ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PrintMenu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D107(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D31
type: string
name: parse_data
arguments: data_tag: D49
type: string
name: parse_data
arguments: data_tag: D42
type: string
name: parse_data
arguments: data_tag: D8
type: string
name: parse_data
arguments: data_tag: D58
type: string
name: parse_data
arguments: data_tag: D106
type: string
name: parse_data
arguments: data_tag: D107
type: string
name: parse_data
arguments: data_tag: D5
type: string
name: parse_data
arguments: data_tag: D23
type: string
name: parse_data
arguments: data_tag: D43
type: string
name: parse_data
arguments: data_tag: D89
type: string

tool: D31: WHAT DO YOU WANT TO DO?



tool: D49: 	1. Expand ROM 0.5 MB


tool: D42: 	2. Expand ROM 1.0 MB


tool: D8: 	3. Expand ROM 1.5 MB


tool: D58: 	4. Expand ROM 2.0 MB


tool: D106: 	5. Expand ROM 2.5 MB


tool: D107: 	6. Expand ROM 3.0 MB


tool: D5: 	7. Custom ROM Expansion


tool: D23: 	8. Quit



tool: D43: WHAT IS YOUR CHOICE? 

tool: D89: %d

assistant: ```c
int PrintMenu() {
  int choice;
  printf("WHAT DO YOU WANT TO DO?\n\n");
  printf("\t1. Expand ROM 0.5 MB\n");
  printf("\t2. Expand ROM 1.0 MB\n");
  printf("\t3. Expand ROM 1.5 MB\n");
  printf("\t4. Expand ROM 2.0 MB\n");
  printf("\t5. Expand ROM 2.5 MB\n");
  printf("\t6. Expand ROM 3.0 MB\n");
  printf("\t7. Custom ROM Expansion\n");
  printf("\t8. Quit\n\n");
  printf("WHAT IS YOUR CHOICE? ");
  scanf("%d", &choice);
  return (choice);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PrintMenu>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    D88(%rip),%rdi
  call   <printf@plt>
  lea    D86(%rip),%rdi
  call   <printf@plt>
  lea    D45(%rip),%rdi
  call   <printf@plt>
  lea    D93(%rip),%rdi
  call   <printf@plt>
  lea    D89(%rip),%rdi
  call   <printf@plt>
  lea    D103(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    D51(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D68(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_guessed: string
name: disassemble_data
arguments: tag: D38
type_guessed: string
name: disassemble_data
arguments: tag: D88
type_guessed: string
name: disassemble_data
arguments: tag: D86
type_guessed: string
name: disassemble_data
arguments: tag: D45
type_guessed: string
name: disassemble_data
arguments: tag: D93
type_guessed: string
name: disassemble_data
arguments: tag: D89
type_guessed: string
name: disassemble_data
arguments: tag: D103
type_guessed: string
name: disassemble_data
arguments: tag: D27
type_guessed: string
name: disassemble_data
arguments: tag: D51
type_guessed: string
name: disassemble_data
arguments: tag: D68
type_guessed: string

tool: D4: WHAT DO YOU WANT TO DO?



tool: D38: 	1. Expand ROM 0.5 MB


tool: D88: 	2. Expand ROM 1.0 MB


tool: D86: 	3. Expand ROM 1.5 MB


tool: D45: 	4. Expand ROM 2.0 MB


tool: D93: 	5. Expand ROM 2.5 MB


tool: D89: 	6. Expand ROM 3.0 MB


tool: D103: 	7. Custom ROM Expansion


tool: D27: 	8. Quit



tool: D51: WHAT IS YOUR CHOICE? 

tool: D68: %d

assistant: ```c
int PrintMenu() {
  int choice;
  printf("WHAT DO YOU WANT TO DO?\n\n");
  printf("\t1. Expand ROM 0.5 MB\n");
  printf("\t2. Expand ROM 1.0 MB\n");
  printf("\t3. Expand ROM 1.5 MB\n");
  printf("\t4. Expand ROM 2.0 MB\n");
  printf("\t5. Expand ROM 2.5 MB\n");
  printf("\t6. Expand ROM 3.0 MB\n");
  printf("\t7. Custom ROM Expansion\n");
  printf("\t8. Quit\n\n");
  printf("WHAT IS YOUR CHOICE? ");
  scanf("%d", &choice);
  return (choice);
}
```
user: Translate the below assembly code into C programming language:

```asm
<PrintMenu>:
  endbr64
  sub    $0x18,%rsp
  lea    D77(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D87(%rip),%rdi
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <printf@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    D63(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D66(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    D76(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D65(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D77
type_guessed: string
name: parse_data
arguments: label: D87
type_guessed: string
name: parse_data
arguments: label: D50
type_guessed: string
name: parse_data
arguments: label: D7
type_guessed: string
name: parse_data
arguments: label: D27
type_guessed: string
name: parse_data
arguments: label: D63
type_guessed: string
name: parse_data
arguments: label: D0
type_guessed: string
name: parse_data
arguments: label: D66
type_guessed: string
name: parse_data
arguments: label: D36
type_guessed: string
name: parse_data
arguments: label: D76
type_guessed: string
name: parse_data
arguments: label: D65
type_guessed: string

tool: D77: WHAT DO YOU WANT TO DO?



tool: D87: 	1. Expand ROM 0.5 MB


tool: D50: 	2. Expand ROM 1.0 MB


tool: D7: 	3. Expand ROM 1.5 MB


tool: D27: 	4. Expand ROM 2.0 MB


tool: D63: 	5. Expand ROM 2.5 MB


tool: D0: 	6. Expand ROM 3.0 MB


tool: D66: 	7. Custom ROM Expansion


tool: D36: 	8. Quit



tool: D76: WHAT IS YOUR CHOICE? 

tool: D65: %d

assistant: ```c
int PrintMenu() {
  int choice;
  printf("WHAT DO YOU WANT TO DO?\n\n");
  printf("\t1. Expand ROM 0.5 MB\n");
  printf("\t2. Expand ROM 1.0 MB\n");
  printf("\t3. Expand ROM 1.5 MB\n");
  printf("\t4. Expand ROM 2.0 MB\n");
  printf("\t5. Expand ROM 2.5 MB\n");
  printf("\t6. Expand ROM 3.0 MB\n");
  printf("\t7. Custom ROM Expansion\n");
  printf("\t8. Quit\n\n");
  printf("WHAT IS YOUR CHOICE? ");
  scanf("%d", &choice);
  return (choice);
}
```
user: What is the c source code of the assembly code below:

```asm
<PrintMenu>:
  endbr64
  sub    $0x18,%rsp
  lea    D10(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D64(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D67(%rip),%rdi
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  lea    D55(%rip),%rdi
  call   <printf@plt>
  lea    D100(%rip),%rdi
  call   <printf@plt>
  lea    D51(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D56(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D10
type_infered: string
name: parse_data
arguments: data_tag: D64
type_infered: string
name: parse_data
arguments: data_tag: D42
type_infered: string
name: parse_data
arguments: data_tag: D67
type_infered: string
name: parse_data
arguments: data_tag: D52
type_infered: string
name: parse_data
arguments: data_tag: D55
type_infered: string
name: parse_data
arguments: data_tag: D100
type_infered: string
name: parse_data
arguments: data_tag: D51
type_infered: string
name: parse_data
arguments: data_tag: D35
type_infered: string
name: parse_data
arguments: data_tag: D16
type_infered: string
name: parse_data
arguments: data_tag: D56
type_infered: string

tool: D10: WHAT DO YOU WANT TO DO?



tool: D64: 	1. Expand ROM 0.5 MB


tool: D42: 	2. Expand ROM 1.0 MB


tool: D67: 	3. Expand ROM 1.5 MB


tool: D52: 	4. Expand ROM 2.0 MB


tool: D55: 	5. Expand ROM 2.5 MB


tool: D100: 	6. Expand ROM 3.0 MB


tool: D51: 	7. Custom ROM Expansion


tool: D35: 	8. Quit



tool: D16: WHAT IS YOUR CHOICE? 

tool: D56: %d

assistant: ```c
int PrintMenu() {
  int choice;
  printf("WHAT DO YOU WANT TO DO?\n\n");
  printf("\t1. Expand ROM 0.5 MB\n");
  printf("\t2. Expand ROM 1.0 MB\n");
  printf("\t3. Expand ROM 1.5 MB\n");
  printf("\t4. Expand ROM 2.0 MB\n");
  printf("\t5. Expand ROM 2.5 MB\n");
  printf("\t6. Expand ROM 3.0 MB\n");
  printf("\t7. Custom ROM Expansion\n");
  printf("\t8. Quit\n\n");
  printf("WHAT IS YOUR CHOICE? ");
  scanf("%d", &choice);
  return (choice);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fib>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L19
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fib>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fib>
  add    %ebx,%eax
  jmp    L0
L19:
  mov    -0x14(%rbp),%eax
L0:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fib(int n) {
  if (n > 1) {
    return fib(n - 1) + fib(n - 2);
  }
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fib>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  cmp    $0x1,%edi
  jg     L16
L17:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  lea    -0x1(%rdi),%edi
  call   <fib>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fib>
  lea    0x0(%rbp,%rax,1),%ebx
  jmp    L17
```

assistant: ```c
int fib(int n) {
  if (n > 1) {
    return fib(n - 1) + fib(n - 2);
  }
  return n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edx
  mov    $0x342,%esi
  mov    $0x340,%edi
  call   <ioperm@plt>
  mov    $0x0,%esi
  mov    $0x340,%edi
  call   <outb@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  ioperm(0x340, 0x342, 1);
  outb(0x340, 0x0);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x342,%esi
  mov    $0x340,%edi
  call   <ioperm@plt>
  mov    $0x0,%esi
  mov    $0x340,%edi
  call   <outb@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  ioperm(0x340, 0x342, 1);
  outb(0x340, 0x0);
  exit(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x342,%esi
  mov    $0x340,%edi
  call   <ioperm@plt>
  xor    %esi,%esi
  mov    $0x340,%edi
  call   <outb@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  ioperm(0x340, 0x342, 1);
  outb(0x340, 0x0);
  exit(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
  mov    $0x1,%edx
  mov    $0x342,%esi
  mov    $0x340,%edi
  call   <ioperm@plt>
  xor    %esi,%esi
  mov    $0x340,%edi
  call   <outb@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  ioperm(0x340, 0x342, 1);
  outb(0x340, 0x0);
  exit(0);
}
```
user: What is the c source code of the assembly code below:

```asm
<fibonacci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  je     L27
  cmpl   $0x2,-0x14(%rbp)
  jne    L2
L27:
  mov    $0x1,%eax
  jmp    L13
L2:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fibonacci>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fibonacci>
  add    %ebx,%eax
L13:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fibonacci(int n) {
  if (n == 1 || n == 2) {
    return 1;
  } else {
    return fibonacci(n - 1) + fibonacci(n - 2);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fibonacci>:
  endbr64
  lea    -0x1(%rdi),%edx
  mov    $0x1,%eax
  cmp    $0x1,%edx
  ja     L2
  ret
L2:
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %edx,%edi
  call   <fibonacci>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fibonacci>
  add    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fibonacci(int n) {
  if (n == 1 || n == 2) {
    return 1;
  } else {
    return fibonacci(n - 1) + fibonacci(n - 2);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<m>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L2
L15:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rbx
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <m>
  mov    %rax,%rdi
  call   <f@plt>
  mov    %rax,%rdx
  mov    %rbx,%rax
  sub    %rdx,%rax
L2:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long long m(int n) {
  if (n == 0) {
    return 0;
  }
  return n - f(m(n - 1));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<m>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  jne    L3
  ret
L3:
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <m>
  mov    %rax,%rdi
  call   <f@plt>
  mov    %rax,%rdx
  movslq %ebx,%rax
  sub    %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
long long m(int n) {
  if (n == 0) {
    return 0;
  }
  return n - f(m(n - 1));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<m>:
  endbr64
  test   %edi,%edi
  jne    L19
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  push   %r13
  mov    %edi,%eax
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  movslq %edi,%rbx
  sub    $0x8,%rsp
  sub    $0x1,%eax
  je     L6
  movslq %eax,%rbp
  mov    %edi,%eax
  xor    %r12d,%r12d
  sub    $0x2,%eax
  jne    L12
L41:
  mov    %r12,%rdi
  call   <f@plt>
  sub    %rax,%rbp
L6:
  mov    %rbp,%rdi
  call   <f@plt>
  add    $0x8,%rsp
  mov    %rax,%r8
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  sub    %r8,%rax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %edi,%r13d
  movslq %eax,%r12
  xor    %r8d,%r8d
  sub    $0x3,%r13d
  je     L24
  sub    $0x4,%edi
  call   <m>
  mov    %rax,%rdi
  call   <f@plt>
  movslq %r13d,%r8
  sub    %rax,%r8
L24:
  mov    %r8,%rdi
  call   <f@plt>
  sub    %rax,%r12
  jmp    L41
```

assistant: ```c
long long m(int n) {
  if (n == 0) {
    return 0;
  }
  return n - f(m(n - 1));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<m>:
  endbr64
  test   %edi,%edi
  jne    L43
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  push   %r13
  mov    %edi,%eax
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  movslq %edi,%rbx
  sub    $0x8,%rsp
  sub    $0x1,%eax
  je     L10
  movslq %eax,%rbp
  mov    %edi,%eax
  xor    %r12d,%r12d
  sub    $0x2,%eax
  jne    L14
L4:
  mov    %r12,%rdi
  call   <f@plt>
  sub    %rax,%rbp
L10:
  mov    %rbp,%rdi
  call   <f@plt>
  add    $0x8,%rsp
  mov    %rax,%r8
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  sub    %r8,%rax
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %edi,%r13d
  movslq %eax,%r12
  xor    %r8d,%r8d
  sub    $0x3,%r13d
  je     L39
  sub    $0x4,%edi
  call   <m>
  mov    %rax,%rdi
  call   <f@plt>
  movslq %r13d,%r8
  sub    %rax,%r8
L39:
  mov    %r8,%rdi
  call   <f@plt>
  sub    %rax,%r12
  jmp    L4
```

assistant: ```c
long long m(int n) {
  if (n == 0) {
    return 0;
  }
  return n - f(m(n - 1));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%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,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L69
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  negl   -0x14(%rbp)
  jmp    L69
L63:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%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,-0x14(%rbp)
L69:
  cmpl   $0x0,-0x14(%rbp)
  jg     L63
  cmpl   $0x0,-0x4(%rbp)
  jns    L5
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
L5:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jge    L4
  jmp    L42
L44:
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x7e,(%rax)
L42:
  mov    -0x18(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L44
L4:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse@plt>
  nop
  leave
  ret
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i = 0;
  int sign = n % 10;
  s[i++] = (sign < 0 ? -sign : sign) + '0';
  n /= 10;
  if (n < 0) {
    sign = n;
    n = -n;
  }
  while (n > 0) {
    s[i++] = n % 10 + '0';
    n /= 10;
  }
  if (sign < 0) {
    s[i++] = '-';
  }
  if (i < w) {
    while (w > i) {
      s[i++] = '~';
    }
  }
  s[i] = '\0';
  reverse(s);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<itoa>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    %rsi,%r8
  mov    %edx,%r10d
  movslq %edi,%r11
  imul   $0x66666667,%r11,%r11
  sar    $0x22,%r11
  cltd
  sub    %edx,%r11d
  mov    %r11d,%ecx
  lea    (%r11,%r11,4),%edx
  add    %edx,%edx
  mov    %edi,%r11d
  sub    %edx,%r11d
  mov    %r11d,%edx
  neg    %edx
  cmovs  %r11d,%edx
  add    $0x30,%edx
  mov    %dl,(%rsi)
  mov    %ecx,%edx
  cmp    $0xfffffff7,%edi
  jge    L38
  mov    %ecx,%r11d
  neg    %ecx
  mov    %ecx,%edx
L38:
  mov    $0x2,%ecx
  test   %edx,%edx
  jle    L76
L16:
  mov    %ecx,%r9d
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  mov    %edx,%esi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,-0x1(%r8,%rcx,1)
  mov    %edx,%esi
  mov    %eax,%edx
  add    $0x1,%rcx
  cmp    $0x9,%esi
  jg     L16
L12:
  test   %r11d,%r11d
  js     L60
L64:
  cmp    %r10d,%r9d
  jge    L20
  movslq %r9d,%rdx
  lea    (%r8,%rdx,1),%rax
  lea    0x1(%r8,%rdx,1),%rdx
  lea    -0x1(%r10),%ecx
  sub    %r9d,%ecx
  add    %rcx,%rdx
L18:
  movb   $0x7e,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L18
  mov    %r10d,%r9d
L20:
  movslq %r9d,%r9
  movb   $0x0,(%r8,%r9,1)
  mov    %r8,%rdi
  call   <reverse@plt>
  add    $0x8,%rsp
  ret
L76:
  mov    $0x1,%r9d
  jmp    L12
L60:
  movslq %r9d,%rax
  movb   $0x2d,(%r8,%rax,1)
  lea    0x1(%r9),%r9d
  jmp    L64
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i = 0;
  int sign = n % 10;
  s[i++] = (sign < 0 ? -sign : sign) + '0';
  n /= 10;
  if (n < 0) {
    sign = n;
    n = -n;
  }
  while (n > 0) {
    s[i++] = n % 10 + '0';
    n /= 10;
  }
  if (sign < 0) {
    s[i++] = '-';
  }
  if (i < w) {
    while (w > i) {
      s[i++] = '~';
    }
  }
  s[i] = '\0';
  reverse(s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<itoa>:
  endbr64
  movslq %edi,%r9
  push   %rbp
  mov    %rsi,%rbp
  mov    %r9,%rax
  imul   $0x66666667,%r9,%r9
  push   %rbx
  mov    %edx,%ebx
  cltd
  sar    $0x22,%r9
  sub    $0x8,%rsp
  sub    %edx,%r9d
  lea    (%r9,%r9,4),%ecx
  mov    %r9d,%edx
  mov    %eax,%r9d
  add    %ecx,%ecx
  sub    %ecx,%r9d
  mov    %r9d,%ecx
  neg    %ecx
  cmovs  %r9d,%ecx
  add    $0x30,%ecx
  mov    %cl,(%rsi)
  cmp    $0xfffffff7,%eax
  jge    L23
  mov    %edx,%r9d
  neg    %edx
L23:
  mov    $0x2,%ecx
  mov    $0xcccccccd,%r10d
  test   %edx,%edx
  jle    L24
  nopw   0x0(%rax,%rax,1)
L47:
  mov    %edx,%eax
  mov    %edx,%esi
  mov    %ecx,%r8d
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,-0x1(%rbp,%rcx,1)
  mov    %edx,%esi
  add    $0x1,%rcx
  mov    %eax,%edx
  cmp    $0x9,%esi
  jg     L47
  movslq %r8d,%rax
L1:
  test   %r9d,%r9d
  jns    L45
  add    $0x1,%r8d
  movb   $0x2d,0x0(%rbp,%rax,1)
  movslq %r8d,%rax
L45:
  cmp    %ebx,%r8d
  jge    L58
  lea    -0x1(%rbx),%edx
  movslq %r8d,%rdi
  mov    $0x7e,%esi
  sub    %r8d,%edx
  add    %rbp,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
  movslq %ebx,%rax
L58:
  movb   $0x0,0x0(%rbp,%rax,1)
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <reverse@plt>
  nopl   0x0(%rax)
L24:
  mov    $0x1,%eax
  mov    $0x1,%r8d
  jmp    L1
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i = 0;
  int sign = n % 10;
  s[i++] = (sign < 0 ? -sign : sign) + '0';
  n /= 10;
  if (n < 0) {
    sign = n;
    n = -n;
  }
  while (n > 0) {
    s[i++] = n % 10 + '0';
    n /= 10;
  }
  if (sign < 0) {
    s[i++] = '-';
  }
  if (i < w) {
    while (w > i) {
      s[i++] = '~';
    }
  }
  s[i] = '\0';
  reverse(s);
}
```