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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <calculate_fib@plt>
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D16
type_guessed: string

tool: D16: %d

assistant: ```c
int main() {
  int n = 0, res = 0;
  res = scanf("%d", &n);
  printf("%d", calculate_fib(n));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <calculate_fib@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
data_type: string

tool: D2: %d

assistant: ```c
int main() {
  int n = 0, res = 0;
  res = scanf("%d", &n);
  printf("%d", calculate_fib(n));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  movl   $0x0,0x4(%rsp)
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <calculate_fib@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d

assistant: ```c
int main() {
  int n = 0, res = 0;
  res = scanf("%d", &n);
  printf("%d", calculate_fib(n));
  return 0;
}
```
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
  lea    D4(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  movl   $0x0,0x4(%rsp)
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <calculate_fib@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type_infered: string

tool: D4: %d

assistant: ```c
int main() {
  int n = 0, res = 0;
  res = scanf("%d", &n);
  printf("%d", calculate_fib(n));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type: string

tool: D0: 
 Testando git...

assistant: ```c
int main() {
  printf("\n Testando git...");
  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
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
datatype: string

tool: D7: 
 Testando git...

assistant: ```c
int main() {
  printf("\n Testando git...");
  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
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D3
datatype: string

tool: D3: 
 Testando git...

assistant: ```c
int main() {
  printf("\n Testando git...");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_infered: string

tool: D3: 
 Testando git...

assistant: ```c
int main() {
  printf("\n Testando git...");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_endian_swap_s1@plt>
  mov    $0x0,%eax
  call   <test_s2@plt>
  mov    $0x0,%eax
  call   <test_s3@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  test_endian_swap_s1();
  test_s2();
  test_s3();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_endian_swap_s1@plt>
  mov    $0x0,%eax
  call   <test_s2@plt>
  mov    $0x0,%eax
  call   <test_s3@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  test_endian_swap_s1();
  test_s2();
  test_s3();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_endian_swap_s1@plt>
  xor    %eax,%eax
  call   <test_s2@plt>
  xor    %eax,%eax
  call   <test_s3@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  test_endian_swap_s1();
  test_s2();
  test_s3();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_endian_swap_s1@plt>
  xor    %eax,%eax
  call   <test_s2@plt>
  xor    %eax,%eax
  call   <test_s3@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  test_endian_swap_s1();
  test_s2();
  test_s3();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L135
L70:
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  cmp    $0x5,%eax
  jne    L36
  addl   $0x1,-0x14(%rbp)
  jmp    L0
L36:
  addl   $0x1,-0x10(%rbp)
L0:
  addl   $0x1,-0x18(%rbp)
L135:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L70
  movl   $0x0,-0xc(%rbp)
L97:
  cmpl   $0x8,-0x14(%rbp)
  jle    L109
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  movslq %edx,%rax
  imul   $0x38e38e39,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L52
  movl   $0x1,-0xc(%rbp)
  jmp    L124
L52:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x38e38e39,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    %eax,-0x14(%rbp)
  jmp    L97
L109:
  nop
L124:
  cmpl   $0x0,-0xc(%rbp)
  je     L66
  cmpl   $0x0,-0x10(%rbp)
  jle    L66
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_answer@plt>
  mov    $0x0,%eax
  jmp    L138
L66:
  cmpl   $0x0,-0x10(%rbp)
  jne    L134
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L105
L134:
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L105:
  mov    $0x0,%eax
L138:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L122
  call   <__stack_chk_fail@plt>
L122:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D10
datatype: string
name: parse_data
arguments: tag: D2
datatype: string
name: parse_data
arguments: tag: D29
datatype: string

tool: D10: %d

tool: D2: -1


tool: D29: 0


assistant: ```c
int main() {
  int i, n;
  scanf("%d", &n);
  int count_5 = 0;
  int count_0 = 0;
  for (i = 0; i < n; i++) {
    int num;
    scanf("%d", &num);
    if (num == 5)
      count_5++;
    else
      count_0++;
  }
  int find_answer = 0;
  while (1) {
    if (count_5 < 9)
      break;
    if (count_5 * 5 % 9 == 0) {
      find_answer = 1;
      break;
    }
    count_5 = count_5 - count_5 % 9;
  }
  if (find_answer && count_0 > 0) {
    print_answer(count_5, count_0);
    return 0;
  }
  if (count_0 == 0)
    printf("-1\n");
  else
    printf("0\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,(%rsp)
  jle    L110
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  lea    D3(%rip),%r13
  jmp    L6
L131:
  add    $0x1,%ebp
L44:
  add    $0x1,%r12d
  cmp    %r12d,(%rsp)
  jle    L104
L6:
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  jne    L131
  add    $0x1,%ebx
  jmp    L44
L104:
  cmp    $0x8,%ebx
  jle    L55
  lea    (%rbx,%rbx,4),%edx
  movslq %edx,%rax
  imul   $0x38e38e39,%rax,%rax
  sar    $0x21,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,8),%eax
  cmp    %eax,%edx
  je     L60
L1:
  movslq %ebx,%rax
  imul   $0x38e38e39,%rax,%rax
  sar    $0x21,%rax
  sar    $0x1f,%ebx
  sub    %ebx,%eax
  lea    (%rax,%rax,8),%ebx
  cmp    $0x8,%ebx
  jle    L55
  lea    (%rbx,%rbx,4),%edx
  movslq %edx,%rax
  imul   $0x38e38e39,%rax,%rax
  sar    $0x21,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,8),%eax
  cmp    %eax,%edx
  jne    L1
L60:
  test   %ebp,%ebp
  jg     L70
L55:
  test   %ebp,%ebp
  je     L110
  lea    D29(%rip),%rdi
  call   <printf@plt>
  jmp    L66
L70:
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <print_answer@plt>
  jmp    L66
L110:
  lea    D9(%rip),%rdi
  call   <printf@plt>
L66:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
data_type: string
name: disassemble_data
arguments: label: D29
data_type: string
name: disassemble_data
arguments: label: D9
data_type: string

tool: D3: %d

tool: D29: 0


tool: D9: -1


assistant: ```c
int main() {
  int i, n;
  scanf("%d", &n);
  int count_5 = 0;
  int count_0 = 0;
  for (i = 0; i < n; i++) {
    int num;
    scanf("%d", &num);
    if (num == 5)
      count_5++;
    else
      count_0++;
  }
  int find_answer = 0;
  while (1) {
    if (count_5 < 9)
      break;
    if (count_5 * 5 % 9 == 0) {
      find_answer = 1;
      break;
    }
    count_5 = count_5 - count_5 % 9;
  }
  if (find_answer && count_0 > 0) {
    print_answer(count_5, count_0);
    return 0;
  }
  if (count_0 == 0)
    printf("-1\n");
  else
    printf("0\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D26(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L117
  xor    %r12d,%r12d
  xor    %ebp,%ebp
  lea    0x4(%rsp),%r14
  xor    %ebx,%ebx
  jmp    L52
  nopw   0x0(%rax,%rax,1)
L114:
  add    $0x1,%r12d
L56:
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jle    L101
L52:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  jne    L114
  add    $0x1,%ebp
  jmp    L56
  nopl   0x0(%rax,%rax,1)
L63:
  imul   $0x38e38e39,%rbp,%rbp
  shr    $0x21,%rbp
  lea    0x0(%rbp,%rbp,8),%ebp
L101:
  cmp    $0x8,%ebp
  jle    L80
  imul   $0x1c71c71d,%ebp,%eax
  cmp    $0x1c71c71c,%eax
  ja     L63
  test   %r12d,%r12d
  jne    L77
L117:
  lea    D18(%rip),%rdi
  call   <printf@plt>
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L92
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L80:
  test   %r12d,%r12d
  je     L117
  lea    D8(%rip),%rdi
  call   <printf@plt>
  jmp    L41
L77:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <print_answer@plt>
  jmp    L41
L92:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D26
type_infered: string
name: extract_symbolic_data
arguments: tag: D18
type_infered: string
name: extract_symbolic_data
arguments: tag: D8
type_infered: string

tool: D26: %d

tool: D18: -1


tool: D8: 0


assistant: ```c
int main() {
  int i, n;
  scanf("%d", &n);
  int count_5 = 0;
  int count_0 = 0;
  for (i = 0; i < n; i++) {
    int num;
    scanf("%d", &num);
    if (num == 5)
      count_5++;
    else
      count_0++;
  }
  int find_answer = 0;
  while (1) {
    if (count_5 < 9)
      break;
    if (count_5 * 5 % 9 == 0) {
      find_answer = 1;
      break;
    }
    count_5 = count_5 - count_5 % 9;
  }
  if (find_answer && count_0 > 0) {
    print_answer(count_5, count_0);
    return 0;
  }
  if (count_0 == 0)
    printf("-1\n");
  else
    printf("0\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L114
  xor    %r12d,%r12d
  xor    %ebp,%ebp
  lea    0x4(%rsp),%r14
  xor    %ebx,%ebx
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L34:
  add    $0x1,%r12d
L65:
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jle    L49
L18:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x5,0x4(%rsp)
  jne    L34
  add    $0x1,%ebp
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L80:
  imul   $0x38e38e39,%rbp,%rbp
  shr    $0x21,%rbp
  lea    0x0(%rbp,%rbp,8),%ebp
L49:
  cmp    $0x8,%ebp
  jle    L118
  imul   $0x1c71c71d,%ebp,%eax
  cmp    $0x1c71c71c,%eax
  ja     L80
  test   %r12d,%r12d
  jne    L22
L114:
  lea    D6(%rip),%rdi
  call   <printf@plt>
L117:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L56
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L118:
  test   %r12d,%r12d
  je     L114
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L117
L22:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <print_answer@plt>
  jmp    L117
L56:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D29
type_infered: string
name: retrieve_data
arguments: data_tag: D6
type_infered: string
name: retrieve_data
arguments: data_tag: D12
type_infered: string

tool: D29: %d

tool: D6: -1


tool: D12: 0


assistant: ```c
int main() {
  int i, n;
  scanf("%d", &n);
  int count_5 = 0;
  int count_0 = 0;
  for (i = 0; i < n; i++) {
    int num;
    scanf("%d", &num);
    if (num == 5)
      count_5++;
    else
      count_0++;
  }
  int find_answer = 0;
  while (1) {
    if (count_5 < 9)
      break;
    if (count_5 * 5 % 9 == 0) {
      find_answer = 1;
      break;
    }
    count_5 = count_5 - count_5 % 9;
  }
  if (find_answer && count_0 > 0) {
    print_answer(count_5, count_0);
    return 0;
  }
  if (count_0 == 0)
    printf("-1\n");
  else
    printf("0\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_escape>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,%rax
  mov    %rsi,%rdx
  jmp    L10
L53:
  movzbl (%rdx),%ecx
  cmp    $0x5c,%cl
  jne    L11
  add    $0x1,%rdx
  movzbl (%rdx),%ecx
  movsbl %cl,%ecx
  cmp    $0x30,%ecx
  je     L7
  cmp    $0x30,%ecx
  jl     L28
  cmp    $0x76,%ecx
  jg     L28
  cmp    $0x62,%ecx
  jl     L28
  sub    $0x62,%ecx
  cmp    $0x14,%ecx
  ja     L28
  mov    %ecx,%ecx
  lea    0x0(,%rcx,4),%rsi
  lea    D17(%rip),%rcx
  mov    (%rsi,%rcx,1),%ecx
  movslq %ecx,%rcx
  lea    D17(%rip),%rsi
  add    %rsi,%rcx
  notrack jmp *%rcx
  movb   $0xa,(%rax)
  jmp    L92
  movb   $0x9,(%rax)
  jmp    L92
  movb   $0x8,(%rax)
  jmp    L92
  movb   $0xd,(%rax)
  jmp    L92
  movb   $0xc,(%rax)
  jmp    L92
  movb   $0xb,(%rax)
  jmp    L92
L7:
  movb   $0x0,(%rax)
  jmp    L92
L28:
  movzbl (%rdx),%ecx
  mov    %cl,(%rax)
  jmp    L92
L11:
  movzbl (%rdx),%ecx
  cmp    $0xa,%cl
  jne    L50
  movb   $0x0,(%rax)
  jmp    L92
L50:
  movzbl (%rdx),%ecx
  mov    %cl,(%rax)
L92:
  add    $0x1,%rdx
  add    $0x1,%rax
L10:
  movzbl (%rdx),%ecx
  test   %cl,%cl
  jne    L53
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void _escape(op, ip) register char *op, *ip;
{
  for (; *ip != 0; ip++, op++) {
    if (*ip == '\\') {
      switch (*++ip) {
      case 'n':
        *op = '\n';
        break;
      case 't':
        *op = '\t';
        break;
      case 'b':
        *op = '\b';
        break;
      case 'r':
        *op = '\r';
        break;
      case 'f':
        *op = '\f';
        break;
      case 'v':
        *op = '\v';
        break;
      case '0':
        *op = 0;
        break;
      default:
        *op = *ip;
        break;
      }
    } else {
      if (*ip == '\n') {
        *op = '\0';
      } else {
        *op = *ip;
      }
    }
  }
  *op = 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_escape>:
  endbr64
  mov    %rdi,%rdx
  movzbl (%rsi),%eax
  test   %al,%al
  je     L108
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%r8d
  lea    D2(%rip),%rdi
  mov    $0x9,%ebp
  mov    $0xb,%ebx
  mov    $0xc,%r11d
  mov    $0xd,%r10d
  mov    $0x8,%r9d
  mov    $0xa,%r12d
  jmp    L100
L117:
  lea    0x1(%rsi),%rcx
  movzbl 0x1(%rsi),%eax
  cmp    $0x30,%al
  je     L63
  lea    -0x62(%rax),%esi
  cmp    $0x14,%sil
  ja     L87
  ja     L87
  movzbl %sil,%esi
  movslq (%rdi,%rsi,4),%rsi
  add    %rdi,%rsi
  notrack jmp *%rsi
  mov    %r12d,%eax
  jmp    L87
  mov    %r9d,%eax
L87:
  mov    %al,(%rdx)
  lea    0x1(%rcx),%rsi
  add    $0x1,%rdx
  movzbl 0x1(%rcx),%eax
  test   %al,%al
  je     L71
L100:
  cmp    $0x5c,%al
  je     L117
  mov    %rsi,%rcx
  cmp    $0xa,%al
  jne    L87
  mov    %r8d,%eax
  jmp    L87
  mov    %r10d,%eax
  jmp    L87
  mov    %r11d,%eax
  jmp    L87
  mov    %ebx,%eax
  jmp    L87
L63:
  mov    %r8d,%eax
  jmp    L87
  mov    %ebp,%eax
  jmp    L87
L71:
  movb   $0x0,(%rdx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L108:
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void _escape(op, ip) register char *op, *ip;
{
  for (; *ip != 0; ip++, op++) {
    if (*ip == '\\') {
      switch (*++ip) {
      case 'n':
        *op = '\n';
        break;
      case 't':
        *op = '\t';
        break;
      case 'b':
        *op = '\b';
        break;
      case 'r':
        *op = '\r';
        break;
      case 'f':
        *op = '\f';
        break;
      case 'v':
        *op = '\v';
        break;
      case '0':
        *op = 0;
        break;
      default:
        *op = *ip;
        break;
      }
    } else {
      if (*ip == '\n') {
        *op = '\0';
      } else {
        *op = *ip;
      }
    }
  }
  *op = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_escape>:
  endbr64
  movzbl (%rsi),%eax
  test   %al,%al
  je     L81
  lea    D1(%rip),%r8
  jmp    L13
  nopl   0x0(%rax)
L98:
  mov    %rsi,%rdx
  cmp    $0xa,%al
  je     L47
L50:
  mov    %al,(%rdi)
  movzbl 0x1(%rdx),%eax
  lea    0x1(%rdx),%rsi
  add    $0x1,%rdi
  test   %al,%al
  je     L81
L13:
  lea    0x1(%rsi),%rdx
  cmp    $0x5c,%al
  jne    L98
  movzbl 0x1(%rsi),%eax
  cmp    $0x30,%al
  je     L47
  lea    -0x62(%rax),%ecx
  cmp    $0x14,%cl
  ja     L50
  movzbl %cl,%ecx
  movslq (%r8,%rcx,4),%rcx
  add    %r8,%rcx
  notrack jmp *%rcx
  cs nopw 0x0(%rax,%rax,1)
L47:
  xor    %eax,%eax
  jmp    L50
  nopl   0x0(%rax)
L81:
  movb   $0x0,(%rdi)
  ret
  nopl   0x0(%rax)
  mov    $0xb,%eax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
  mov    $0xd,%eax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
  mov    $0xc,%eax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
  mov    $0x8,%eax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
  mov    $0xa,%eax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
  mov    $0x9,%eax
  jmp    L50
```

assistant: ```c
void _escape(op, ip) register char *op, *ip;
{
  for (; *ip != 0; ip++, op++) {
    if (*ip == '\\') {
      switch (*++ip) {
      case 'n':
        *op = '\n';
        break;
      case 't':
        *op = '\t';
        break;
      case 'b':
        *op = '\b';
        break;
      case 'r':
        *op = '\r';
        break;
      case 'f':
        *op = '\f';
        break;
      case 'v':
        *op = '\v';
        break;
      case '0':
        *op = 0;
        break;
      default:
        *op = *ip;
        break;
      }
    } else {
      if (*ip == '\n') {
        *op = '\0';
      } else {
        *op = *ip;
      }
    }
  }
  *op = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_escape>:
  endbr64
  movzbl (%rsi),%eax
  test   %al,%al
  je     L13
  lea    D8(%rip),%r8
  jmp    L9
  nopl   0x0(%rax)
L22:
  mov    %rsi,%rdx
  cmp    $0xa,%al
  je     L129
L33:
  mov    %al,(%rdi)
  movzbl 0x1(%rdx),%eax
  lea    0x1(%rdx),%rsi
  add    $0x1,%rdi
  test   %al,%al
  je     L13
L9:
  lea    0x1(%rsi),%rdx
  cmp    $0x5c,%al
  jne    L22
  movzbl 0x1(%rsi),%eax
  cmp    $0x30,%al
  je     L129
  lea    -0x62(%rax),%ecx
  cmp    $0x14,%cl
  ja     L33
  movzbl %cl,%ecx
  movslq (%r8,%rcx,4),%rcx
  add    %r8,%rcx
  notrack jmp *%rcx
  cs nopw 0x0(%rax,%rax,1)
L129:
  xor    %eax,%eax
  jmp    L33
  nopl   0x0(%rax)
L13:
  movb   $0x0,(%rdi)
  ret
  nopl   0x0(%rax)
  mov    $0xb,%eax
  jmp    L33
  nopw   0x0(%rax,%rax,1)
  mov    $0xd,%eax
  jmp    L33
  nopw   0x0(%rax,%rax,1)
  mov    $0xc,%eax
  jmp    L33
  nopw   0x0(%rax,%rax,1)
  mov    $0x8,%eax
  jmp    L33
  nopw   0x0(%rax,%rax,1)
  mov    $0xa,%eax
  jmp    L33
  nopw   0x0(%rax,%rax,1)
  mov    $0x9,%eax
  jmp    L33
```

assistant: ```c
void _escape(op, ip) register char *op, *ip;
{
  for (; *ip != 0; ip++, op++) {
    if (*ip == '\\') {
      switch (*++ip) {
      case 'n':
        *op = '\n';
        break;
      case 't':
        *op = '\t';
        break;
      case 'b':
        *op = '\b';
        break;
      case 'r':
        *op = '\r';
        break;
      case 'f':
        *op = '\f';
        break;
      case 'v':
        *op = '\v';
        break;
      case '0':
        *op = 0;
        break;
      default:
        *op = *ip;
        break;
      }
    } else {
      if (*ip == '\n') {
        *op = '\0';
      } else {
        *op = *ip;
      }
    }
  }
  *op = 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_init_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L19
  mov    $0xffffffff,%eax
  jmp    L6
L19:
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
int ft_init_line(char **line) {
  *line = (char *)malloc(sizeof(char) * 1);
  if (!*line)
    return (-1);
  return (0);
}
```