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
<place_commas_in_double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <commas_in_double@plt>
  lea    D6(%rip),%rax
  leave
  ret
```

assistant: ```c
char *place_commas_in_double(double d) {
  static char destination[64];
  commas_in_double(destination, d);
  return destination;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<place_commas_in_double>:
  endbr64
  push   %rbx
  lea    D1(%rip),%rbx
  mov    %rbx,%rdi
  call   <commas_in_double@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *place_commas_in_double(double d) {
  static char destination[64];
  commas_in_double(destination, d);
  return destination;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<place_commas_in_double>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  mov    %r12,%rdi
  call   <commas_in_double@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *place_commas_in_double(double d) {
  static char destination[64];
  commas_in_double(destination, d);
  return destination;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<place_commas_in_double>:
  endbr64
  push   %r12
  lea    D0(%rip),%r12
  mov    %r12,%rdi
  call   <commas_in_double@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *place_commas_in_double(double d) {
  static char destination[64];
  commas_in_double(destination, d);
  return destination;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xe0(%rbp),%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xe8(%rbp)
  movl   $0x1,-0xe4(%rbp)
  jmp    L0
L55:
  mov    -0xe8(%rbp),%eax
  cltq
  movzbl -0xe0(%rbp,%rax,1),%edx
  mov    -0xe8(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L14
  movl   $0x0,-0xe4(%rbp)
L14:
  addl   $0x1,-0xe8(%rbp)
L0:
  cmpl   $0x1,-0xe4(%rbp)
  jne    L74
  mov    -0xe8(%rbp),%eax
  cltq
  movzbl -0xe0(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L55
  mov    -0xe8(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L55
L74:
  cmpl   $0x1,-0xe4(%rbp)
  jne    L56
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L48
L56:
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L48:
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L61
  call   <__stack_chk_fail@plt>
L61:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D64
type_infered: string
name: extract_symbolic_data
arguments: data_label: D65
type_infered: string
name: extract_symbolic_data
arguments: data_label: D55
type_infered: string
name: extract_symbolic_data
arguments: data_label: D27
type_infered: string
name: extract_symbolic_data
arguments: data_label: D18
type_infered: string
name: extract_symbolic_data
arguments: data_label: D28
type_infered: string
name: extract_symbolic_data
arguments: data_label: D5
type_infered: string

tool: D64: Inserire la prima parola 

tool: D65: %s

tool: D55: 
Inserire la seconda parola 

tool: D27: 
Le due parole sono uguali

tool: D18: 
Le due parole sono diverse

tool: D28: 


tool: D5: PAUSE

assistant: ```c
int main() {
  char parola1[100];
  char parola2[100];
  int i;
  int uguali;
  printf("Inserire la prima parola ");
  scanf("%s", parola1);
  printf("\nInserire la seconda parola ");
  scanf("%s", parola2);
  i = 0;
  uguali = 1;
  while ((uguali == 1) && (parola1[i] != '\0' || parola2[i] != '\0')) {
    if (parola1[i] != parola2[i]) {
      uguali = 0;
    }
    i++;
  }
  if (uguali == 1) {
    printf("\nLe due parole sono uguali");
  } else {
    printf("\nLe due parole sono diverse");
  }
  printf("\n");
  system("PAUSE");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D66(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D69(%rip),%rdi
  call   <printf@plt>
  lea    0x70(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  lea    0x6f(%rsp),%rsi
  jmp    L22
L29:
  add    $0x1,%rax
  cmp    %dl,(%rsi,%rax,1)
  jne    L6
L22:
  movzbl (%rsp,%rax,1),%edx
  test   %dl,%dl
  jne    L29
  movslq %eax,%rcx
  cmpb   $0x0,0x70(%rsp,%rcx,1)
  je     L36
L6:
  lea    D59(%rip),%rdi
  call   <printf@plt>
L34:
  lea    D44(%rip),%rdi
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <system@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0xe0,%rsp
  pop    %rbx
  ret
L36:
  lea    D60(%rip),%rdi
  call   <printf@plt>
  jmp    L34
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D16
type_infered: string
name: read_data
arguments: tag: D66
type_infered: string
name: read_data
arguments: tag: D69
type_infered: string
name: read_data
arguments: tag: D59
type_infered: string
name: read_data
arguments: tag: D44
type_infered: string
name: read_data
arguments: tag: D50
type_infered: string
name: read_data
arguments: tag: D60
type_infered: string

tool: D16: Inserire la prima parola 

tool: D66: %s

tool: D69: 
Inserire la seconda parola 

tool: D59: 
Le due parole sono diverse

tool: D44: 


tool: D50: PAUSE

tool: D60: 
Le due parole sono uguali

assistant: ```c
int main() {
  char parola1[100];
  char parola2[100];
  int i;
  int uguali;
  printf("Inserire la prima parola ");
  scanf("%s", parola1);
  printf("\nInserire la seconda parola ");
  scanf("%s", parola2);
  i = 0;
  uguali = 1;
  while ((uguali == 1) && (parola1[i] != '\0' || parola2[i] != '\0')) {
    if (parola1[i] != parola2[i]) {
      uguali = 0;
    }
    i++;
  }
  if (uguali == 1) {
    printf("\nLe due parole sono uguali");
  } else {
    printf("\nLe due parole sono diverse");
  }
  printf("\n");
  system("PAUSE");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  lea    D63(%rip),%rdi
  lea    D20(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x70(%rsp),%rbx
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
L39:
  movzbl 0x0(%rbp,%rax,1),%edx
  movzbl (%rbx,%rax,1),%ecx
  test   %dl,%dl
  jne    L25
  test   %cl,%cl
  je     L2
L29:
  lea    D52(%rip),%rdi
  call   <printf@plt>
L47:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <system@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0xe0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  add    $0x1,%rax
  cmp    %cl,%dl
  je     L39
  jmp    L29
L2:
  lea    D21(%rip),%rdi
  call   <printf@plt>
  jmp    L47
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D63
type_guessed: string
name: disassemble_data
arguments: tag: D20
type_guessed: string
name: disassemble_data
arguments: tag: D11
type_guessed: string
name: disassemble_data
arguments: tag: D52
type_guessed: string
name: disassemble_data
arguments: tag: D16
type_guessed: string
name: disassemble_data
arguments: tag: D15
type_guessed: string
name: disassemble_data
arguments: tag: D21
type_guessed: string

tool: D63: Inserire la prima parola 

tool: D20: %s

tool: D11: 
Inserire la seconda parola 

tool: D52: 
Le due parole sono diverse

tool: D16: 


tool: D15: PAUSE

tool: D21: 
Le due parole sono uguali

assistant: ```c
int main() {
  char parola1[100];
  char parola2[100];
  int i;
  int uguali;
  printf("Inserire la prima parola ");
  scanf("%s", parola1);
  printf("\nInserire la seconda parola ");
  scanf("%s", parola2);
  i = 0;
  uguali = 1;
  while ((uguali == 1) && (parola1[i] != '\0' || parola2[i] != '\0')) {
    if (parola1[i] != parola2[i]) {
      uguali = 0;
    }
    i++;
  }
  if (uguali == 1) {
    printf("\nLe due parole sono uguali");
  } else {
    printf("\nLe due parole sono diverse");
  }
  printf("\n");
  system("PAUSE");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  lea    D58(%rip),%rdi
  lea    D63(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x70(%rsp),%rbx
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D54(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
L38:
  movzbl 0x0(%rbp,%rax,1),%edx
  movzbl (%rbx,%rax,1),%ecx
  test   %dl,%dl
  jne    L1
  test   %cl,%cl
  je     L8
L22:
  lea    D60(%rip),%rdi
  call   <printf@plt>
L13:
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <system@plt>
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0xe0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rax
  cmp    %cl,%dl
  je     L38
  jmp    L22
L8:
  lea    D29(%rip),%rdi
  call   <printf@plt>
  jmp    L13
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D58
data_type: string
name: parse_data
arguments: label: D63
data_type: string
name: parse_data
arguments: label: D54
data_type: string
name: parse_data
arguments: label: D60
data_type: string
name: parse_data
arguments: label: D25
data_type: string
name: parse_data
arguments: label: D6
data_type: string
name: parse_data
arguments: label: D29
data_type: string

tool: D58: Inserire la prima parola 

tool: D63: %s

tool: D54: 
Inserire la seconda parola 

tool: D60: 
Le due parole sono diverse

tool: D25: 


tool: D6: PAUSE

tool: D29: 
Le due parole sono uguali

assistant: ```c
int main() {
  char parola1[100];
  char parola2[100];
  int i;
  int uguali;
  printf("Inserire la prima parola ");
  scanf("%s", parola1);
  printf("\nInserire la seconda parola ");
  scanf("%s", parola2);
  i = 0;
  uguali = 1;
  while ((uguali == 1) && (parola1[i] != '\0' || parola2[i] != '\0')) {
    if (parola1[i] != parola2[i]) {
      uguali = 0;
    }
    i++;
  }
  if (uguali == 1) {
    printf("\nLe due parole sono uguali");
  } else {
    printf("\nLe due parole sono diverse");
  }
  printf("\n");
  system("PAUSE");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L17
L6:
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <putc@plt>
L17:
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <getc@plt>
  mov    %eax,-0x4(%rbp)
  mov    D26(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int c;
  while (EOF != (c = getc(stdin)))
    putc(c, stdout);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    D27(%rip),%rbx
  jmp    L1
L6:
  mov    D7(%rip),%rax
  mov    (%rax),%esi
  call   <putc@plt>
L1:
  mov    (%rbx),%edi
  call   <getc@plt>
  mov    %eax,%edi
  mov    D22(%rip),%rax
  cmp    %edi,(%rax)
  jne    L6
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
  int c;
  while (EOF != (c = getc(stdin)))
    putc(c, stdout);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D1(%rip),%rbp
  mov    D25(%rip),%rbx
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L12:
  mov    D28(%rip),%rax
  mov    (%rax),%esi
  call   <putc@plt>
L19:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L12
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int c;
  while (EOF != (c = getc(stdin)))
    putc(c, stdout);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D3(%rip),%rbp
  mov    D7(%rip),%rbx
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L11:
  mov    D29(%rip),%rax
  mov    (%rax),%esi
  call   <putc@plt>
L15:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L11
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int c;
  while (EOF != (c = getc(stdin)))
    putc(c, stdout);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<mt_deactivate_alarm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <timer_delete@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L13
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  neg    %eax
  jmp    L0
L13:
  mov    $0x0,%eax
L0:
  leave
  ret
```

assistant: ```c
int mt_deactivate_alarm(timer_t timer) {
  int ret;
  ret = timer_delete(timer);
  if (ret)
    return ((ret > 0) ? -ret : ret);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mt_deactivate_alarm>:
  endbr64
  sub    $0x8,%rsp
  call   <timer_delete@plt>
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  neg    %edx
  test   %eax,%eax
  cmovne %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int mt_deactivate_alarm(timer_t timer) {
  int ret;
  ret = timer_delete(timer);
  if (ret)
    return ((ret > 0) ? -ret : ret);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mt_deactivate_alarm>:
  endbr64
  sub    $0x8,%rsp
  call   <timer_delete@plt>
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  neg    %edx
  test   %eax,%eax
  cmovne %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int mt_deactivate_alarm(timer_t timer) {
  int ret;
  ret = timer_delete(timer);
  if (ret)
    return ((ret > 0) ? -ret : ret);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mt_deactivate_alarm>:
  endbr64
  sub    $0x8,%rsp
  call   <timer_delete@plt>
  mov    %eax,%edx
  neg    %edx
  cmovs  %eax,%edx
  neg    %edx
  test   %eax,%eax
  cmovne %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int mt_deactivate_alarm(timer_t timer) {
  int ret;
  ret = timer_delete(timer);
  if (ret)
    return ((ret > 0) ? -ret : ret);
  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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <stack_create@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L13
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L12
L13:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_pop@plt>
  movl   $0x5,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <stack_push@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_top@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x9,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <stack_push@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_top@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_pop@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_top@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <stack_destroy@plt>
  mov    $0x0,%eax
L12:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D50
datatype: string
name: parse_data
arguments: data_tag: D24
datatype: string
name: parse_data
arguments: data_tag: D36
datatype: string
name: parse_data
arguments: data_tag: D19
datatype: string

tool: D50: creating stack failed

tool: D24: push '%d' into stack


tool: D36: top of the stack is %d


tool: D19: pop the top item of the stack

assistant: ```c
int main(void) {
  int val;
  void *instance;
  int *top;
  instance = stack_create();
  if (instance == NULL) {
    printf("creating stack failed\n");
    return -1;
  }
  stack_pop(instance);
  val = 5;
  stack_push(instance, val);
  printf("push '%d' into stack\n", val);
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  val = 9;
  stack_push(instance, val);
  printf("push '%d' into stack\n", val);
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  stack_pop(instance);
  printf("pop the top item of the stack\n");
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  stack_destroy(instance);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <stack_create@plt>
  test   %rax,%rax
  je     L17
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <stack_pop@plt>
  mov    $0x5,%esi
  mov    %rbx,%rdi
  call   <stack_push@plt>
  mov    $0x5,%esi
  lea    D22(%rip),%r12
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <stack_top@plt>
  mov    (%rax),%esi
  lea    D31(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %rbx,%rdi
  call   <stack_push@plt>
  mov    $0x9,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <stack_top@plt>
  mov    (%rax),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <stack_pop@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rdi
  call   <stack_top@plt>
  mov    (%rax),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <stack_destroy@plt>
  mov    $0x0,%eax
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L16
```

tool_calls: name: parse_data
arguments: tag: D22
data_type: string
name: parse_data
arguments: tag: D31
data_type: string
name: parse_data
arguments: tag: D1
data_type: string
name: parse_data
arguments: tag: D35
data_type: string

tool: D22: push '%d' into stack


tool: D31: top of the stack is %d


tool: D1: pop the top item of the stack

tool: D35: creating stack failed

assistant: ```c
int main(void) {
  int val;
  void *instance;
  int *top;
  instance = stack_create();
  if (instance == NULL) {
    printf("creating stack failed\n");
    return -1;
  }
  stack_pop(instance);
  val = 5;
  stack_push(instance, val);
  printf("push '%d' into stack\n", val);
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  val = 9;
  stack_push(instance, val);
  printf("push '%d' into stack\n", val);
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  stack_pop(instance);
  printf("pop the top item of the stack\n");
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  stack_destroy(instance);
  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   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  call   <stack_create@plt>
  test   %rax,%rax
  je     L19
  mov    %rax,%rbp
  mov    %rax,%rdi
  lea    D33(%rip),%r13
  call   <stack_pop@plt>
  mov    $0x5,%esi
  mov    %rbp,%rdi
  lea    D5(%rip),%r12
  call   <stack_push@plt>
  mov    $0x5,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <stack_top@plt>
  mov    %r12,%rdi
  mov    (%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <stack_push@plt>
  mov    $0x9,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <stack_top@plt>
  mov    %r12,%rdi
  mov    (%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <stack_pop@plt>
  lea    D32(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <stack_top@plt>
  mov    %r12,%rdi
  mov    (%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <stack_destroy@plt>
  xor    %eax,%eax
L12:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L12
```

tool_calls: name: parse_data
arguments: tag: D33
type_infered: string
name: parse_data
arguments: tag: D5
type_infered: string
name: parse_data
arguments: tag: D32
type_infered: string
name: parse_data
arguments: tag: D2
type_infered: string

tool: D33: push '%d' into stack


tool: D5: top of the stack is %d


tool: D32: pop the top item of the stack

tool: D2: creating stack failed

assistant: ```c
int main(void) {
  int val;
  void *instance;
  int *top;
  instance = stack_create();
  if (instance == NULL) {
    printf("creating stack failed\n");
    return -1;
  }
  stack_pop(instance);
  val = 5;
  stack_push(instance, val);
  printf("push '%d' into stack\n", val);
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  val = 9;
  stack_push(instance, val);
  printf("push '%d' into stack\n", val);
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  stack_pop(instance);
  printf("pop the top item of the stack\n");
  top = stack_top(instance);
  printf("top of the stack is %d\n", *top);
  stack_destroy(instance);
  return 0;
}
```