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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D72(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D72(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D72(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  imul   %eax,%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  add    $0xa5,%eax
  movslq %eax,%rdx
  imul   $0x3159721f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D38
type_infered: string
name: disassemble_data
arguments: data_label: D72
type_infered: string
name: disassemble_data
arguments: data_label: D32
type_infered: string
name: disassemble_data
arguments: data_label: D64
type_infered: string
name: disassemble_data
arguments: data_label: D20
type_infered: string
name: disassemble_data
arguments: data_label: D28
type_infered: string

tool: D38: Enter height of box: 

tool: D72: %d

tool: D32: Enter length of box: 

tool: D64: Enter width of box: 

tool: D20: Volume (cubic inches): %d


tool: D28: Dimensional weigth (pounds): %d


assistant: ```c
int main(void) {
  int height, length, width, volume, weigth;
  printf("Enter height of box: ");
  scanf("%d", &height);
  printf("Enter length of box: ");
  scanf("%d", &length);
  printf("Enter width of box: ");
  scanf("%d", &width);
  volume = height * length * width;
  weigth = (volume + 165) / 166;
  printf("Volume (cubic inches): %d\n", volume);
  printf("Dimensional weigth (pounds): %d\n", weigth);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D55(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D15(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ebx
  imul   0x10(%rsp),%ebx
  imul   0x14(%rsp),%ebx
  mov    %ebx,%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0xa5,%ebx
  movslq %ebx,%rsi
  imul   $0x3159721f,%rsi,%rsi
  sar    $0x25,%rsi
  sar    $0x1f,%ebx
  sub    %ebx,%esi
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D55
type_infered: string
name: extract_symbolic_data
arguments: data_label: D15
type_infered: string
name: extract_symbolic_data
arguments: data_label: D53
type_infered: string
name: extract_symbolic_data
arguments: data_label: D48
type_infered: string
name: extract_symbolic_data
arguments: data_label: D23
type_infered: string
name: extract_symbolic_data
arguments: data_label: D54
type_infered: string

tool: D55: Enter height of box: 

tool: D15: %d

tool: D53: Enter length of box: 

tool: D48: Enter width of box: 

tool: D23: Volume (cubic inches): %d


tool: D54: Dimensional weigth (pounds): %d


assistant: ```c
int main(void) {
  int height, length, width, volume, weigth;
  printf("Enter height of box: ");
  scanf("%d", &height);
  printf("Enter length of box: ");
  scanf("%d", &length);
  printf("Enter width of box: ");
  scanf("%d", &width);
  volume = height * length * width;
  weigth = (volume + 165) / 166;
  printf("Volume (cubic inches): %d\n", volume);
  printf("Dimensional weigth (pounds): %d\n", weigth);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D21(%rip),%rdi
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D48(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ebx
  imul   0x10(%rsp),%ebx
  xor    %eax,%eax
  imul   0x14(%rsp),%ebx
  lea    D17(%rip),%rdi
  mov    %ebx,%esi
  add    $0xa5,%ebx
  call   <printf@plt>
  movslq %ebx,%rsi
  sar    $0x1f,%ebx
  xor    %eax,%eax
  imul   $0x3159721f,%rsi,%rsi
  lea    D52(%rip),%rdi
  sar    $0x25,%rsi
  sub    %ebx,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D21
type_guessed: string
name: parse_data
arguments: label: D7
type_guessed: string
name: parse_data
arguments: label: D48
type_guessed: string
name: parse_data
arguments: label: D43
type_guessed: string
name: parse_data
arguments: label: D17
type_guessed: string
name: parse_data
arguments: label: D52
type_guessed: string

tool: D21: Enter height of box: 

tool: D7: %d

tool: D48: Enter length of box: 

tool: D43: Enter width of box: 

tool: D17: Volume (cubic inches): %d


tool: D52: Dimensional weigth (pounds): %d


assistant: ```c
int main(void) {
  int height, length, width, volume, weigth;
  printf("Enter height of box: ");
  scanf("%d", &height);
  printf("Enter length of box: ");
  scanf("%d", &length);
  printf("Enter width of box: ");
  scanf("%d", &width);
  volume = height * length * width;
  weigth = (volume + 165) / 166;
  printf("Volume (cubic inches): %d\n", volume);
  printf("Dimensional weigth (pounds): %d\n", weigth);
  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   %rbp
  lea    D33(%rip),%rdi
  lea    D34(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ebx
  imul   0x10(%rsp),%ebx
  xor    %eax,%eax
  imul   0x14(%rsp),%ebx
  lea    D40(%rip),%rdi
  mov    %ebx,%esi
  add    $0xa5,%ebx
  call   <printf@plt>
  movslq %ebx,%rsi
  sar    $0x1f,%ebx
  xor    %eax,%eax
  imul   $0x3159721f,%rsi,%rsi
  lea    D16(%rip),%rdi
  sar    $0x25,%rsi
  sub    %ebx,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D33
type_guessed: string
name: extract_symbolic_data
arguments: label: D34
type_guessed: string
name: extract_symbolic_data
arguments: label: D17
type_guessed: string
name: extract_symbolic_data
arguments: label: D42
type_guessed: string
name: extract_symbolic_data
arguments: label: D40
type_guessed: string
name: extract_symbolic_data
arguments: label: D16
type_guessed: string

tool: D33: Enter height of box: 

tool: D34: %d

tool: D17: Enter length of box: 

tool: D42: Enter width of box: 

tool: D40: Volume (cubic inches): %d


tool: D16: Dimensional weigth (pounds): %d


assistant: ```c
int main(void) {
  int height, length, width, volume, weigth;
  printf("Enter height of box: ");
  scanf("%d", &height);
  printf("Enter length of box: ");
  scanf("%d", &length);
  printf("Enter width of box: ");
  scanf("%d", &width);
  volume = height * length * width;
  weigth = (volume + 165) / 166;
  printf("Volume (cubic inches): %d\n", volume);
  printf("Dimensional weigth (pounds): %d\n", weigth);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<twist_l_on_cornerperm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <cornerperm_unpack@plt>
  lea    -0x30(%rbp),%rax
  mov    $0x2,%r8d
  mov    $0x3,%ecx
  mov    $0x5,%edx
  mov    $0x6,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <four_cycle@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <cornerperm_pack@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int twist_l_on_cornerperm(cperm)
int cperm;
{
  int temp_arr[8];
  cornerperm_unpack(cperm, temp_arr);
  four_cycle(temp_arr, 6, 5, 3, 2);
  return cornerperm_pack(temp_arr);
}
```
user: Translate the below assembly code into C programming language:

```asm
<twist_l_on_cornerperm>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  call   <cornerperm_unpack@plt>
  mov    $0x2,%r8d
  mov    $0x3,%ecx
  mov    $0x5,%edx
  mov    $0x6,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <four_cycle@plt>
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <cornerperm_pack@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x30,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int twist_l_on_cornerperm(cperm)
int cperm;
{
  int temp_arr[8];
  cornerperm_unpack(cperm, temp_arr);
  four_cycle(temp_arr, 6, 5, 3, 2);
  return cornerperm_pack(temp_arr);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<twist_l_on_cornerperm>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <cornerperm_unpack@plt>
  mov    $0x5,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x2,%r8d
  mov    $0x3,%ecx
  mov    $0x6,%esi
  call   <four_cycle@plt>
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <cornerperm_pack@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x30,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int twist_l_on_cornerperm(cperm)
int cperm;
{
  int temp_arr[8];
  cornerperm_unpack(cperm, temp_arr);
  four_cycle(temp_arr, 6, 5, 3, 2);
  return cornerperm_pack(temp_arr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<twist_l_on_cornerperm>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <cornerperm_unpack@plt>
  mov    $0x5,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x2,%r8d
  mov    $0x3,%ecx
  mov    $0x6,%esi
  call   <four_cycle@plt>
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <cornerperm_pack@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x30,%rsp
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int twist_l_on_cornerperm(cperm)
int cperm;
{
  int temp_arr[8];
  cornerperm_unpack(cperm, temp_arr);
  four_cycle(temp_arr, 6, 5, 3, 2);
  return cornerperm_pack(temp_arr);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L29
  movl   $0xffffffff,-0x14(%rbp)
  negl   -0x24(%rbp)
  jmp    L29
L11:
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L29:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x18(%rbp)
  jle    L11
  movl   $0x0,-0x10(%rbp)
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L20
  mov    -0x1c(%rbp),%eax
  add    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
  jmp    L35
L20:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L35
L66:
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  lea    0x30(%rax),%ecx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x10(%rbp)
L35:
  cmpl   $0x0,-0x18(%rbp)
  jg     L66
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ft_itoa(int nb) {
  int length;
  int div;
  int signe;
  char *res;
  int i;
  int ori_nbr;
  length = 0;
  div = 1;
  signe = 1;
  ori_nbr = nb;
  if (nb < 0) {
    signe = -1;
    nb = -nb;
  }
  while (div <= (nb / 10)) {
    div *= 10;
    length++;
  }
  i = 0;
  if (signe == -1) {
    res = (char *)malloc(sizeof(char) * (length + 2));
    res[length + 1] = '\0';
    res[0] = '-';
  } else {
    res = (char *)malloc(sizeof(char) * (length + 1));
    res[length] = '\0';
  }
  while (div >= 1) {
    res[i] = (nb / div) + 48;
    nb -= (nb / div) * div;
    div /= 10;
    i++;
  }
  nb = ori_nbr;
  return (res);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_itoa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  test   %edi,%edi
  js     L10
  movslq %edi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    $0x1,%ecx
  mov    $0x1,%ebx
  mov    $0x0,%r12d
  cmp    $0x9,%edi
  jle    L23
L4:
  mov    $0x1,%ebx
  mov    $0x0,%r12d
L19:
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  mov    %eax,%ebx
  add    $0x1,%r12d
  cmp    %edx,%eax
  jle    L19
  cmp    $0xffffffff,%ecx
  je     L60
L23:
  lea    0x1(%r12),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movslq %r12d,%r12
  movb   $0x0,(%rax,%r12,1)
L58:
  mov    %rdi,%rcx
  test   %ebx,%ebx
  jle    L73
L80:
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  add    $0x30,%eax
  mov    %al,(%rcx)
  mov    %edx,%ebp
  mov    %ebx,%edx
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ebx
  add    $0x1,%rcx
  cmp    $0x9,%edx
  jg     L80
L73:
  mov    %rdi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  mov    %edi,%eax
  neg    %eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  cmp    $0xfffffff7,%edi
  jge    L40
  mov    %eax,%ebp
  mov    $0xffffffff,%ecx
  jmp    L4
L40:
  mov    %eax,%ebp
  mov    $0x1,%ebx
  mov    $0x0,%r12d
L60:
  lea    0x2(%r12),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movslq %r12d,%r12
  movb   $0x0,0x1(%rax,%r12,1)
  movb   $0x2d,(%rax)
  jmp    L58
```

assistant: ```c
char *ft_itoa(int nb) {
  int length;
  int div;
  int signe;
  char *res;
  int i;
  int ori_nbr;
  length = 0;
  div = 1;
  signe = 1;
  ori_nbr = nb;
  if (nb < 0) {
    signe = -1;
    nb = -nb;
  }
  while (div <= (nb / 10)) {
    div *= 10;
    length++;
  }
  i = 0;
  if (signe == -1) {
    res = (char *)malloc(sizeof(char) * (length + 2));
    res[length + 1] = '\0';
    res[0] = '-';
  } else {
    res = (char *)malloc(sizeof(char) * (length + 1));
    res[length] = '\0';
  }
  while (div >= 1) {
    res[i] = (nb / div) + 48;
    nb -= (nb / div) * div;
    div /= 10;
    i++;
  }
  nb = ori_nbr;
  return (res);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_itoa>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  test   %edi,%edi
  js     L32
  movslq %edi,%rax
  mov    %edi,%edx
  xor    %ebp,%ebp
  mov    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  mov    $0x1,%ebx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    $0x1,%edx
  cmp    $0x9,%edi
  jle    L2
L49:
  mov    $0x1,%ebx
  xor    %ecx,%ecx
  cs nopw 0x0(%rax,%rax,1)
L40:
  lea    (%rbx,%rbx,4),%ebx
  mov    %ecx,%edi
  add    $0x1,%ecx
  add    %ebx,%ebx
  cmp    %eax,%ebx
  jle    L40
  movslq %ecx,%rbp
  lea    0x2(%rdi),%r8d
  cmp    $0xffffffff,%edx
  je     L9
L2:
  mov    %r8d,%edi
  call   <malloc@plt>
  movb   $0x0,(%rax,%rbp,1)
  mov    %rax,%r8
L81:
  mov    %r8,%rsi
  mov    $0xcccccccd,%edi
  nopl   0x0(%rax)
L47:
  mov    %r12d,%eax
  add    $0x1,%rsi
  cltd
  idiv   %ebx
  mov    %ebx,%ebx
  imul   %rdi,%rbx
  shr    $0x23,%rbx
  lea    0x30(%rax),%ecx
  mov    %edx,%r12d
  mov    %cl,-0x1(%rsi)
  test   %ebx,%ebx
  jne    L47
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L32:
  mov    %edi,%edx
  mov    $0xcccccccd,%ecx
  neg    %edx
  mov    %edx,%eax
  mov    %edx,%r12d
  imul   %rcx,%rax
  shr    $0x23,%rax
  cmp    $0xfffffff7,%edi
  jge    L77
  mov    $0xffffffff,%edx
  jmp    L49
  nopl   (%rax)
L9:
  add    $0x3,%edi
  add    $0x1,%rbp
L50:
  call   <malloc@plt>
  movb   $0x0,(%rax,%rbp,1)
  mov    %rax,%r8
  movb   $0x2d,(%rax)
  jmp    L81
L77:
  mov    $0x1,%ebp
  mov    $0x2,%edi
  mov    $0x1,%ebx
  jmp    L50
```

assistant: ```c
char *ft_itoa(int nb) {
  int length;
  int div;
  int signe;
  char *res;
  int i;
  int ori_nbr;
  length = 0;
  div = 1;
  signe = 1;
  ori_nbr = nb;
  if (nb < 0) {
    signe = -1;
    nb = -nb;
  }
  while (div <= (nb / 10)) {
    div *= 10;
    length++;
  }
  i = 0;
  if (signe == -1) {
    res = (char *)malloc(sizeof(char) * (length + 2));
    res[length + 1] = '\0';
    res[0] = '-';
  } else {
    res = (char *)malloc(sizeof(char) * (length + 1));
    res[length] = '\0';
  }
  while (div >= 1) {
    res[i] = (nb / div) + 48;
    nb -= (nb / div) * div;
    div /= 10;
    i++;
  }
  nb = ori_nbr;
  return (res);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_itoa>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  test   %edi,%edi
  js     L46
  movslq %edi,%rax
  mov    %edi,%edx
  xor    %ebp,%ebp
  mov    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  mov    $0x1,%ebx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    $0x1,%edx
  cmp    $0x9,%edi
  jle    L70
L18:
  mov    $0x1,%ebx
  xor    %ecx,%ecx
  cs nopw 0x0(%rax,%rax,1)
L0:
  lea    (%rbx,%rbx,4),%ebx
  mov    %ecx,%edi
  add    $0x1,%ecx
  add    %ebx,%ebx
  cmp    %eax,%ebx
  jle    L0
  movslq %ecx,%rbp
  lea    0x2(%rdi),%r8d
  cmp    $0xffffffff,%edx
  je     L3
L70:
  mov    %r8d,%edi
  call   <malloc@plt>
  movb   $0x0,(%rax,%rbp,1)
  mov    %rax,%r8
L71:
  mov    %r8,%rsi
  mov    $0xcccccccd,%edi
  nopl   0x0(%rax)
L11:
  mov    %r12d,%eax
  add    $0x1,%rsi
  cltd
  idiv   %ebx
  mov    %ebx,%ebx
  imul   %rdi,%rbx
  shr    $0x23,%rbx
  lea    0x30(%rax),%ecx
  mov    %edx,%r12d
  mov    %cl,-0x1(%rsi)
  test   %ebx,%ebx
  jne    L11
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L46:
  mov    %edi,%edx
  mov    $0xcccccccd,%ecx
  neg    %edx
  mov    %edx,%eax
  mov    %edx,%r12d
  imul   %rcx,%rax
  shr    $0x23,%rax
  cmp    $0xfffffff7,%edi
  jge    L53
  mov    $0xffffffff,%edx
  jmp    L18
  nopl   (%rax)
L3:
  add    $0x3,%edi
  add    $0x1,%rbp
L27:
  call   <malloc@plt>
  movb   $0x0,(%rax,%rbp,1)
  mov    %rax,%r8
  movb   $0x2d,(%rax)
  jmp    L71
L53:
  mov    $0x1,%ebp
  mov    $0x2,%edi
  mov    $0x1,%ebx
  jmp    L27
```

assistant: ```c
char *ft_itoa(int nb) {
  int length;
  int div;
  int signe;
  char *res;
  int i;
  int ori_nbr;
  length = 0;
  div = 1;
  signe = 1;
  ori_nbr = nb;
  if (nb < 0) {
    signe = -1;
    nb = -nb;
  }
  while (div <= (nb / 10)) {
    div *= 10;
    length++;
  }
  i = 0;
  if (signe == -1) {
    res = (char *)malloc(sizeof(char) * (length + 2));
    res[length + 1] = '\0';
    res[0] = '-';
  } else {
    res = (char *)malloc(sizeof(char) * (length + 1));
    res[length] = '\0';
  }
  while (div >= 1) {
    res[i] = (nb / div) + 48;
    nb -= (nb / div) * div;
    div /= 10;
    i++;
  }
  nb = ori_nbr;
  return (res);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ms_gcd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  jmp    L11
L15:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L11:
  cmpl   $0x0,-0x18(%rbp)
  jne    L15
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ms_gcd(int m, int n) {
  while (n != 0) {
    int t = m % n;
    m = n;
    n = t;
  }
  return m;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ms_gcd>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  je     L28
L5:
  mov    %esi,%ecx
  cltd
  idiv   %esi
  mov    %edx,%esi
  mov    %ecx,%eax
  test   %edx,%edx
  jne    L5
L27:
  mov    %ecx,%eax
  ret
L28:
  mov    %edi,%ecx
  jmp    L27
```

assistant: ```c
int ms_gcd(int m, int n) {
  while (n != 0) {
    int t = m % n;
    m = n;
    n = t;
  }
  return m;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ms_gcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edx
  test   %esi,%esi
  je     L14
  nopl   0x0(%rax)
L11:
  mov    %edx,%r8d
  cltd
  idiv   %r8d
  mov    %r8d,%eax
  test   %edx,%edx
  jne    L11
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %edi,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ms_gcd(int m, int n) {
  while (n != 0) {
    int t = m % n;
    m = n;
    n = t;
  }
  return m;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ms_gcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edx
  test   %esi,%esi
  je     L3
  nopl   0x0(%rax)
L5:
  mov    %edx,%r8d
  cltd
  idiv   %r8d
  mov    %r8d,%eax
  test   %edx,%edx
  jne    L5
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %edi,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ms_gcd(int m, int n) {
  while (n != 0) {
    int t = m % n;
    m = n;
    n = t;
  }
  return m;
}
```