ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sys_audiodevnametonumber>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xa000(%rsp),%r11
L60:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L60
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xa018(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  movl   $0x0,(%rsp)
  mov    %edi,%ebp
  movl   $0x0,0x4(%rsp)
  lea    0x5010(%rsp),%r12
  lea    0x10(%rsp),%r14
  mov    %rsi,%rbx
  mov    %rsp,%rsi
  push   $0x400
  mov    %r12,%rdx
  mov    %r14,%rdi
  push   $0x14
  lea    0x1c(%rsp),%r9
  lea    0x18(%rsp),%r8
  call   <sys_get_audio_devs@plt>
  pop    %rax
  pop    %rdx
  test   %ebp,%ebp
  jne    L21
  mov    (%rsp),%r13d
  test   %r13d,%r13d
  jle    L72
  mov    %rbx,%rdi
  xor    %r12d,%r12d
  call   <strlen@plt>
  mov    %rax,%rbp
  xchg   %ax,%ax
L87:
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %rbx,%rdi
  cmp    %rbp,%rax
  cmova  %rbp,%rax
  mov    %rax,%rdx
  call   <strncmp@plt>
  test   %eax,%eax
  je     L80
  add    $0x1,%r12d
  add    $0x400,%r14
  cmp    %r13d,%r12d
  jne    L87
L72:
  mov    $0xffffffff,%r12d
L80:
  mov    0xa018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L91
  add    $0xa020,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L21:
  mov    0x4(%rsp),%r13d
  test   %r13d,%r13d
  jle    L72
  mov    %rbx,%rdi
  mov    %r12,%r14
  xor    %r12d,%r12d
  call   <strlen@plt>
  mov    %rax,%rbp
  nopl   0x0(%rax,%rax,1)
L93:
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %rbx,%rdi
  cmp    %rbp,%rax
  cmova  %rbp,%rax
  mov    %rax,%rdx
  call   <strncmp@plt>
  test   %eax,%eax
  je     L80
  add    $0x1,%r12d
  add    $0x400,%r14
  cmp    %r12d,%r13d
  jne    L93
  jmp    L72
L91:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sys_audiodevnametonumber(int output, const char *name) {
  char indevlist[20 * 1024], outdevlist[20 * 1024];
  int nindevs = 0, noutdevs = 0, i, canmulti, cancallback;
  sys_get_audio_devs(indevlist, &nindevs, outdevlist, &noutdevs, &canmulti, &cancallback, 20, 1024);
  if (output) {
    for (i = 0; i < noutdevs; i++) {
      unsigned long comp = strlen(name);
      if (comp > strlen(outdevlist + i * 1024))
        comp = strlen(outdevlist + i * 1024);
      if (!strncmp(name, outdevlist + i * 1024, comp))
        return (i);
    }
  } else {
    for (i = 0; i < nindevs; i++) {
      unsigned long comp = strlen(name);
      if (comp > strlen(indevlist + i * 1024))
        comp = strlen(indevlist + i * 1024);
      if (!strncmp(name, indevlist + i * 1024, comp))
        return (i);
    }
  }
  return (-1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sortScores>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x44(%rbp),%eax
  sub    $0x1,%eax
  mov    %ax,-0x38(%rbp)
  jmp    L27
L5:
  movw   $0x0,-0x36(%rbp)
  jmp    L17
L38:
  movzwl -0x36(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  movzwl -0x36(%rbp),%eax
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x50(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jg     L18
  movzwl -0x36(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x34(%rbp)
  movzwl -0x36(%rbp),%eax
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  movzwl -0x36(%rbp),%edx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x50(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  movzwl -0x36(%rbp),%eax
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x50(%rbp),%rax
  add    %rax,%rdx
  mov    -0x34(%rbp),%eax
  mov    %eax,(%rdx)
  movzwl -0x36(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movzwl -0x36(%rbp),%eax
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  movzwl -0x36(%rbp),%eax
  lea    0x0(,%rax,8),%rcx
  mov    -0x58(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movzwl -0x36(%rbp),%eax
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  lea    -0x30(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L18:
  movzwl -0x36(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x36(%rbp)
L17:
  movzwl -0x36(%rbp),%eax
  cmp    -0x38(%rbp),%ax
  jb     L38
  movzwl -0x38(%rbp),%eax
  sub    $0x1,%eax
  mov    %ax,-0x38(%rbp)
L27:
  cmpw   $0x0,-0x38(%rbp)
  jne    L5
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
void sortScores(const int nbScores, int *arrayScores, char **arrayNames) {
  unsigned short i, j;
  int tmpScore;
  char tmpName[35];
  for (i = nbScores - 1; i >= 1; i--) {
    for (j = 0; j < i; j++) {
      if (arrayScores[j] <= arrayScores[j + 1]) {
        tmpScore = arrayScores[j];
        arrayScores[j] = arrayScores[j + 1];
        arrayScores[j + 1] = tmpScore;
        strcpy(tmpName, arrayNames[j]);
        strcpy(arrayNames[j], arrayNames[j + 1]);
        strcpy(arrayNames[j + 1], tmpName);
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sortScores>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rsi,(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%di
  je     L8
  lea    -0x2(%rdi),%r14d
  lea    0x8(%rsi),%r15
  lea    0x10(%rsp),%r13
  jmp    L56
L84:
  add    $0x4,%rbx
  add    $0x8,%rbp
  cmp    %r12,%rbx
  je     L15
L74:
  mov    -0x4(%rbx),%eax
  mov    (%rbx),%edx
  cmp    %edx,%eax
  jg     L84
  mov    %edx,-0x4(%rbx)
  mov    %eax,(%rbx)
  mov    -0x8(%rbp),%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rdi
  mov    0x0(%rbp),%rsi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    0x0(%rbp),%rdi
  call   <strcpy@plt>
  jmp    L84
L15:
  test   %r14w,%r14w
  je     L8
L14:
  sub    $0x1,%r14d
L56:
  cmp    $0xffff,%r14w
  je     L14
  mov    (%rsp),%rax
  lea    0x4(%rax),%rbx
  mov    0x8(%rsp),%rax
  lea    0x8(%rax),%rbp
  movzwl %r14w,%eax
  lea    (%r15,%rax,4),%r12
  jmp    L74
L8:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortScores(const int nbScores, int *arrayScores, char **arrayNames) {
  unsigned short i, j;
  int tmpScore;
  char tmpName[35];
  for (i = nbScores - 1; i >= 1; i--) {
    for (j = 0; j < i; j++) {
      if (arrayScores[j] <= arrayScores[j + 1]) {
        tmpScore = arrayScores[j];
        arrayScores[j] = arrayScores[j + 1];
        arrayScores[j + 1] = tmpScore;
        strcpy(tmpName, arrayNames[j]);
        strcpy(arrayNames[j], arrayNames[j + 1]);
        strcpy(arrayNames[j + 1], tmpName);
      }
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sortScores>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  sub    $0x1,%r12w
  je     L57
  mov    %rsi,%rbp
  mov    %rdx,%r14
  mov    %rsp,%r13
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %ebx,%ebx
  jmp    L55
  nopl   0x0(%rax)
L45:
  add    $0x1,%rbx
  cmp    %bx,%r12w
  jbe    L28
L55:
  mov    0x0(%rbp,%rbx,4),%eax
  mov    0x4(%rbp,%rbx,4),%edx
  cmp    %edx,%eax
  jg     L45
  mov    %edx,0x0(%rbp,%rbx,4)
  mov    (%r14,%rbx,8),%rsi
  mov    %r13,%rdi
  mov    %eax,0x4(%rbp,%rbx,4)
  call   <strcpy@plt>
  mov    0x8(%r14,%rbx,8),%rsi
  mov    (%r14,%rbx,8),%rdi
  call   <strcpy@plt>
  mov    0x8(%r14,%rbx,8),%rdi
  mov    %r13,%rsi
  add    $0x1,%rbx
  call   <strcpy@plt>
  cmp    %bx,%r12w
  ja     L55
L28:
  sub    $0x1,%r12w
  jne    L0
L57:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L44:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortScores(const int nbScores, int *arrayScores, char **arrayNames) {
  unsigned short i, j;
  int tmpScore;
  char tmpName[35];
  for (i = nbScores - 1; i >= 1; i--) {
    for (j = 0; j < i; j++) {
      if (arrayScores[j] <= arrayScores[j + 1]) {
        tmpScore = arrayScores[j];
        arrayScores[j] = arrayScores[j + 1];
        arrayScores[j + 1] = tmpScore;
        strcpy(tmpName, arrayNames[j]);
        strcpy(arrayNames[j], arrayNames[j + 1]);
        strcpy(arrayNames[j + 1], tmpName);
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sortScores>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  sub    $0x1,%bp
  je     L6
  mov    %rsi,%r12
  mov    %rdx,%r14
  mov    %rsp,%r13
  nopl   0x0(%rax)
L61:
  xor    %ebx,%ebx
  jmp    L41
  nopl   0x0(%rax)
L26:
  add    $0x1,%rbx
  cmp    %bx,%bp
  jbe    L24
L41:
  movq   (%r12,%rbx,4),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%edx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%edx
  jg     L26
  mov    (%r14,%rbx,8),%rsi
  movq   %xmm1,(%r12,%rbx,4)
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    0x8(%r14,%rbx,8),%rsi
  mov    (%r14,%rbx,8),%rdi
  call   <strcpy@plt>
  mov    0x8(%r14,%rbx,8),%rdi
  mov    %r13,%rsi
  add    $0x1,%rbx
  call   <strcpy@plt>
  cmp    %bx,%bp
  ja     L41
L24:
  sub    $0x1,%bp
  jne    L61
L6:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortScores(const int nbScores, int *arrayScores, char **arrayNames) {
  unsigned short i, j;
  int tmpScore;
  char tmpName[35];
  for (i = nbScores - 1; i >= 1; i--) {
    for (j = 0; j < i; j++) {
      if (arrayScores[j] <= arrayScores[j + 1]) {
        tmpScore = arrayScores[j];
        arrayScores[j] = arrayScores[j + 1];
        arrayScores[j + 1] = tmpScore;
        strcpy(tmpName, arrayNames[j]);
        strcpy(arrayNames[j], arrayNames[j + 1]);
        strcpy(arrayNames[j + 1], tmpName);
      }
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<str_append_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L0
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  jmp    L11
L0:
  mov    $0x1,%eax
L11:
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <xrealloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L37
  mov    $0x0,%eax
  jmp    L28
L37:
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x1c(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L28:
  leave
  ret
```

assistant: ```c
char *str_append_char(char *source, char c) {
  int strsiz;
  char *nptr;
  strsiz = source == NULL ? 1 : strlen(source) + 1;
  nptr = (char *)xrealloc((void *)source, strsiz + 1);
  if (nptr == NULL)
    return NULL;
  nptr[strsiz - 1] = c;
  nptr[strsiz] = '\0';
  return nptr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<str_append_char>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x1,%ebx
  test   %rdi,%rdi
  je     L7
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
L7:
  lea    0x1(%rbx),%esi
  mov    %r12,%rdi
  call   <xrealloc@plt>
  test   %rax,%rax
  je     L14
  movslq %ebx,%rbx
  mov    %bpl,-0x1(%rax,%rbx,1)
  movb   $0x0,(%rax,%rbx,1)
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  mov    $0x0,%eax
  jmp    L15
```

assistant: ```c
char *str_append_char(char *source, char c) {
  int strsiz;
  char *nptr;
  strsiz = source == NULL ? 1 : strlen(source) + 1;
  nptr = (char *)xrealloc((void *)source, strsiz + 1);
  if (nptr == NULL)
    return NULL;
  nptr[strsiz - 1] = c;
  nptr[strsiz] = '\0';
  return nptr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<str_append_char>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  test   %rdi,%rdi
  je     L26
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  lea    0x2(%rax),%esi
L13:
  mov    %rbp,%rdi
  call   <xrealloc@plt>
  test   %rax,%rax
  je     L6
  movslq %ebx,%rbx
  mov    %r12b,-0x1(%rax,%rbx,1)
  movb   $0x0,(%rax,%rbx,1)
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L26:
  mov    $0x2,%esi
  mov    $0x1,%ebx
  jmp    L13
```

assistant: ```c
char *str_append_char(char *source, char c) {
  int strsiz;
  char *nptr;
  strsiz = source == NULL ? 1 : strlen(source) + 1;
  nptr = (char *)xrealloc((void *)source, strsiz + 1);
  if (nptr == NULL)
    return NULL;
  nptr[strsiz - 1] = c;
  nptr[strsiz] = '\0';
  return nptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<str_append_char>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  test   %rdi,%rdi
  je     L23
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  lea    0x2(%rax),%esi
L18:
  mov    %rbp,%rdi
  call   <xrealloc@plt>
  test   %rax,%rax
  je     L15
  movslq %ebx,%rbx
  mov    %r12b,-0x1(%rax,%rbx,1)
  movb   $0x0,(%rax,%rbx,1)
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L23:
  mov    $0x2,%esi
  mov    $0x1,%ebx
  jmp    L18
```

assistant: ```c
char *str_append_char(char *source, char c) {
  int strsiz;
  char *nptr;
  strsiz = source == NULL ? 1 : strlen(source) + 1;
  nptr = (char *)xrealloc((void *)source, strsiz + 1);
  if (nptr == NULL)
    return NULL;
  nptr[strsiz - 1] = c;
  nptr[strsiz] = '\0';
  return nptr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<quad_idx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  lea    D6(%rip),%rax
  mov    (%rdx,%rax,1),%rax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
data_type: string

tool: 

assistant: ```c
size_t quad_idx(size_t a, size_t b) {
  enum { xx = 0, yy, zz, xy, xz, yz };
  static const size_t idx[] = {xx, xy, xz, xy, yy, yz, xz, yz, zz};
  return idx[a * 3 + b];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<quad_idx>:
  endbr64
  lea    (%rdi,%rdi,2),%rax
  add    %rsi,%rax
  lea    D0(%rip),%rdx
  mov    (%rdx,%rax,8),%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type: string

tool: 

assistant: ```c
size_t quad_idx(size_t a, size_t b) {
  enum { xx = 0, yy, zz, xy, xz, yz };
  static const size_t idx[] = {xx, xy, xz, xy, yy, yz, xz, yz, zz};
  return idx[a * 3 + b];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quad_idx>:
  endbr64
  lea    (%rdi,%rdi,2),%rax
  lea    D2(%rip),%rdx
  add    %rsi,%rax
  mov    (%rdx,%rax,8),%rax
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type: string

tool: 

assistant: ```c
size_t quad_idx(size_t a, size_t b) {
  enum { xx = 0, yy, zz, xy, xz, yz };
  static const size_t idx[] = {xx, xy, xz, xy, yy, yz, xz, yz, zz};
  return idx[a * 3 + b];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<quad_idx>:
  endbr64
  lea    (%rdi,%rdi,2),%rax
  lea    D8(%rip),%rdx
  add    %rsi,%rax
  mov    (%rdx,%rax,8),%rax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
type_infered: string

tool: 

assistant: ```c
size_t quad_idx(size_t a, size_t b) {
  enum { xx = 0, yy, zz, xy, xz, yz };
  static const size_t idx[] = {xx, xy, xz, xy, yy, yz, xz, yz, zz};
  return idx[a * 3 + b];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rdi
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rsi
  lea    -0x10(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <measure@plt>
  mov    -0x10(%rbp),%rax
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D26
type_infered: string
name: read_data
arguments: data_label: D25
type_infered: string
name: read_data
arguments: data_label: D34
type_infered: string
name: read_data
arguments: data_label: D18
type_infered: string
name: read_data
arguments: data_label: D39
type_infered: string

tool: D26: Total value of inches is: %d


tool: D25: Value of inch is: %d


tool: D34: Value of feet is: %d


tool: D18: Value of yard is: %d


tool: D39: Value of mile is: %d


assistant: ```c
int main() {
  int inch, feet, yard, mile;
  long inches;
  measure(&inches, &inch, &feet, &yard, &mile);
  printf("Total value of inches is: %d\n", inches);
  printf("Value of inch is: %d\n", inch);
  printf("Value of feet is: %d\n", feet);
  printf("Value of yard is: %d\n", yard);
  printf("Value of mile is: %d\n", mile);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rcx
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    0x10(%rsp),%rdi
  lea    0xc(%rsp),%r8
  call   <measure@plt>
  mov    0x10(%rsp),%esi
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp),%esi
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D45(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D39
type_guessed: string
name: disassemble_data
arguments: data_tag: D26
type_guessed: string
name: disassemble_data
arguments: data_tag: D45
type_guessed: string
name: disassemble_data
arguments: data_tag: D11
type_guessed: string
name: disassemble_data
arguments: data_tag: D30
type_guessed: string

tool: D39: Total value of inches is: %d


tool: D26: Value of inch is: %d


tool: D45: Value of feet is: %d


tool: D11: Value of yard is: %d


tool: D30: Value of mile is: %d


assistant: ```c
int main() {
  int inch, feet, yard, mile;
  long inches;
  measure(&inches, &inch, &feet, &yard, &mile);
  printf("Total value of inches is: %d\n", inches);
  printf("Value of inch is: %d\n", inch);
  printf("Value of feet is: %d\n", feet);
  printf("Value of yard is: %d\n", yard);
  printf("Value of mile is: %d\n", mile);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    0x8(%rsp),%rcx
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r8
  lea    0x10(%rsp),%rdi
  call   <measure@plt>
  mov    0x10(%rsp),%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D11
type_guessed: string
name: retrieve_data
arguments: tag: D7
type_guessed: string
name: retrieve_data
arguments: tag: D13
type_guessed: string
name: retrieve_data
arguments: tag: D25
type_guessed: string
name: retrieve_data
arguments: tag: D33
type_guessed: string

tool: D11: Total value of inches is: %d


tool: D7: Value of inch is: %d


tool: D13: Value of feet is: %d


tool: D25: Value of yard is: %d


tool: D33: Value of mile is: %d


assistant: ```c
int main() {
  int inch, feet, yard, mile;
  long inches;
  measure(&inches, &inch, &feet, &yard, &mile);
  printf("Total value of inches is: %d\n", inches);
  printf("Value of inch is: %d\n", inch);
  printf("Value of feet is: %d\n", feet);
  printf("Value of yard is: %d\n", yard);
  printf("Value of mile is: %d\n", mile);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    0x8(%rsp),%rcx
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r8
  lea    0x10(%rsp),%rdi
  call   <measure@plt>
  mov    0x10(%rsp),%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    (%rsp),%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D47(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D13
data_type: string
name: unpack_data
arguments: tag: D11
data_type: string
name: unpack_data
arguments: tag: D47
data_type: string
name: unpack_data
arguments: tag: D39
data_type: string
name: unpack_data
arguments: tag: D32
data_type: string

tool: D13: Total value of inches is: %d


tool: D11: Value of inch is: %d


tool: D47: Value of feet is: %d


tool: D39: Value of yard is: %d


tool: D32: Value of mile is: %d


assistant: ```c
int main() {
  int inch, feet, yard, mile;
  long inches;
  measure(&inches, &inch, &feet, &yard, &mile);
  printf("Total value of inches is: %d\n", inches);
  printf("Value of inch is: %d\n", inch);
  printf("Value of feet is: %d\n", feet);
  printf("Value of yard is: %d\n", yard);
  printf("Value of mile is: %d\n", mile);
  return 0;
}
```