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
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r12
  lea    0x10(%rsp),%rbp
  lea    D10(%rip),%rbx
  lea    D0(%rip),%r13
  jmp    L131
L104:
  sub    %edx,%eax
  sub    %eax,%esi
L33:
  mov    %r13,%rdi
  call   <printf@plt>
L131:
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  mov    %r12,%r8
  mov    %rbp,%rcx
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x4,%eax
  jne    L63
  cmpl   $0x0,0x8(%rsp)
  jne    L96
  mov    0x10(%rsp),%eax
  or     0xc(%rsp),%eax
  or     0x14(%rsp),%eax
  je     L63
  movl   $0x18,0x8(%rsp)
L96:
  cmpl   $0x0,0x10(%rsp)
  jne    L9
  movl   $0x18,0x10(%rsp)
L9:
  mov    0x10(%rsp),%esi
  mov    0x8(%rsp),%eax
  cmp    %eax,%esi
  jle    L26
  sub    %eax,%esi
  imul   $0x3c,%esi,%esi
  mov    0xc(%rsp),%eax
  mov    0x14(%rsp),%edx
  cmp    %edx,%eax
  jg     L104
  sub    %eax,%edx
  add    %edx,%esi
  jmp    L33
L26:
  je     L71
  sub    %eax,%esi
  lea    0x18(%rsi),%esi
  imul   $0x3c,%esi,%esi
  mov    0x14(%rsp),%edx
  mov    0xc(%rsp),%eax
  cmp    %eax,%edx
  jl     L107
  sub    %eax,%edx
  add    %edx,%esi
  jmp    L33
L71:
  mov    0x14(%rsp),%edx
  mov    0xc(%rsp),%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    0x5a0(%rcx),%esi
  cmp    %eax,%edx
  cmovge %ecx,%esi
  jmp    L33
L107:
  sub    %edx,%eax
  sub    %eax,%esi
  jmp    L33
L63:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L133
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L133:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D10
data_type: string
name: disassemble_data
arguments: tag: D0
data_type: string

tool: D10: %d %d %d %d

tool: D0: %d


assistant: ```c
int main() {
  int h1, m1, h2, m2, h, m, total, hour;
  while (scanf("%d %d %d %d", &h1, &m1, &h2, &m2) == 4) {
    if (h1 == 0 && h2 == 0 && m1 == 0 && m2 == 0)
      break;
    if (h1 == 0) {
      h1 = 24;
    }
    if (h2 == 0) {
      h2 = 24;
    }
    total = 24 * 60;
    if (h2 > h1) {
      h = (h2 - h1) * 60;
      if (m1 <= m2) {
        m = m2 - m1;
        hour = h + m;
      } else {
        m = m1 - m2;
        hour = h - m;
      }
    } else if (h2 == h1) {
      if (m2 >= m1) {
        m = m2 - m1;
        hour = m;
      } else {
        m = m1 - m2;
        hour = total - m;
      }
    } else {
      h = (24 - (h1 - h2)) * 60;
      if (m2 >= m1) {
        m = m2 - m1;
        hour = h + m;
      } else {
        m = m1 - m2;
        hour = h - m;
      }
    }
    printf("%d\n", hour);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D0(%rip),%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D17(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    0x8(%rsp),%rbp
  jmp    L79
  cs nopw 0x0(%rax,%rax,1)
L61:
  sub    %ecx,%esi
  imul   $0x3c,%esi,%esi
  cmp    %edx,%eax
  jg     L39
L84:
  sub    %eax,%edx
  add    %edx,%esi
L97:
  mov    %r15,%rdi
  call   <printf@plt>
L79:
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x4,%eax
  jne    L107
  mov    0x8(%rsp),%ecx
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%eax
  mov    0x14(%rsp),%edx
  test   %ecx,%ecx
  jne    L11
  mov    %eax,%ecx
  or     %esi,%ecx
  or     %edx,%ecx
  je     L107
  movl   $0x18,0x8(%rsp)
  mov    $0x18,%ecx
L11:
  test   %esi,%esi
  jne    L101
  movl   $0x18,0x10(%rsp)
  mov    $0x18,%esi
L101:
  cmp    %ecx,%esi
  jg     L61
  je     L86
  sub    %ecx,%esi
  add    $0x18,%esi
  imul   $0x3c,%esi,%esi
  cmp    %edx,%eax
  jle    L84
L39:
  sub    %edx,%eax
  sub    %eax,%esi
  jmp    L97
L86:
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%esi
  add    $0x5a0,%esi
  cmp    %edx,%eax
  cmovle %ecx,%esi
  jmp    L97
  cs nopw 0x0(%rax,%rax,1)
L107:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_guessed: string
name: disassemble_data
arguments: tag: D17
type_guessed: string

tool: D0: %d


tool: D17: %d %d %d %d

assistant: ```c
int main() {
  int h1, m1, h2, m2, h, m, total, hour;
  while (scanf("%d %d %d %d", &h1, &m1, &h2, &m2) == 4) {
    if (h1 == 0 && h2 == 0 && m1 == 0 && m2 == 0)
      break;
    if (h1 == 0) {
      h1 = 24;
    }
    if (h2 == 0) {
      h2 = 24;
    }
    total = 24 * 60;
    if (h2 > h1) {
      h = (h2 - h1) * 60;
      if (m1 <= m2) {
        m = m2 - m1;
        hour = h + m;
      } else {
        m = m1 - m2;
        hour = h - m;
      }
    } else if (h2 == h1) {
      if (m2 >= m1) {
        m = m2 - m1;
        hour = m;
      } else {
        m = m1 - m2;
        hour = total - m;
      }
    } else {
      h = (24 - (h1 - h2)) * 60;
      if (m2 >= m1) {
        m = m2 - m1;
        hour = h + m;
      } else {
        m = m1 - m2;
        hour = h - m;
      }
    }
    printf("%d\n", hour);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D8(%rip),%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D14(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    0x8(%rsp),%rbp
  jmp    L105
  cs nopw 0x0(%rax,%rax,1)
L113:
  sub    %ecx,%esi
  imul   $0x3c,%esi,%esi
  cmp    %eax,%edx
  jl     L73
L76:
  sub    %eax,%edx
  add    %edx,%esi
L53:
  mov    %r15,%rdi
  call   <printf@plt>
L105:
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x4,%eax
  jne    L108
  mov    0x8(%rsp),%ecx
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%eax
  mov    0x14(%rsp),%edx
  test   %ecx,%ecx
  jne    L83
  mov    %eax,%ecx
  or     %esi,%ecx
  or     %edx,%ecx
  je     L108
  movl   $0x18,0x8(%rsp)
  mov    $0x18,%ecx
L83:
  test   %esi,%esi
  jne    L117
  movl   $0x18,0x10(%rsp)
  mov    $0x18,%esi
L117:
  cmp    %ecx,%esi
  jg     L113
  je     L49
  sub    %ecx,%esi
  add    $0x18,%esi
  imul   $0x3c,%esi,%esi
  cmp    %eax,%edx
  jge    L76
L73:
  sub    %edx,%eax
  sub    %eax,%esi
  jmp    L53
L49:
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%esi
  add    $0x5a0,%esi
  cmp    %eax,%edx
  cmovge %ecx,%esi
  jmp    L53
  cs nopw 0x0(%rax,%rax,1)
L108:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
data_type: string
name: read_data
arguments: label: D14
data_type: string

tool: D8: %d


tool: D14: %d %d %d %d

assistant: ```c
int main() {
  int h1, m1, h2, m2, h, m, total, hour;
  while (scanf("%d %d %d %d", &h1, &m1, &h2, &m2) == 4) {
    if (h1 == 0 && h2 == 0 && m1 == 0 && m2 == 0)
      break;
    if (h1 == 0) {
      h1 = 24;
    }
    if (h2 == 0) {
      h2 = 24;
    }
    total = 24 * 60;
    if (h2 > h1) {
      h = (h2 - h1) * 60;
      if (m1 <= m2) {
        m = m2 - m1;
        hour = h + m;
      } else {
        m = m1 - m2;
        hour = h - m;
      }
    } else if (h2 == h1) {
      if (m2 >= m1) {
        m = m2 - m1;
        hour = m;
      } else {
        m = m1 - m2;
        hour = total - m;
      }
    } else {
      h = (24 - (h1 - h2)) * 60;
      if (m2 >= m1) {
        m = m2 - m1;
        hour = h + m;
      } else {
        m = m1 - m2;
        hour = h - m;
      }
    }
    printf("%d\n", hour);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x24,-0x14(%rbp)
  mov    $0x4,%edi
  call   <Malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x24,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  je     L16
  mov    $0x1,%eax
  jmp    L19
L16:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <Free@plt>
  mov    $0x100001,%edi
  call   <Malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L17
  mov    $0x1,%eax
  jmp    L19
L17:
  mov    $0x0,%eax
L19:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int r = 36;
  int *p = (int *)Malloc(sizeof(int));
  *p = 36;
  if (r != *p)
    return 1;
  Free(p);
  void *q = Malloc(1024 * 1024 + 1);
  if (q == 0)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <Malloc@plt>
  mov    %rax,%rdi
  movl   $0x24,(%rax)
  call   <Free@plt>
  mov    $0x100001,%edi
  call   <Malloc@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int r = 36;
  int *p = (int *)Malloc(sizeof(int));
  *p = 36;
  if (r != *p)
    return 1;
  Free(p);
  void *q = Malloc(1024 * 1024 + 1);
  if (q == 0)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <Malloc@plt>
  movl   $0x24,(%rax)
  mov    %rax,%rdi
  call   <Free@plt>
  mov    $0x100001,%edi
  call   <Malloc@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int r = 36;
  int *p = (int *)Malloc(sizeof(int));
  *p = 36;
  if (r != *p)
    return 1;
  Free(p);
  void *q = Malloc(1024 * 1024 + 1);
  if (q == 0)
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%edi
  call   <Malloc@plt>
  movl   $0x24,(%rax)
  mov    %rax,%rdi
  call   <Free@plt>
  mov    $0x100001,%edi
  call   <Malloc@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int r = 36;
  int *p = (int *)Malloc(sizeof(int));
  *p = 36;
  if (r != *p)
    return 1;
  Free(p);
  void *q = Malloc(1024 * 1024 + 1);
  if (q == 0)
    return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_sqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x32,-0x4(%rbp)
  jmp    L1
L42:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <ft_putnbr@plt>
  mov    $0xa,%edi
  call   <ft_putchar@plt>
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x14(%rbp)
  jge    L51
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L51:
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x14(%rbp)
  jle    L46
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L46:
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L16
  mov    -0x10(%rbp),%eax
  jmp    L18
L16:
  subl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x0,-0x4(%rbp)
  jg     L42
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int ft_sqrt(int nb) {
  int x;
  int min;
  int max;
  int n;
  min = 0;
  max = nb;
  n = 50;
  while (n > 0) {
    ft_putnbr(x);
    ft_putchar('\n');
    x = (max + min) / 2;
    if ((x * x) > nb)
      max = x;
    if ((x * x) < nb)
      min = x;
    if ((x * x) == nb)
      return (x);
    n--;
  }
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_sqrt>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %edi,%ebx
  mov    $0x32,%ebp
  mov    $0x0,%r13d
  jmp    L2
L40:
  mov    %eax,%r13d
L22:
  sub    $0x1,%ebp
  je     L45
L2:
  mov    %ecx,%edi
  call   <ft_putnbr@plt>
  mov    $0xa,%edi
  call   <ft_putchar@plt>
  lea    0x0(%r13,%rbx,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%ecx
  mov    %eax,%edx
  imul   %eax,%edx
  cmp    %r12d,%edx
  cmovg  %eax,%ebx
  jl     L40
  jne    L22
L33:
  mov    %ecx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L45:
  mov    $0x0,%ecx
  jmp    L33
```

assistant: ```c
int ft_sqrt(int nb) {
  int x;
  int min;
  int max;
  int n;
  min = 0;
  max = nb;
  n = 50;
  while (n > 0) {
    ft_putnbr(x);
    ft_putchar('\n');
    x = (max + min) / 2;
    if ((x * x) > nb)
      max = x;
    if ((x * x) < nb)
      min = x;
    if ((x * x) == nb)
      return (x);
    n--;
  }
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_sqrt>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %edi,%r12d
  xor    %edi,%edi
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x32,%ebx
  sub    $0x8,%rsp
  jmp    L32
  nop
L40:
  jge    L59
  mov    %edi,%ebp
L29:
  sub    $0x1,%ebx
  je     L1
L32:
  call   <ft_putnbr@plt>
  mov    $0xa,%edi
  call   <ft_putchar@plt>
  lea    0x0(%rbp,%r12,1),%eax
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,%eax
  imul   %edi,%eax
  cmp    %r13d,%eax
  jle    L40
  mov    %edi,%r12d
  sub    $0x1,%ebx
  jne    L32
L1:
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbx
  mov    %edi,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L59:
  jne    L29
  add    $0x8,%rsp
  mov    %edi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_sqrt(int nb) {
  int x;
  int min;
  int max;
  int n;
  min = 0;
  max = nb;
  n = 50;
  while (n > 0) {
    ft_putnbr(x);
    ft_putchar('\n');
    x = (max + min) / 2;
    if ((x * x) > nb)
      max = x;
    if ((x * x) < nb)
      min = x;
    if ((x * x) == nb)
      return (x);
    n--;
  }
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_sqrt>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %edi,%r12d
  xor    %edi,%edi
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x32,%ebx
  sub    $0x8,%rsp
  jmp    L26
  nop
L18:
  jge    L33
  mov    %edi,%ebp
L21:
  sub    $0x1,%ebx
  je     L25
L26:
  call   <ft_putnbr@plt>
  mov    $0xa,%edi
  call   <ft_putchar@plt>
  lea    0x0(%rbp,%r12,1),%eax
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,%eax
  imul   %edi,%eax
  cmp    %r13d,%eax
  jle    L18
  mov    %edi,%r12d
  sub    $0x1,%ebx
  jne    L26
L25:
  add    $0x8,%rsp
  xor    %edi,%edi
  pop    %rbx
  mov    %edi,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L33:
  jne    L21
  add    $0x8,%rsp
  mov    %edi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ft_sqrt(int nb) {
  int x;
  int min;
  int max;
  int n;
  min = 0;
  max = nb;
  n = 50;
  while (n > 0) {
    ft_putnbr(x);
    ft_putchar('\n');
    x = (max + min) / 2;
    if ((x * x) > nb)
      max = x;
    if ((x * x) < nb)
      min = x;
    if ((x * x) == nb)
      return (x);
    n--;
  }
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dmat_iset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  jmp    L9
L5:
  mov    -0x10(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
L9:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L5
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dmat_iset(int n, const int val, int *dst) {
  while (n-- != 0)
    *dst++ = val;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<dmat_iset>:
  endbr64
  test   %edi,%edi
  je     L1
  mov    %edi,%edi
  lea    (%rdx,%rdi,4),%rax
L5:
  add    $0x4,%rdx
  mov    %esi,-0x4(%rdx)
  cmp    %rax,%rdx
  jne    L5
L1:
  ret
```

assistant: ```c
void dmat_iset(int n, const int val, int *dst) {
  while (n-- != 0)
    *dst++ = val;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<dmat_iset>:
  endbr64
  test   %edi,%edi
  je     L7
  mov    %edi,%edi
  lea    (%rdx,%rdi,4),%rax
  xchg   %ax,%ax
L16:
  add    $0x4,%rdx
  mov    %esi,-0x4(%rdx)
  cmp    %rax,%rdx
  jne    L16
L7:
  ret
```

assistant: ```c
void dmat_iset(int n, const int val, int *dst) {
  while (n-- != 0)
    *dst++ = val;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dmat_iset>:
  endbr64
  lea    -0x1(%rdi),%r8d
  test   %edi,%edi
  je     L8
  cmp    $0x2,%r8d
  jbe    L0
  mov    %edi,%ecx
  movd   %esi,%xmm1
  mov    %rdx,%rax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm1,%xmm0
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdx,%rcx,1),%rcx
  nop
L49:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L49
  mov    %edi,%eax
  and    $0xfffffffc,%eax
  mov    %eax,%ecx
  sub    %eax,%r8d
  lea    (%rdx,%rcx,4),%rdx
  cmp    %eax,%edi
  je     L8
L0:
  mov    %esi,(%rdx)
  test   %r8d,%r8d
  je     L8
  mov    %esi,0x4(%rdx)
  cmp    $0x1,%r8d
  je     L8
  mov    %esi,0x8(%rdx)
L8:
  ret
```

assistant: ```c
void dmat_iset(int n, const int val, int *dst) {
  while (n-- != 0)
    *dst++ = val;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x38,-0x60(%rbp)
  movl   $0xffffffe9,-0x5c(%rbp)
  movl   $0x12,-0x58(%rbp)
  movl   $0xffffffd3,-0x54(%rbp)
  movl   $0x0,-0x50(%rbp)
  movl   $0x3,-0x4c(%rbp)
  movl   $0x36,-0x48(%rbp)
  movl   $0xfffffff7,-0x44(%rbp)
  movl   $0x20,-0x40(%rbp)
  movl   $0x4e,-0x3c(%rbp)
  movl   $0x6d,-0x38(%rbp)
  movl   $0x17,-0x34(%rbp)
  movl   $0x1,-0x30(%rbp)
  movl   $0x43,-0x2c(%rbp)
  movl   $0xd,-0x28(%rbp)
  movl   $0x33,-0x24(%rbp)
  movl   $0xfffffff8,-0x20(%rbp)
  movl   $0x47,-0x1c(%rbp)
  movl   $0x5c,-0x18(%rbp)
  movl   $0x6,-0x14(%rbp)
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x14,%edi
  call   <heapSort@plt>
  movl   $0x0,-0x64(%rbp)
  jmp    L14
L11:
  mov    -0x64(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x64(%rbp)
L14:
  cmpl   $0x13,-0x64(%rbp)
  jle    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

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

tool: D2: %d  

assistant: ```c
int main(int argc, char **argv) {
  int i;
  int test[20] = {56, -23, 18, -45, 0, 3, 54, -9, 32, 78, 109, 23, 1, 67, 13, 51, -8, 71, 92, 6};
  heapSort(20, test);
  for (i = 0; i < 20; i++) {
    printf("%d  ", test[i]);
  }
  printf("\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movl   $0x38,(%rsp)
  movl   $0xffffffe9,0x4(%rsp)
  movl   $0x12,0x8(%rsp)
  movl   $0xffffffd3,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x3,0x14(%rsp)
  movl   $0x36,0x18(%rsp)
  movl   $0xfffffff7,0x1c(%rsp)
  movl   $0x20,0x20(%rsp)
  movl   $0x4e,0x24(%rsp)
  movl   $0x6d,0x28(%rsp)
  movl   $0x17,0x2c(%rsp)
  movl   $0x1,0x30(%rsp)
  movl   $0x43,0x34(%rsp)
  movl   $0xd,0x38(%rsp)
  movl   $0x33,0x3c(%rsp)
  movl   $0xfffffff8,0x40(%rsp)
  movl   $0x47,0x44(%rsp)
  movl   $0x5c,0x48(%rsp)
  movl   $0x6,0x4c(%rsp)
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  mov    $0x14,%edi
  call   <heapSort@plt>
  lea    0x50(%rsp),%r12
  lea    D6(%rip),%rbp
L2:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L2
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
type: string

tool: D6: %d  

assistant: ```c
int main(int argc, char **argv) {
  int i;
  int test[20] = {56, -23, 18, -45, 0, 3, 54, -9, 32, 78, 109, 23, 1, 67, 13, 51, -8, 71, 92, 6};
  heapSort(20, test);
  for (i = 0; i < 20; i++) {
    printf("%d  ", test[i]);
  }
  printf("\n");
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x14,%edi
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  movabs $0xffffffe900000038,%rax
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  mov    %rbx,%rsi
  mov    %rax,(%rsp)
  movabs $0xffffffd300000012,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x300000000,%rax
  mov    %rax,0x10(%rsp)
  movabs $0xfffffff700000036,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x4e00000020,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x170000006d,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x4300000001,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x330000000d,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x47fffffff8,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x60000005c,%rax
  mov    %rax,0x48(%rsp)
  call   <heapSort@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type: string

tool: D8: %d  

assistant: ```c
int main(int argc, char **argv) {
  int i;
  int test[20] = {56, -23, 18, -45, 0, 3, 54, -9, 32, 78, 109, 23, 1, 67, 13, 51, -8, 71, 92, 6};
  heapSort(20, test);
  for (i = 0; i < 20; i++) {
    printf("%d  ", test[i]);
  }
  printf("\n");
  return 0;
}
```