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   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x30(%rbp)
  movl   $0x3,-0x2c(%rbp)
  movl   $0x1,-0x28(%rbp)
  movl   $0x7,-0x24(%rbp)
  movl   $0x4,-0x20(%rbp)
  movl   $0x1,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x14,-0x14(%rbp)
  movl   $0x8,-0x34(%rbp)
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <bead_sort@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L29
L28:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L29:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L28
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D18
type_guessed: string
name: unpack_data
arguments: label: D16
type_guessed: string

tool: D18: The sorted array is:

tool: D16: %d


assistant: ```c
int main() {
  int i, x[] = {5, 3, 1, 7, 4, 1, 1, 20};
  int len = sizeof(x) / sizeof(x[0]);
  bead_sort(x, len);
  printf("The sorted array is:\n");
  for (i = 0; i < len; i++)
    printf("%d\n", x[i]);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x5,(%rsp)
  movl   $0x3,0x4(%rsp)
  movl   $0x1,0x8(%rsp)
  movl   $0x7,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x1,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x14,0x1c(%rsp)
  mov    %rsp,%rbx
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <bead_sort@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%r12
  lea    D8(%rip),%rbp
L6:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L6
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D13
datatype: string
name: unpack_data
arguments: data_tag: D8
datatype: string

tool: D13: The sorted array is:

tool: D8: %d


assistant: ```c
int main() {
  int i, x[] = {5, 3, 1, 7, 4, 1, 1, 20};
  int len = sizeof(x) / sizeof(x[0]);
  bead_sort(x, len);
  printf("The sorted array is:\n");
  for (i = 0; i < len; i++)
    printf("%d\n", x[i]);
  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    $0x8,%esi
  push   %rbp
  lea    D14(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x300000005,%rax
  mov    %rsp,%rbx
  lea    0x20(%rsp),%r12
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x700000001,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x100000004,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x1400000001,%rax
  mov    %rax,0x18(%rsp)
  call   <bead_sort@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  nopl   (%rax)
L1:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L1
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D14
type_infered: string
name: unpack_data
arguments: tag: D3
type_infered: string

tool: D14: %d


tool: D3: The sorted array is:

assistant: ```c
int main() {
  int i, x[] = {5, 3, 1, 7, 4, 1, 1, 20};
  int len = sizeof(x) / sizeof(x[0]);
  bead_sort(x, len);
  printf("The sorted array is:\n");
  for (i = 0; i < len; i++)
    printf("%d\n", x[i]);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x8,%esi
  push   %rbp
  lea    D28(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D10(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x20(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D8(%rip),%xmm0
  mov    %rbx,%rdi
  movaps %xmm0,0x10(%rsp)
  call   <bead_sort@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  nopl   (%rax)
L16:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L16
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D28
type: string
name: read_data
arguments: tag: D10
type: byte[16]
name: read_data
arguments: tag: D8
type: byte[16]
name: read_data
arguments: tag: D36
type: string

tool: D28: %d


tool: {"D10": [0x05,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D8": [0x04,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: D36: The sorted array is:

assistant: ```c
int main() {
  int i, x[] = {5, 3, 1, 7, 4, 1, 1, 20};
  int len = sizeof(x) / sizeof(x[0]);
  bead_sort(x, len);
  printf("The sorted array is:\n");
  for (i = 0; i < len; i++)
    printf("%d\n", x[i]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<screen_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <DisplayOfScreen@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L38
L37:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <ScreenOfDisplay@plt>
  cmp    %rax,-0x18(%rbp)
  jne    L17
  mov    -0xc(%rbp),%eax
  jmp    L23
L17:
  addl   $0x1,-0xc(%rbp)
L38:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <ScreenCount@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L37
  mov    $0x0,%eax
  call   <abort@plt>
  mov    $0x0,%eax
L23:
  leave
  ret
```

assistant: ```c
int screen_number(Screen *screen) {
  Display *dpy = DisplayOfScreen(screen);
  int i;
  for (i = 0; i < ScreenCount(dpy); i++)
    if (ScreenOfDisplay(dpy, i) == screen)
      return i;
  abort();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<screen_number>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <DisplayOfScreen@plt>
  mov    %rax,%rbp
  mov    $0x0,%ebx
  jmp    L7
L33:
  add    $0x1,%ebx
L7:
  mov    %rbp,%rdi
  call   <ScreenCount@plt>
  cmp    %ebx,%eax
  jle    L28
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <ScreenOfDisplay@plt>
  cmp    %rax,%r12
  jne    L33
  jmp    L14
L28:
  mov    $0x0,%eax
  call   <abort@plt>
  mov    $0x0,%ebx
L14:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int screen_number(Screen *screen) {
  Display *dpy = DisplayOfScreen(screen);
  int i;
  for (i = 0; i < ScreenCount(dpy); i++)
    if (ScreenOfDisplay(dpy, i) == screen)
      return i;
  abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<screen_number>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <DisplayOfScreen@plt>
  mov    %rax,%rbx
  jmp    L13
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <ScreenOfDisplay@plt>
  cmp    %rax,%rbp
  je     L5
  add    $0x1,%r12d
L13:
  mov    %rbx,%rdi
  call   <ScreenCount@plt>
  cmp    %r12d,%eax
  jg     L17
  xor    %eax,%eax
  xor    %r12d,%r12d
  call   <abort@plt>
L5:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int screen_number(Screen *screen) {
  Display *dpy = DisplayOfScreen(screen);
  int i;
  for (i = 0; i < ScreenCount(dpy); i++)
    if (ScreenOfDisplay(dpy, i) == screen)
      return i;
  abort();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<screen_number>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <DisplayOfScreen@plt>
  mov    %rax,%rbx
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <ScreenOfDisplay@plt>
  cmp    %rax,%rbp
  je     L3
  add    $0x1,%r12d
L25:
  mov    %rbx,%rdi
  call   <ScreenCount@plt>
  cmp    %r12d,%eax
  jg     L6
  xor    %eax,%eax
  xor    %r12d,%r12d
  call   <abort@plt>
L3:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int screen_number(Screen *screen) {
  Display *dpy = DisplayOfScreen(screen);
  int i;
  for (i = 0; i < ScreenCount(dpy); i++)
    if (ScreenOfDisplay(dpy, i) == screen)
      return i;
  abort();
  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
<ft_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L25
  addl   $0x1,-0x10(%rbp)
  jmp    L25
L74:
  mov    -0xc(%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,-0xc(%rbp)
L25:
  cmpl   $0x9,-0xc(%rbp)
  jg     L68
  cmpl   $0xfffffff7,-0xc(%rbp)
  jge    L19
L68:
  addl   $0x1,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  jne    L74
L19:
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L57
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L78
L88:
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%edi
  call   <mod@plt>
  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    -0xc(%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,-0xc(%rbp)
L78:
  subl   $0x1,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  jns    L88
  cmpl   $0x0,-0x14(%rbp)
  jns    L57
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
L57:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *res;
  int tn;
  i = 1;
  tn = n;
  if (n < 0)
    i++;
  while ((tn > 9 || tn < -9) && ++i)
    tn /= 10;
  res = (char *)malloc(sizeof(char) * (i + 1));
  if (res) {
    res[i] = 0;
    tn = n;
    while (--i >= 0) {
      res[i] = mod(tn % 10) + 48;
      tn /= 10;
    }
    if (n < 0)
      res[0] = '-';
  }
  return (res);
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_itoa>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %edi,%ebx
  shr    $0x1f,%ebx
  add    $0x1,%ebx
  lea    0x9(%rdi),%eax
  cmp    $0x12,%eax
  jbe    L93
  add    $0x1,%ebx
  mov    %edi,%edx
L87:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  add    $0x9,%eax
  cmp    $0x12,%eax
  jbe    L12
  add    $0x1,%ebx
  jne    L87
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L61
L45:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L45
  movslq %ebx,%rax
  movb   $0x0,(%r12,%rax,1)
  sub    $0x1,%ebx
  js     L102
L56:
  movslq %ebx,%rbx
  mov    %ebp,%edi
L98:
  movslq %edi,%r13
  imul   $0x66666667,%r13,%r13
  sar    $0x22,%r13
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%r13d
  lea    0x0(%r13,%r13,4),%eax
  add    %eax,%eax
  sub    %eax,%edi
  call   <mod@plt>
  add    $0x30,%eax
  mov    %al,(%r12,%rbx,1)
  mov    %r13d,%edi
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jns    L98
L102:
  test   %ebp,%ebp
  jns    L45
  movb   $0x2d,(%r12)
  jmp    L45
L93:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L45
  movslq %ebx,%rax
  movb   $0x0,(%r12,%rax,1)
  sub    $0x1,%ebx
  jmp    L56
L61:
  movb   $0x0,(%r12)
  jmp    L102
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *res;
  int tn;
  i = 1;
  tn = n;
  if (n < 0)
    i++;
  while ((tn > 9 || tn < -9) && ++i)
    tn /= 10;
  res = (char *)malloc(sizeof(char) * (i + 1));
  if (res) {
    res[i] = 0;
    tn = n;
    while (--i >= 0) {
      res[i] = mod(tn % 10) + 48;
      tn /= 10;
    }
    if (n < 0)
      res[0] = '-';
  }
  return (res);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ft_itoa>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  sar    $0x1f,%edi
  push   %rbx
  mov    %ebp,%ebx
  not    %edi
  lea    0x9(%rbp),%eax
  sar    $0x1f,%ebx
  add    $0x3,%edi
  not    %ebx
  add    $0x2,%ebx
  cmp    $0x12,%eax
  jbe    L26
  mov    %ebp,%edx
  nopw   0x0(%rax,%rax,1)
L42:
  movslq %edx,%rax
  mov    %edx,%ecx
  mov    %ebx,%edi
  add    $0x1,%ebx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x9(%rax),%eax
  cmp    $0x12,%eax
  ja     L42
  add    $0x2,%edi
L26:
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L45
  movslq %ebx,%rax
  sub    $0x1,%ebx
  mov    %ebp,%edi
  movb   $0x0,(%r12,%rax,1)
  movslq %ebx,%r14
  lea    -0x2(%r12,%rax,1),%r13
  mov    %ebx,%ebx
  add    %r12,%r14
  sub    %rbx,%r13
  nopl   0x0(%rax)
L12:
  movslq %edi,%rbx
  mov    %edi,%eax
  sub    $0x1,%r14
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  sub    %eax,%edi
  call   <mod@plt>
  mov    %ebx,%edi
  add    $0x30,%eax
  mov    %al,0x1(%r14)
  cmp    %r13,%r14
  jne    L12
  test   %ebp,%ebp
  js     L20
L45:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  movb   $0x2d,(%r12)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *res;
  int tn;
  i = 1;
  tn = n;
  if (n < 0)
    i++;
  while ((tn > 9 || tn < -9) && ++i)
    tn /= 10;
  res = (char *)malloc(sizeof(char) * (i + 1));
  if (res) {
    res[i] = 0;
    tn = n;
    while (--i >= 0) {
      res[i] = mod(tn % 10) + 48;
      tn /= 10;
    }
    if (n < 0)
      res[0] = '-';
  }
  return (res);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_itoa>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  sar    $0x1f,%edi
  push   %rbx
  mov    %ebp,%ebx
  not    %edi
  lea    0x9(%rbp),%eax
  sar    $0x1f,%ebx
  add    $0x3,%edi
  not    %ebx
  add    $0x2,%ebx
  cmp    $0x12,%eax
  jbe    L3
  mov    %ebp,%edx
  nopw   0x0(%rax,%rax,1)
L41:
  movslq %edx,%rax
  mov    %edx,%ecx
  mov    %ebx,%edi
  add    $0x1,%ebx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x9(%rax),%eax
  cmp    $0x12,%eax
  ja     L41
  add    $0x2,%edi
L3:
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L32
  movslq %ebx,%rax
  sub    $0x1,%ebx
  mov    %ebp,%edi
  movb   $0x0,(%r12,%rax,1)
  movslq %ebx,%r14
  lea    -0x2(%r12,%rax,1),%r13
  mov    %ebx,%ebx
  add    %r12,%r14
  sub    %rbx,%r13
  nopl   0x0(%rax)
L30:
  movslq %edi,%rbx
  mov    %edi,%eax
  sub    $0x1,%r14
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%eax
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  sub    %eax,%edi
  call   <mod@plt>
  mov    %ebx,%edi
  add    $0x30,%eax
  mov    %al,0x1(%r14)
  cmp    %r14,%r13
  jne    L30
  test   %ebp,%ebp
  js     L49
L32:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L49:
  movb   $0x2d,(%r12)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *res;
  int tn;
  i = 1;
  tn = n;
  if (n < 0)
    i++;
  while ((tn > 9 || tn < -9) && ++i)
    tn /= 10;
  res = (char *)malloc(sizeof(char) * (i + 1));
  if (res) {
    res[i] = 0;
    tn = n;
    while (--i >= 0) {
      res[i] = mod(tn % 10) + 48;
      tn /= 10;
    }
    if (n < 0)
      res[0] = '-';
  }
  return (res);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_leap_year>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L33
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L1
L33:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L21
L1:
  mov    $0x1,%eax
  jmp    L9
L21:
  mov    $0x0,%eax
L9:
  pop    %rbp
  ret
```

assistant: ```c
int is_leap_year(int year) {
  if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<is_leap_year>:
  endbr64
  test   $0x3,%dil
  jne    L16
  movslq %edi,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x64,%eax,%eax
  mov    $0x1,%edx
  cmp    %eax,%edi
  jne    L4
L16:
  movslq %edi,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x27,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x190,%eax,%eax
  cmp    %eax,%edi
  sete   %dl
  movzbl %dl,%edx
L4:
  mov    %edx,%eax
  ret
```

assistant: ```c
int is_leap_year(int year) {
  if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    return 1;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_leap_year>:
  endbr64
  test   $0x3,%dil
  jne    L4
  imul   $0xc28f5c29,%edi,%eax
  mov    $0x1,%r8d
  add    $0x51eb850,%eax
  ror    $0x2,%eax
  cmp    $0x28f5c28,%eax
  ja     L11
L4:
  imul   $0xc28f5c29,%edi,%edi
  xor    %r8d,%r8d
  add    $0x51eb850,%edi
  ror    $0x4,%edi
  cmp    $0xa3d70a,%edi
  setbe  %r8b
L11:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_leap_year(int year) {
  if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<is_leap_year>:
  endbr64
  test   $0x3,%dil
  jne    L9
  imul   $0xc28f5c29,%edi,%eax
  mov    $0x1,%r8d
  add    $0x51eb850,%eax
  ror    $0x2,%eax
  cmp    $0x28f5c28,%eax
  ja     L19
L9:
  imul   $0xc28f5c29,%edi,%edi
  xor    %r8d,%r8d
  add    $0x51eb850,%edi
  ror    $0x4,%edi
  cmp    $0xa3d70a,%edi
  setbe  %r8b
L19:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_leap_year(int year) {
  if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    return 1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<peer_closesocket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <shutdown@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  nop
  leave
  ret
```

assistant: ```c
void peer_closesocket(int sfd) {
  shutdown(sfd, SHUT_RDWR);
  close(sfd);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<peer_closesocket>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    D5(%rip),%rax
  mov    (%rax),%esi
  call   <shutdown@plt>
  mov    %ebx,%edi
  call   <close@plt>
  pop    %rbx
  ret
```

assistant: ```c
void peer_closesocket(int sfd) {
  shutdown(sfd, SHUT_RDWR);
  close(sfd);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<peer_closesocket>:
  endbr64
  mov    D7(%rip),%rax
  push   %rbp
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <shutdown@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <close@plt>
```

assistant: ```c
void peer_closesocket(int sfd) {
  shutdown(sfd, SHUT_RDWR);
  close(sfd);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<peer_closesocket>:
  endbr64
  mov    D0(%rip),%rax
  push   %rbp
  mov    %edi,%ebp
  mov    (%rax),%esi
  call   <shutdown@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <close@plt>
```

assistant: ```c
void peer_closesocket(int sfd) {
  shutdown(sfd, SHUT_RDWR);
  close(sfd);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movl   $0x64636261,-0x2d(%rbp)
  movb   $0x0,-0x29(%rbp)
  lea    -0x2d(%rbp),%rcx
  add    $0x3,%rcx
  lea    -0x2d(%rbp),%rdx
  add    $0x2,%rdx
  lea    -0x2d(%rbp),%rax
  add    $0x1,%rax
  lea    -0x2d(%rbp),%rsi
  mov    %rcx,%r8
  mov    %rdx,%rcx
  mov    %rax,%rdx
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x74,-0x32(%rbp)
  movb   $0x30,-0x31(%rbp)
  movb   $0x31,-0x30(%rbp)
  movb   $0x36,-0x2f(%rbp)
  movb   $0x35,-0x2e(%rbp)
  lea    -0x2e(%rbp),%rdi
  lea    -0x2f(%rbp),%rsi
  lea    -0x30(%rbp),%rcx
  lea    -0x31(%rbp),%rdx
  lea    -0x32(%rbp),%rax
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x2f(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  mov    %eax,%r13d
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  mov    %eax,%r12d
  lea    -0x31(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  mov    %eax,%ebx
  lea    -0x32(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <atoi@plt>
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  add    $0x38,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D25
type: string
name: retrieve_data
arguments: label: D12
type: string
name: retrieve_data
arguments: label: D15
type: string

tool: D25: %#x	%#x	%#x	%#x


tool: D12: %#x	%#x	%#x	%#x	%#x


tool: D15: %i	%i	%i	%i


assistant: ```c
int main(int argc, char *argv[]) {
  char a[] = "abcd";
  printf("%#x\t%#x\t%#x\t%#x\n", a, a + 1, a + 2, a + 3);
  char cp = 't';
  char cp0 = '0';
  char cp1 = '1';
  char cp2 = '6';
  char cp3 = '5';
  printf("%#x\t%#x\t%#x\t%#x\t%#x\n", &cp, &cp0, &cp1, &cp2, &cp3);
  printf("%i\t%i\t%i\t%i\n", atoi(&cp), atoi(&cp0), atoi(&cp1), atoi(&cp2));
  return 0;
}
```