ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<timeval_diff_msec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  imul   $0x3e8,%eax,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long long timeval_diff_msec(struct timeval *start_ts, struct timeval *end_ts) {
  unsigned long long msec;
  msec = (end_ts->tv_sec - start_ts->tv_sec) * 1000;
  msec += (end_ts->tv_usec - start_ts->tv_usec) / 1000;
  return msec;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<timeval_diff_msec>:
  endbr64
  mov    (%rsi),%edx
  sub    (%rdi),%edx
  imul   $0x3e8,%edx,%edx
  movslq %edx,%rdx
  mov    0x4(%rsi),%ecx
  sub    0x4(%rdi),%ecx
  movslq %ecx,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  cltq
  add    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long timeval_diff_msec(struct timeval *start_ts, struct timeval *end_ts) {
  unsigned long long msec;
  msec = (end_ts->tv_sec - start_ts->tv_sec) * 1000;
  msec += (end_ts->tv_usec - start_ts->tv_usec) / 1000;
  return msec;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<timeval_diff_msec>:
  endbr64
  mov    0x4(%rsi),%ecx
  sub    0x4(%rdi),%ecx
  movslq %ecx,%rax
  mov    (%rsi),%edx
  sub    (%rdi),%edx
  sar    $0x1f,%ecx
  imul   $0x10624dd3,%rax,%rax
  imul   $0x3e8,%edx,%edx
  sar    $0x26,%rax
  sub    %ecx,%eax
  movslq %edx,%rdx
  cltq
  add    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long timeval_diff_msec(struct timeval *start_ts, struct timeval *end_ts) {
  unsigned long long msec;
  msec = (end_ts->tv_sec - start_ts->tv_sec) * 1000;
  msec += (end_ts->tv_usec - start_ts->tv_usec) / 1000;
  return msec;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timeval_diff_msec>:
  endbr64
  mov    0x4(%rsi),%ecx
  sub    0x4(%rdi),%ecx
  movslq %ecx,%rax
  mov    (%rsi),%edx
  sub    (%rdi),%edx
  sar    $0x1f,%ecx
  imul   $0x10624dd3,%rax,%rax
  imul   $0x3e8,%edx,%edx
  sar    $0x26,%rax
  sub    %ecx,%eax
  movslq %edx,%rdx
  cltq
  add    %rdx,%rax
  ret
```

assistant: ```c
unsigned long long timeval_diff_msec(struct timeval *start_ts, struct timeval *end_ts) {
  unsigned long long msec;
  msec = (end_ts->tv_sec - start_ts->tv_sec) * 1000;
  msec += (end_ts->tv_usec - start_ts->tv_usec) / 1000;
  return msec;
}
```
user: Translate the below assembly code into C programming language:

```asm
<def_target_pink>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rcx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    (%rsi),%r12d
  mov    0x4(%rsi),%ebx
  mov    0x8(%rsi),%eax
  mov    %r12d,0xc(%rsp)
  mov    %ebx,0x10(%rsp)
  mov    %eax,0x14(%rsp)
  cmp    $0x1,%eax
  je     L16
  cmp    $0x2,%eax
  je     L128
  cmp    $0x3,%eax
  je     L48
  cmp    $0x4,%eax
  je     L76
L3:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  sub    $0x4,%r12d
  mov    %r12d,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L3
  mov    %r12d,0x0(%rbp)
  mov    %ebx,0x4(%rbp)
  jmp    L3
L128:
  add    $0x4,%ebx
  mov    %ebx,0x10(%rsp)
  lea    0xc(%rsp),%rsi
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L3
  mov    %r12d,0x0(%rbp)
  mov    %ebx,0x4(%rbp)
  jmp    L3
L48:
  add    $0x4,%r12d
  mov    %r12d,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L3
  mov    %r12d,0x0(%rbp)
  mov    %ebx,0x4(%rbp)
  jmp    L3
L76:
  sub    $0x4,%ebx
  mov    %ebx,0x10(%rsp)
  lea    0xc(%rsp),%rsi
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L3
  mov    %r12d,0x0(%rbp)
  mov    %ebx,0x4(%rbp)
  jmp    L3
L75:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void def_target_pink(char maze[23][30], int ghost[3], int pacman[3], int target[2]) {
  int result, copia[3], i = ghost[0], j = ghost[1], d = ghost[2];
  copia[0] = ghost[0];
  copia[1] = ghost[1];
  copia[2] = ghost[2];
  if (d == 1) {
    copia[0] -= 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i - 4;
      target[1] = j;
    }
  } else if (d == 2) {
    copia[1] += 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i;
      target[1] = j + 4;
    }
  } else if (d == 3) {
    copia[0] += 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i + 4;
      target[1] = j;
    }
  } else if (d == 4) {
    copia[1] -= 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i;
      target[1] = j - 4;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<def_target_pink>:
  endbr64
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    (%rsi),%ebx
  mov    0x4(%rsi),%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsi),%eax
  mov    %ebx,0xc(%rsp)
  mov    %ebp,0x10(%rsp)
  mov    %eax,0x14(%rsp)
  cmp    $0x1,%eax
  je     L104
  cmp    $0x2,%eax
  je     L30
  cmp    $0x3,%eax
  je     L1
  cmp    $0x4,%eax
  je     L59
L80:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L104:
  sub    $0x4,%ebx
L69:
  lea    0xc(%rsp),%rsi
  mov    %ebx,0xc(%rsp)
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L80
L9:
  mov    %ebx,(%r12)
  mov    %ebp,0x4(%r12)
  jmp    L80
  nopl   0x0(%rax)
L59:
  sub    $0x4,%ebp
L24:
  lea    0xc(%rsp),%rsi
  mov    %ebp,0x10(%rsp)
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L80
  jmp    L9
L30:
  add    $0x4,%ebp
  jmp    L24
  nopl   (%rax)
L1:
  add    $0x4,%ebx
  jmp    L69
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void def_target_pink(char maze[23][30], int ghost[3], int pacman[3], int target[2]) {
  int result, copia[3], i = ghost[0], j = ghost[1], d = ghost[2];
  copia[0] = ghost[0];
  copia[1] = ghost[1];
  copia[2] = ghost[2];
  if (d == 1) {
    copia[0] -= 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i - 4;
      target[1] = j;
    }
  } else if (d == 2) {
    copia[1] += 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i;
      target[1] = j + 4;
    }
  } else if (d == 3) {
    copia[0] += 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i + 4;
      target[1] = j;
    }
  } else if (d == 4) {
    copia[1] -= 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i;
      target[1] = j - 4;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<def_target_pink>:
  endbr64
  push   %r14
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x28,%rsp
  movq   (%rsi),%xmm0
  mov    (%rsi),%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsi),%eax
  mov    0x4(%rsi),%r9d
  movd   %xmm0,%edx
  movq   %xmm0,0x8(%rsp)
  pshufd $0xe5,%xmm0,%xmm1
  mov    %eax,0x10(%rsp)
  cmp    $0x1,%eax
  je     L13
  movd   %xmm1,%ecx
  cmp    $0x2,%eax
  je     L106
  cmp    $0x3,%eax
  je     L40
  cmp    $0x4,%eax
  je     L62
L54:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L92
  add    $0x28,%rsp
  pop    %rbx
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  sub    $0x4,%edx
  movd   %r9d,%xmm6
  movd   %edx,%xmm2
  punpckldq %xmm6,%xmm2
  movq   %xmm2,%r14
L9:
  lea    0x8(%rsp),%rsi
  mov    %edx,0x8(%rsp)
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L54
L76:
  mov    %r14,(%rbx)
  jmp    L54
  nopl   (%rax)
L62:
  sub    $0x4,%ecx
  movd   %r8d,%xmm5
  movd   %ecx,%xmm6
  punpckldq %xmm6,%xmm5
  movq   %xmm5,%r14
L34:
  lea    0x8(%rsp),%rsi
  mov    %ecx,0xc(%rsp)
  call   <validate_position@plt>
  cmp    $0x1,%eax
  jne    L54
  jmp    L76
  nopw   0x0(%rax,%rax,1)
L106:
  add    $0x4,%ecx
  movd   %r8d,%xmm3
  movd   %ecx,%xmm7
  punpckldq %xmm7,%xmm3
  movq   %xmm3,%r14
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L40:
  add    $0x4,%edx
  movd   %r9d,%xmm7
  movd   %edx,%xmm4
  punpckldq %xmm7,%xmm4
  movq   %xmm4,%r14
  jmp    L9
L92:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void def_target_pink(char maze[23][30], int ghost[3], int pacman[3], int target[2]) {
  int result, copia[3], i = ghost[0], j = ghost[1], d = ghost[2];
  copia[0] = ghost[0];
  copia[1] = ghost[1];
  copia[2] = ghost[2];
  if (d == 1) {
    copia[0] -= 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i - 4;
      target[1] = j;
    }
  } else if (d == 2) {
    copia[1] += 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i;
      target[1] = j + 4;
    }
  } else if (d == 3) {
    copia[0] += 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i + 4;
      target[1] = j;
    }
  } else if (d == 4) {
    copia[1] -= 4;
    result = validate_position(maze, copia);
    if (result == 1) {
      target[0] = i;
      target[1] = j - 4;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edx
  mov    $0x2,%esi
  mov    $0x0,%edi
  call   <__builtin_arm_prefetch@plt>
  mov    $0x2,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <__builtin_arm_prefetch@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test4() {
  __builtin_arm_prefetch(0, 2, 0);
  __builtin_arm_prefetch(0, 0, 2);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test4>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edx
  mov    $0x2,%esi
  mov    $0x0,%edi
  call   <__builtin_arm_prefetch@plt>
  mov    $0x2,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <__builtin_arm_prefetch@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test4() {
  __builtin_arm_prefetch(0, 2, 0);
  __builtin_arm_prefetch(0, 0, 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<test4>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  mov    $0x2,%esi
  xor    %edi,%edi
  call   <__builtin_arm_prefetch@plt>
  mov    $0x2,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1050 <__builtin_arm_prefetch@plt>
```

assistant: ```c
void test4() {
  __builtin_arm_prefetch(0, 2, 0);
  __builtin_arm_prefetch(0, 0, 2);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test4>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  mov    $0x2,%esi
  xor    %edi,%edi
  call   <__builtin_arm_prefetch@plt>
  mov    $0x2,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1050 <__builtin_arm_prefetch@plt>
```

assistant: ```c
void test4() {
  __builtin_arm_prefetch(0, 2, 0);
  __builtin_arm_prefetch(0, 0, 2);
}
```
user: What is the c source code of the assembly code below:

```asm
<tree_get_parent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L20
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L0
L20:
  mov    $0x0,%eax
  jmp    L19
L38:
  mov    -0x8(%rbp),%rax
  mov    -0x8(%rax),%rax
  mov    %rax,-0x8(%rbp)
L0:
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rax
  mov    (%rax),%rax
  sub    $0x10,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  je     L38
  mov    -0x8(%rbp),%rax
  mov    -0x8(%rax),%rax
L19:
  pop    %rbp
  ret
```

assistant: ```c
void *tree_get_parent(void *mem) {
  if (!mem || (!(((void **)(mem))[-1])))
    return NULL;
  while (!((((void **)((((void **)(mem))[-1])))[-2]) != (mem)))
    mem = (((void **)(mem))[-1]);
  return (((void **)(mem))[-1]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tree_get_parent>:
  endbr64
  mov    %rdi,%rax
  test   %rdi,%rdi
  je     L20
  mov    -0x8(%rdi),%rdx
  test   %rdx,%rdx
  je     L23
L0:
  mov    %rax,%rdx
  mov    -0x8(%rax),%rax
  cmp    %rdx,-0x10(%rax)
  je     L0
  ret
L23:
  mov    %rdx,%rax
L20:
  ret
```

assistant: ```c
void *tree_get_parent(void *mem) {
  if (!mem || (!(((void **)(mem))[-1])))
    return NULL;
  while (!((((void **)((((void **)(mem))[-1])))[-2]) != (mem)))
    mem = (((void **)(mem))[-1]);
  return (((void **)(mem))[-1]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tree_get_parent>:
  endbr64
  test   %rdi,%rdi
  je     L0
  mov    -0x8(%rdi),%rax
  test   %rax,%rax
  jne    L13
  jmp    L6
  nopl   0x0(%rax)
L33:
  mov    %rax,%rdi
  mov    -0x8(%rax),%rax
L13:
  cmp    %rdi,-0x10(%rax)
  je     L33
L6:
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *tree_get_parent(void *mem) {
  if (!mem || (!(((void **)(mem))[-1])))
    return NULL;
  while (!((((void **)((((void **)(mem))[-1])))[-2]) != (mem)))
    mem = (((void **)(mem))[-1]);
  return (((void **)(mem))[-1]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tree_get_parent>:
  endbr64
  test   %rdi,%rdi
  je     L27
  mov    -0x8(%rdi),%rax
  test   %rax,%rax
  je     L10
  cmp    -0x10(%rax),%rdi
  jne    L10
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %rax,%rdx
  mov    -0x8(%rax),%rax
  cmp    %rdx,-0x10(%rax)
  je     L6
  ret
  xchg   %ax,%ax
L27:
  xor    %eax,%eax
L10:
  ret
```

assistant: ```c
void *tree_get_parent(void *mem) {
  if (!mem || (!(((void **)(mem))[-1])))
    return NULL;
  while (!((((void **)((((void **)(mem))[-1])))[-2]) != (mem)))
    mem = (((void **)(mem))[-1]);
  return (((void **)(mem))[-1]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_palindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <is_palindrome_aux@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int is_palindrome(struct list_node *head) {
  struct list_node *out;
  return is_palindrome_aux(head, head, &out);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<is_palindrome>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    %rdi,%rsi
  call   <is_palindrome_aux@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_palindrome(struct list_node *head) {
  struct list_node *out;
  return is_palindrome_aux(head, head, &out);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_palindrome>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  call   <is_palindrome_aux@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_palindrome(struct list_node *head) {
  struct list_node *out;
  return is_palindrome_aux(head, head, &out);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<is_palindrome>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  call   <is_palindrome_aux@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_palindrome(struct list_node *head) {
  struct list_node *out;
  return is_palindrome_aux(head, head, &out);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<report>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %rdi,-0xa8(%rbp)
  mov    %esi,-0xac(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0xac(%rbp)
  jle    L24
  mov    -0xac(%rbp),%eax
  mov    %eax,%edi
  call   <strerror@plt>
  mov    %eax,%edx
  mov    -0xa8(%rbp),%rcx
  lea    -0x90(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  jmp    L40
L24:
  mov    -0xa8(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L40:
  movl   $0x0,-0x94(%rbp)
  cmpl   $0x0,-0x94(%rbp)
  je     L38
  mov    D29(%rip),%rax
  mov    (%rax),%edx
  lea    -0x90(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <fputs@plt>
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <fputc@plt>
  jmp    L35
L38:
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  lea    -0x90(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <syslog@plt>
L35:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
void report(char *logmsg, int errnum) {
  char buf[127];
  int debug;
  if (errnum > 0) {
    sprintf(buf, logmsg, strerror(errnum));
  } else {
    strcpy(buf, logmsg);
  }
  debug = 0;
  if (debug) {
    fputs(buf, stderr);
    fputc('\n', stderr);
  } else {
    syslog(LOG_WARNING, buf);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<report>:
  endbr64
  push   %rbx
  sub    $0x90,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L13
  mov    %esi,%edi
  call   <strerror@plt>
  mov    %eax,%edx
  mov    %rsp,%rdi
  mov    %rbx,%rsi
  call   <sprintf@plt>
L1:
  mov    %rsp,%rsi
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <syslog@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x90,%rsp
  pop    %rbx
  ret
L13:
  mov    %rsp,%rdi
  mov    %rbx,%rsi
  call   <strcpy@plt>
  jmp    L1
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void report(char *logmsg, int errnum) {
  char buf[127];
  int debug;
  if (errnum > 0) {
    sprintf(buf, logmsg, strerror(errnum));
  } else {
    strcpy(buf, logmsg);
  }
  debug = 0;
  if (debug) {
    fputs(buf, stderr);
    fputc('\n', stderr);
  } else {
    syslog(LOG_WARNING, buf);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<report>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L15
  mov    %esi,%edi
  mov    %rsp,%r12
  call   <strerror@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    %eax,%edx
  call   <sprintf@plt>
L17:
  mov    D8(%rip),%rax
  mov    %r12,%rsi
  mov    (%rax),%edi
  call   <syslog@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x98,%rsp
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L15:
  mov    %rsp,%r12
  mov    %rdi,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L17
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void report(char *logmsg, int errnum) {
  char buf[127];
  int debug;
  if (errnum > 0) {
    sprintf(buf, logmsg, strerror(errnum));
  } else {
    strcpy(buf, logmsg);
  }
  debug = 0;
  if (debug) {
    fputs(buf, stderr);
    fputc('\n', stderr);
  } else {
    syslog(LOG_WARNING, buf);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<report>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L7
  mov    %esi,%edi
  mov    %rsp,%r12
  call   <strerror@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    %eax,%edx
  call   <sprintf@plt>
L15:
  mov    D1(%rip),%rax
  mov    %r12,%rsi
  mov    (%rax),%edi
  call   <syslog@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x98,%rsp
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L7:
  mov    %rsp,%r12
  mov    %rdi,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L15
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void report(char *logmsg, int errnum) {
  char buf[127];
  int debug;
  if (errnum > 0) {
    sprintf(buf, logmsg, strerror(errnum));
  } else {
    strcpy(buf, logmsg);
  }
  debug = 0;
  if (debug) {
    fputs(buf, stderr);
    fputc('\n', stderr);
  } else {
    syslog(LOG_WARNING, buf);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<kaijyou>:
  endbr64
  test   %edi,%edi
  js     L0
  mov    D11(%rip),%eax
  cmp    %edi,%eax
  jge    L19
  movslq %eax,%rcx
  lea    D41(%rip),%rdx
  movsd  (%rdx,%rcx,8),%xmm0
  add    $0x1,%eax
  cmp    %eax,%edi
  jl     L36
  cltq
L30:
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%edi
  jge    L30
L36:
  mov    %edi,D37(%rip)
  ret
L19:
  movslq %edi,%rdi
  lea    D41(%rip),%rax
  movsd  (%rax,%rdi,8),%xmm0
  ret
L0:
  movsd  D37(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D37
datatype: f64

tool: D37: 1

assistant: ```c
double kaijyou(int n) {
  static double bank[1000] = {1, 1};
  static int nmax = 1;
  int i;
  double a;
  if (n < 0) {
    return 1.0;
  } else if (n <= nmax) {
    return bank[n];
  }
  a = bank[nmax];
  for (i = nmax + 1; i <= n; i++) {
    a *= i;
    bank[i] = a;
  }
  nmax = n;
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<kaijyou>:
  endbr64
  test   %edi,%edi
  js     L6
  mov    D37(%rip),%eax
  cmp    %edi,%eax
  jge    L5
  movslq %eax,%rcx
  lea    D29(%rip),%rdx
  add    $0x1,%eax
  movsd  (%rdx,%rcx,8),%xmm0
  cmp    %eax,%edi
  jl     L16
  cltq
  nopw   0x0(%rax,%rax,1)
L31:
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%edi
  jge    L31
L16:
  mov    %edi,D11(%rip)
  ret
L5:
  movslq %edi,%rdi
  lea    D29(%rip),%rax
  movsd  (%rax,%rdi,8),%xmm0
  ret
L6:
  movsd  D11(%rip),%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D11
datatype: f64

tool: D11: 1

assistant: ```c
double kaijyou(int n) {
  static double bank[1000] = {1, 1};
  static int nmax = 1;
  int i;
  double a;
  if (n < 0) {
    return 1.0;
  } else if (n <= nmax) {
    return bank[n];
  }
  a = bank[nmax];
  for (i = nmax + 1; i <= n; i++) {
    a *= i;
    bank[i] = a;
  }
  nmax = n;
  return a;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<kaijyou>:
  endbr64
  test   %edi,%edi
  js     L27
  mov    D32(%rip),%eax
  cmp    %edi,%eax
  jge    L30
  movslq %eax,%rcx
  lea    D14(%rip),%rdx
  add    $0x1,%eax
  movsd  (%rdx,%rcx,8),%xmm0
  cmp    %eax,%edi
  jl     L7
  cltq
  nopw   0x0(%rax,%rax,1)
L34:
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%edi
  jge    L34
L7:
  mov    %edi,D46(%rip)
  ret
L30:
  movslq %edi,%rdi
  lea    D14(%rip),%rax
  movsd  (%rax,%rdi,8),%xmm0
  ret
L27:
  movsd  D46(%rip),%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: tag: D46
datatype: f64

tool: D46: 1

assistant: ```c
double kaijyou(int n) {
  static double bank[1000] = {1, 1};
  static int nmax = 1;
  int i;
  double a;
  if (n < 0) {
    return 1.0;
  } else if (n <= nmax) {
    return bank[n];
  }
  a = bank[nmax];
  for (i = nmax + 1; i <= n; i++) {
    a *= i;
    bank[i] = a;
  }
  nmax = n;
  return a;
}
```