14,%rcx
  mov    $0x1,%esi
  sub    %rbx,%rdx
  call   <fread@plt>
  test   %eax,%eax
  jle    L61
  cltq
  mov    %r13,%rbp
  add    %rax,%rbx
  cmp    $0xb,%rbx
  ja     L87
  jmp    L73
  nopw   0x0(%rax,%rax,1)
L44:
  lea    0x10(%rbp),%r9
  lea    0x11(%rbp,%rax,1),%rbp
  movsbl %cl,%ecx
  mov    %r12,%rdi
  lea    0x1(%rax),%eax
  cltq
  sub    %rax,%rbx
  sub    $0x10,%rbx
  call   <printf@plt>
  cmp    $0xb,%rbx
  jbe    L73
L87:
  movslq 0xc(%rbp),%rax
  mov    0x0(%rbp),%esi
  mov    0x4(%rbp),%edx
  mov    0x8(%rbp),%ecx
  lea    0xc(%rax),%rdi
  mov    %rax,%r8
  cmp    %rbx,%rdi
  jbe    L44
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <memmove@plt>
  cmp    $0x10,%rbx
  ja     L40
  nop
L73:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <memmove@plt>
  jmp    L40
L61:
  mov    0x10008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  add    $0x10010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_infered: string

tool: D9: %i %i %c %u %s


assistant: ```c
void parse_inputstream(FILE *input_file) {
  char buf[64 * 1024];
  ssize_t buf_size = sizeof(buf);
  ssize_t buf_offset = 0;
  int read;
  while ((read = fread(buf + buf_offset, 1, buf_size - buf_offset, input_file)) > 0) {
    size_t buf_alive = buf_offset + read;
    const char *bufp = buf;
    while (buf_alive >= 3 * sizeof(uint64_t)) {
      int64_t timestamp_secs = ((int64_t *)bufp)[0];
      uint64_t size = ((uint64_t *)bufp)[1];
      char event_type = (char)((uint64_t *)bufp)[2];
      uint64_t len_of_path = ((uint64_t *)bufp)[3];
      if (3 * sizeof(uint64_t) + len_of_path > buf_alive) {
        buf_offset = buf_alive;
        memmove(buf, bufp, buf_alive);
        break;
      }
      const char *path = bufp + 4 * sizeof(uint64_t);
      bufp += len_of_path + 1 + 4 * sizeof(uint64_t);
      buf_alive -= len_of_path + 1 + 4 * sizeof(uint64_t);
      printf("%i %i %c %u %s\n", timestamp_secs, size, event_type, len_of_path, path);
    }
    if (4 * sizeof(uint64_t) >= buf_alive) {
      buf_offset = buf_alive;
      memmove(buf, bufp, buf_alive);
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<csum_tcpudp_nofold>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %ecx,%eax
  mov    %r8d,-0x1c(%rbp)
  mov    %dx,-0x14(%rbp)
  mov    %ax,-0x18(%rbp)
  movzwl -0x14(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ntohs@plt>
  shl    $0x10,%eax
  mov    %eax,%edx
  movzwl -0x18(%rbp),%eax
  shl    $0x8,%eax
  add    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    -0x10(%rbp),%eax
  adc    -0x8(%rbp),%eax
  adc    %edx,%eax
  adc    $0x0,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned long csum_tcpudp_nofold(unsigned long saddr, unsigned long daddr, unsigned short len, unsigned short proto, unsigned int sum) {
  __asm__("addl %1, %0	;\n"
          "adcl %2, %0	;\n"
          "adcl %3, %0	;\n"
          "adcl $0, %0	;\n"
          : "=r"(sum)
          : "g"(daddr), "g"(saddr), "g"((ntohs(len) << 16) + proto * 256), "0"(sum));
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fib>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L0
  movq   $0x0,-0x18(%rbp)
  jmp    L30
L0:
  cmpl   $0x1,-0x24(%rbp)
  jne    L2
  movq   $0x1,-0x18(%rbp)
  jmp    L30
L2:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fib>
  mov    %rax,%rbx
  mov    -0x24(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fib>
  add    %rbx,%rax
  mov    %rax,-0x18(%rbp)
L30:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long fib(int n) {
  long result;
  if (n == 0)
    result = 0;
  else if (n == 1)
    result = 1;
  else
    result = fib(n - 1) + fib(n - 2);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fib>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  je     L20
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jne    L19
L23:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  lea    -0x1(%rdi),%edi
  call   <fib>
  mov    %rax,%rbp
  lea    -0x2(%rbx),%edi
  call   <fib>
  add    %rbp,%rax
  jmp    L23
L20:
  ret
```

assistant: ```c
long fib(int n) {
  long result;
  if (n == 0)
    result = 0;
  else if (n == 1)
    result = 1;
  else
    result = fib(n - 1) + fib(n - 2);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <os_process_arguments@plt>
  mov    $0x0,%eax
  call   <init_memory@plt>
  mov    $0x0,%eax
  call   <os_init_screen@plt>
  mov    $0x0,%eax
  call   <init_undo@plt>
  mov    $0x0,%eax
  call   <z_restart@plt>
  mov    $0x0,%eax
  call   <interpret@plt>
  mov    $0x0,%eax
  call   <reset_memory@plt>
  mov    $0x0,%eax
  call   <os_reset_screen@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  os_process_arguments(argc, argv);
  init_memory();
  os_init_screen();
  init_undo();
  z_restart();
  interpret();
  reset_memory();
  os_reset_screen();
  return 0;
}
```