ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x44(%rbp)
  jmp    L20
L96:
  mov    -0x44(%rbp),%eax
  cltq
  movl   $0x0,-0x40(%rbp,%rax,4)
  addl   $0x1,-0x44(%rbp)
L20:
  cmpl   $0xa,-0x44(%rbp)
  jle    L96
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x44(%rbp)
  jmp    L51
L94:
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x4c(%rbp),%eax
  test   %eax,%eax
  jle    L85
  mov    -0x4c(%rbp),%eax
  cmp    $0xa,%eax
  jle    L66
L85:
  mov    -0x4c(%rbp),%eax
  mov    %eax,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L50
L66:
  mov    -0x4c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x40(%rbp,%rax,4)
L50:
  addl   $0x1,-0x44(%rbp)
L51:
  mov    -0x48(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  jle    L94
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x44(%rbp)
  jmp    L3
L75:
  mov    -0x44(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%edx
  mov    -0x44(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x44(%rbp)
L3:
  cmpl   $0xa,-0x44(%rbp)
  jle    L75
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D50
type: string
name: disassemble_data
arguments: tag: D26
type: string
name: disassemble_data
arguments: tag: D35
type: string
name: disassemble_data
arguments: tag: D34
type: string
name: disassemble_data
arguments: tag: D6
type: string
name: disassemble_data
arguments: tag: D41
type: string
name: disassemble_data
arguments: tag: D25
type: string

tool: D50: Cuantas respuestas deseas introducir? 

tool: D26: %i

tool: D35: Introduce tus respuestas

tool: D34: Respuesta incorrecta: %i


tool: D6: 

Calificacion	Numero de Respuesta

tool: D41: ----- ----------------

tool: D25: %4i%14i


assistant: ```c
int main(void) {
  int i, respuesta, contadores[11], contador, total;
  for (i = 1; i <= 10; i++) {
    contadores[i] = 0;
  }
  printf("Cuantas respuestas deseas introducir? ");
  scanf("%i", &total);
  printf("Introduce tus respuestas\n");
  for (i = 1; i <= total; i++) {
    scanf("%i", &respuesta);
    if (respuesta < 1 || respuesta > 10) {
      printf("Respuesta incorrecta: %i\n", respuesta);
    } else {
      contadores[respuesta]++;
    }
  }
  printf("\n\nCalificacion	Numero de Respuesta\n");
  printf("----- ----------------\n");
  for (i = 1; i <= 10; i++) {
    printf("%4i%14i\n", i, contadores[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   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rax
  lea    0x3c(%rsp),%rdx
L6:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L6
  lea    D75(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L77
  mov    $0x1,%ebx
  lea    0x8(%rsp),%r12
  lea    D33(%rip),%rbp
  lea    D23(%rip),%r13
  jmp    L51
L9:
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L7:
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jl     L77
L51:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  lea    -0x1(%rsi),%eax
  cmp    $0x9,%eax
  ja     L9
  movslq %esi,%rsi
  addl   $0x1,0x10(%rsp,%rsi,4)
  jmp    L7
L77:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  lea    D28(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%ebx
  lea    D12(%rip),%rbp
L26:
  mov    0x10(%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0xb,%rbx
  jne    L26
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D75
type_guessed: string
name: disassemble_data
arguments: data_label: D33
type_guessed: string
name: disassemble_data
arguments: data_label: D10
type_guessed: string
name: disassemble_data
arguments: data_label: D23
type_guessed: string
name: disassemble_data
arguments: data_label: D26
type_guessed: string
name: disassemble_data
arguments: data_label: D28
type_guessed: string
name: disassemble_data
arguments: data_label: D12
type_guessed: string

tool: D75: Cuantas respuestas deseas introducir? 

tool: D33: %i

tool: D10: Introduce tus respuestas

tool: D23: Respuesta incorrecta: %i


tool: D26: 

Calificacion	Numero de Respuesta

tool: D28: ----- ----------------

tool: D12: %4i%14i


assistant: ```c
int main(void) {
  int i, respuesta, contadores[11], contador, total;
  for (i = 1; i <= 10; i++) {
    contadores[i] = 0;
  }
  printf("Cuantas respuestas deseas introducir? ");
  scanf("%i", &total);
  printf("Introduce tus respuestas\n");
  for (i = 1; i <= total; i++) {
    scanf("%i", &respuesta);
    if (respuesta < 1 || respuesta > 10) {
      printf("Respuesta incorrecta: %i\n", respuesta);
    } else {
      contadores[respuesta]++;
    }
  }
  printf("\n\nCalificacion	Numero de Respuesta\n");
  printf("----- ----------------\n");
  for (i = 1; i <= 10; i++) {
    printf("%4i%14i\n", i, contadores[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   %r13
  pxor   %xmm0,%xmm0
  lea    D14(%rip),%rdi
  push   %r12
  push   %rbp
  lea    D53(%rip),%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movups %xmm0,0x14(%rsp)
  movups %xmm0,0x24(%rsp)
  movq   $0x0,0x34(%rsp)
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L32
  mov    $0x1,%ebx
  lea    0x8(%rsp),%r12
  lea    D5(%rip),%r13
  jmp    L43
  nopl   0x0(%rax)
L62:
  addl   $0x1,0x10(%rsp,%rsi,4)
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jl     L32
L43:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movslq 0x8(%rsp),%rsi
  lea    -0x1(%rsi),%eax
  cmp    $0x9,%eax
  jbe    L62
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jge    L43
L32:
  lea    D38(%rip),%rdi
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r12
  call   <puts@plt>
  lea    D68(%rip),%rdi
  lea    D32(%rip),%rbp
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L1:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xb,%rbx
  jne    L1
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D53
type: string
name: extract_symbolic_data
arguments: data_tag: D36
type: string
name: extract_symbolic_data
arguments: data_tag: D5
type: string
name: extract_symbolic_data
arguments: data_tag: D38
type: string
name: extract_symbolic_data
arguments: data_tag: D68
type: string
name: extract_symbolic_data
arguments: data_tag: D32
type: string

tool: D14: Cuantas respuestas deseas introducir? 

tool: D53: %i

tool: D36: Introduce tus respuestas

tool: D5: Respuesta incorrecta: %i


tool: D38: 

Calificacion	Numero de Respuesta

tool: D68: ----- ----------------

tool: D32: %4i%14i


assistant: ```c
int main(void) {
  int i, respuesta, contadores[11], contador, total;
  for (i = 1; i <= 10; i++) {
    contadores[i] = 0;
  }
  printf("Cuantas respuestas deseas introducir? ");
  scanf("%i", &total);
  printf("Introduce tus respuestas\n");
  for (i = 1; i <= total; i++) {
    scanf("%i", &respuesta);
    if (respuesta < 1 || respuesta > 10) {
      printf("Respuesta incorrecta: %i\n", respuesta);
    } else {
      contadores[respuesta]++;
    }
  }
  printf("\n\nCalificacion	Numero de Respuesta\n");
  printf("----- ----------------\n");
  for (i = 1; i <= 10; i++) {
    printf("%4i%14i\n", i, contadores[i]);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  lea    D64(%rip),%rdi
  push   %r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movups %xmm0,0x14(%rsp)
  movups %xmm0,0x24(%rsp)
  movq   $0x0,0x34(%rsp)
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L65
  mov    $0x1,%ebx
  lea    0x8(%rsp),%r12
  lea    D67(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movslq 0x8(%rsp),%rsi
  lea    -0x1(%rsi),%eax
  cmp    $0x9,%eax
  ja     L63
  addl   $0x1,0x10(%rsp,%rsi,4)
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jge    L64
L65:
  lea    D34(%rip),%rdi
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r12
  call   <puts@plt>
  lea    D31(%rip),%rdi
  lea    D18(%rip),%rbp
  call   <puts@plt>
  nop
L36:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xb,%rbx
  jne    L36
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L63:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    0xc(%rsp),%ebx
  jle    L64
  jmp    L65
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D64
type: string
name: parse_data
arguments: data_label: D8
type: string
name: parse_data
arguments: data_label: D4
type: string
name: parse_data
arguments: data_label: D67
type: string
name: parse_data
arguments: data_label: D34
type: string
name: parse_data
arguments: data_label: D31
type: string
name: parse_data
arguments: data_label: D18
type: string

tool: D64: Cuantas respuestas deseas introducir? 

tool: D8: %i

tool: D4: Introduce tus respuestas

tool: D67: Respuesta incorrecta: %i


tool: D34: 

Calificacion	Numero de Respuesta

tool: D31: ----- ----------------

tool: D18: %4i%14i


assistant: ```c
int main(void) {
  int i, respuesta, contadores[11], contador, total;
  for (i = 1; i <= 10; i++) {
    contadores[i] = 0;
  }
  printf("Cuantas respuestas deseas introducir? ");
  scanf("%i", &total);
  printf("Introduce tus respuestas\n");
  for (i = 1; i <= total; i++) {
    scanf("%i", &respuesta);
    if (respuesta < 1 || respuesta > 10) {
      printf("Respuesta incorrecta: %i\n", respuesta);
    } else {
      contadores[respuesta]++;
    }
  }
  printf("\n\nCalificacion	Numero de Respuesta\n");
  printf("----- ----------------\n");
  for (i = 1; i <= 10; i++) {
    printf("%4i%14i\n", i, contadores[i]);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    $0x0,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x34(%rbp)
  jmp    L19
L18:
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L19:
  cmpl   $0x9,-0x34(%rbp)
  jle    L18
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D21
type_infered: string
name: extract_symbolic_data
arguments: tag: D11
type_infered: string

tool: D21: %d

tool: D11: N[%d] = %d


assistant: ```c
int main(void) {
  int n[10], i;
  scanf("%d", &n[0]);
  printf("N[%d] = %d\n", 0, n[0]);
  for (i = 1; i < 10; i++) {
    n[i] = n[i - 1] * 2;
    printf("N[%d] = %d\n", i, n[i]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%edx
  mov    $0x0,%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rbp
  mov    $0x1,%ebx
  lea    D25(%rip),%r12
L19:
  mov    -0x4(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    %edx,0x0(%rbp)
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    $0xa,%ebx
  jne    L19
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D11
type_infered: string
name: extract_symbolic_data
arguments: data_label: D25
type_infered: string

tool: D11: %d

tool: D25: N[%d] = %d


assistant: ```c
int main(void) {
  int n[10], i;
  scanf("%d", &n[0]);
  printf("N[%d] = %d\n", 0, n[0]);
  for (i = 1; i < 10; i++) {
    n[i] = n[i - 1] * 2;
    printf("N[%d] = %d\n", i, n[i]);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D18(%rip),%rdi
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rbp
  call   <scanf@plt>
  mov    (%rsp),%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L8:
  mov    -0x4(%rbp),%eax
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  lea    (%rax,%rax,1),%edx
  mov    %edx,-0x4(%rbp)
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L8
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D10
type: string

tool: D18: %d

tool: D10: N[%d] = %d


assistant: ```c
int main(void) {
  int n[10], i;
  scanf("%d", &n[0]);
  printf("N[%d] = %d\n", 0, n[0]);
  for (i = 1; i < 10; i++) {
    n[i] = n[i - 1] * 2;
    printf("N[%d] = %d\n", i, n[i]);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  lea    D18(%rip),%rdi
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rbp
  call   <scanf@plt>
  mov    (%rsp),%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L13:
  mov    -0x4(%rbp),%eax
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  lea    (%rax,%rax,1),%edx
  mov    %edx,-0x4(%rbp)
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L13
  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: parse_data
arguments: label: D18
type: string
name: parse_data
arguments: label: D0
type: string

tool: D18: %d

tool: D0: N[%d] = %d


assistant: ```c
int main(void) {
  int n[10], i;
  scanf("%d", &n[0]);
  printf("N[%d] = %d\n", 0, n[0]);
  for (i = 1; i < 10; i++) {
    n[i] = n[i - 1] * 2;
    printf("N[%d] = %d\n", i, n[i]);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L12
L9:
  lea    D27(%rip),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <print_results@plt>
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L12:
  cmpl   $0x2,-0xc(%rbp)
  jle    L9
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D27
datatype: string
name: disassemble_data
arguments: data_label: D7
datatype: string
name: disassemble_data
arguments: data_label: D13
datatype: string

tool: D27: log1pexp

tool: D7: # c::%s


tool: D13: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "log1pexp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x0,%ebx
  lea    D1(%rip),%r13
  lea    D21(%rip),%r12
  lea    D18(%rip),%rbp
L7:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L7
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
datatype: string
name: unpack_data
arguments: label: D21
datatype: string
name: unpack_data
arguments: label: D18
datatype: string

tool: D1: log1pexp

tool: D21: # c::%s


tool: D18: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "log1pexp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D21(%rip),%r13
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  lea    D12(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L7:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L7
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: data_label: D21
type_infered: string
name: parse_data
arguments: data_label: D6
type_infered: string
name: parse_data
arguments: data_label: D12
type_infered: string

tool: D21: log1pexp

tool: D6: # c::%s


tool: D12: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "log1pexp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D8(%rip),%r13
  push   %r12
  lea    D21(%rip),%r12
  push   %rbp
  lea    D10(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L5:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L5
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
datatype: string
name: disassemble_data
arguments: data_tag: D21
datatype: string
name: disassemble_data
arguments: data_tag: D10
datatype: string

tool: D8: log1pexp

tool: D21: # c::%s


tool: D10: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::%s\n", "log1pexp");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<out_s16le>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  jmp    L2
L17:
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    (%rax),%edx
  mov    -0x28(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L2:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jne    L17
  mov    -0x20(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char *out_s16le(int *src, unsigned char *dst, unsigned n, unsigned shift) {
  while (n--) {
    int val = *src++ >> shift;
    *dst++ = val & 0xFF;
    *dst++ = (val >> 8) & 0xFF;
  }
  return dst;
}
```