ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<atol>:
  endbr64
  movzbl (%rdi),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x4,%dl
  jbe    L0
  cmp    $0x20,%al
  jne    L55
L0:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x4,%dl
  jbe    L0
  cmp    $0x20,%al
  je     L0
L55:
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L34
  mov    $0x0,%edx
L36:
  lea    (%rdx,%rdx,4),%rdx
  add    $0x1,%rdi
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%rdx
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  jbe    L36
L57:
  mov    %rdx,%rax
  ret
L34:
  mov    $0x0,%edx
  jmp    L57
```

assistant: ```c
long int atol(const char *str) {
  long int result = 0;
  while (((*str) == ' ' || (*str) == '\t' || (*str) == '\r' || (*str) == '\n' || (*str) == '\f' || (*str) == '\v'))
    str++;
  while (((*str) >= '0' && (*str) <= '9'))
    result = (result * 10) + (*(str++) - '0');
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atol>:
  endbr64
  jmp    L9
  cs nopw 0x0(%rax,%rax,1)
L7:
  add    $0x1,%rdi
L9:
  movzbl (%rdi),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x4,%dl
  jbe    L7
  cmp    $0x20,%al
  je     L7
  lea    -0x30(%rax),%edx
  xor    %r8d,%r8d
  cmp    $0x9,%dl
  ja     L42
  xchg   %ax,%ax
L5:
  sub    $0x30,%eax
  lea    (%r8,%r8,4),%rdx
  add    $0x1,%rdi
  movsbq %al,%rax
  lea    (%rax,%rdx,2),%r8
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L5
L42:
  mov    %r8,%rax
  ret
```

assistant: ```c
long int atol(const char *str) {
  long int result = 0;
  while (((*str) == ' ' || (*str) == '\t' || (*str) == '\r' || (*str) == '\n' || (*str) == '\f' || (*str) == '\v'))
    str++;
  while (((*str) >= '0' && (*str) <= '9'))
    result = (result * 10) + (*(str++) - '0');
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<atol>:
  endbr64
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L6:
  add    $0x1,%rdi
L31:
  movzbl (%rdi),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x4,%dl
  jbe    L6
  cmp    $0x20,%al
  je     L6
  lea    -0x30(%rax),%edx
  xor    %r8d,%r8d
  cmp    $0x9,%dl
  ja     L46
  xchg   %ax,%ax
L44:
  sub    $0x30,%eax
  lea    (%r8,%r8,4),%rdx
  add    $0x1,%rdi
  movsbq %al,%rax
  lea    (%rax,%rdx,2),%r8
  movzbl (%rdi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L44
L46:
  mov    %r8,%rax
  ret
```

assistant: ```c
long int atol(const char *str) {
  long int result = 0;
  while (((*str) == ' ' || (*str) == '\t' || (*str) == '\r' || (*str) == '\n' || (*str) == '\f' || (*str) == '\v'))
    str++;
  while (((*str) >= '0' && (*str) <= '9'))
    result = (result * 10) + (*(str++) - '0');
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<power2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%ecx
  shll   %cl,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <displayBits@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
type: string

tool: D8: %u


assistant: ```c
void power2(int number, int pow) {
  number <<= pow - 1;
  printf("%u\n", number);
  displayBits(number);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<power2>:
  endbr64
  push   %rbx
  lea    -0x1(%rsi),%ecx
  mov    %edi,%ebx
  shl    %cl,%ebx
  mov    %ebx,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  call   <displayBits@plt>
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D1
datatype: string

tool: D1: %u


assistant: ```c
void power2(int number, int pow) {
  number <<= pow - 1;
  printf("%u\n", number);
  displayBits(number);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<power2>:
  endbr64
  push   %rbp
  lea    -0x1(%rsi),%ecx
  shl    %cl,%edi
  mov    %edi,%ebp
  mov    %edi,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <displayBits@plt>
```

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

tool: D7: %u


assistant: ```c
void power2(int number, int pow) {
  number <<= pow - 1;
  printf("%u\n", number);
  displayBits(number);
}
```
user: Translate the below assembly code into C programming language:

```asm
<power2>:
  endbr64
  push   %rbp
  lea    -0x1(%rsi),%ecx
  shl    %cl,%edi
  mov    %edi,%ebp
  mov    %edi,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <displayBits@plt>
```

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

tool: D7: %u


assistant: ```c
void power2(int number, int pow) {
  number <<= pow - 1;
  printf("%u\n", number);
  displayBits(number);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L40
L24:
  movl   $0x0,-0x34(%rbp)
  jmp    L45
L107:
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rcx
  mov    -0x34(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rsi,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x34(%rbp)
L45:
  cmpl   $0x2,-0x34(%rbp)
  jle    L107
  addl   $0x1,-0x38(%rbp)
L40:
  cmpl   $0x2,-0x38(%rbp)
  jle    L24
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L68
L58:
  movl   $0x0,-0x34(%rbp)
  jmp    L51
L44:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jne    L10
  mov    -0x34(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    -0x30(%rbp,%rax,4),%eax
  mov    $0x20,%edx
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L71
L10:
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L71:
  addl   $0x1,-0x34(%rbp)
L51:
  cmpl   $0x2,-0x34(%rbp)
  jle    L44
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x38(%rbp)
L68:
  cmpl   $0x2,-0x38(%rbp)
  jle    L58
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D49
data_type: string
name: extract_symbolic_data
arguments: data_label: D42
data_type: string
name: extract_symbolic_data
arguments: data_label: D30
data_type: string
name: extract_symbolic_data
arguments: data_label: D54
data_type: string
name: extract_symbolic_data
arguments: data_label: D21
data_type: string
name: extract_symbolic_data
arguments: data_label: D19
data_type: string

tool: D49: Matriz Quadrada 3x3:


tool: D42: mat[%d][%d] = 

tool: D30: %i

tool: D54: Mostrando a Diagonal Prncipal:


tool: D21: %5d %3c

tool: D19:     -   

assistant: ```c
int main() {
  int mat[3][3], i, j;
  printf("Matriz Quadrada 3x3:\n\n");
  for (i = 0; i < 3; i++)
    for (j = 0; j < 3; j++) {
      printf("mat[%d][%d] = ", i + 1, j + 1);
      scanf("%i", &mat[i][j]);
    }
  printf("Mostrando a Diagonal Prncipal:\n\n");
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      if (i == j)
        printf("%5d %3c", mat[i][j], ' ');
      else
        printf("    -   ");
    }
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D57(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%r15d
  mov    $0x1,%r12d
  lea    D41(%rip),%r14
  jmp    L21
L65:
  add    $0x1,%r12d
  add    $0x3,%r15
  cmp    $0x4,%r12d
  je     L60
L21:
  mov    %rsp,%r13
  lea    0x0(%r13,%r15,4),%rbp
  mov    $0x0,%ebx
L3:
  add    $0x1,%ebx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    $0x3,%ebx
  jne    L3
  jmp    L65
L60:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebp
  lea    D7(%rip),%r12
  lea    D36(%rip),%r14
  jmp    L18
L82:
  mov    0x0(%r13,%rbx,4),%esi
  mov    $0x20,%edx
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L97
L98:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebp
  add    $0xc,%r13
  cmp    $0x3,%ebp
  je     L51
L18:
  mov    $0x0,%ebx
L96:
  cmp    %ebx,%ebp
  je     L82
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L97:
  add    $0x1,%rbx
  cmp    $0x3,%rbx
  jne    L96
  jmp    L98
L51:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D57
type_guessed: string
name: retrieve_data
arguments: label: D41
type_guessed: string
name: retrieve_data
arguments: label: D2
type_guessed: string
name: retrieve_data
arguments: label: D35
type_guessed: string
name: retrieve_data
arguments: label: D7
type_guessed: string
name: retrieve_data
arguments: label: D36
type_guessed: string

tool: D57: Matriz Quadrada 3x3:


tool: D41: mat[%d][%d] = 

tool: D2: %i

tool: D35: Mostrando a Diagonal Prncipal:


tool: D7:     -   

tool: D36: %5d %3c

assistant: ```c
int main() {
  int mat[3][3], i, j;
  printf("Matriz Quadrada 3x3:\n\n");
  for (i = 0; i < 3; i++)
    for (j = 0; j < 3; j++) {
      printf("mat[%d][%d] = ", i + 1, j + 1);
      scanf("%i", &mat[i][j]);
    }
  printf("Mostrando a Diagonal Prncipal:\n\n");
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      if (i == j)
        printf("%5d %3c", mat[i][j], ' ');
      else
        printf("    -   ");
    }
    printf("\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D44(%rip),%rdi
  xor    %r15d,%r15d
  push   %r14
  push   %r13
  lea    D45(%rip),%r13
  push   %r12
  lea    D53(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  movq   $0x0,(%rsp)
  mov    %rax,0x8(%rsp)
L8:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%rcx
  add    $0x1,%r15d
  xor    %r14d,%r14d
  mov    %rax,%rbp
  lea    (%rax,%rcx,4),%rbx
L5:
  add    $0x1,%r14d
  mov    %r15d,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %r14d,%edx
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    $0x3,%r14d
  jne    L5
  addq   $0x3,(%rsp)
  cmp    $0x3,%r15d
  jne    L8
  lea    D24(%rip),%rdi
  xor    %r12d,%r12d
  lea    D12(%rip),%r13
  call   <puts@plt>
  lea    D41(%rip),%r14
L43:
  xor    %ebx,%ebx
L19:
  cmp    %ebx,%r12d
  je     L26
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L25:
  add    $0x1,%rbx
  cmp    $0x3,%rbx
  jne    L19
  mov    $0xa,%edi
  add    $0x1,%r12d
  add    $0xc,%rbp
  call   <putchar@plt>
  cmp    $0x3,%r12d
  jne    L43
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L26:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    $0x20,%edx
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L25
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D44
type: string
name: read_data
arguments: data_tag: D45
type: string
name: read_data
arguments: data_tag: D53
type: string
name: read_data
arguments: data_tag: D24
type: string
name: read_data
arguments: data_tag: D12
type: string
name: read_data
arguments: data_tag: D41
type: string

tool: D44: Matriz Quadrada 3x3:


tool: D45: mat[%d][%d] = 

tool: D53: %i

tool: D24: Mostrando a Diagonal Prncipal:


tool: D12:     -   

tool: D41: %5d %3c

assistant: ```c
int main() {
  int mat[3][3], i, j;
  printf("Matriz Quadrada 3x3:\n\n");
  for (i = 0; i < 3; i++)
    for (j = 0; j < 3; j++) {
      printf("mat[%d][%d] = ", i + 1, j + 1);
      scanf("%i", &mat[i][j]);
    }
  printf("Mostrando a Diagonal Prncipal:\n\n");
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      if (i == j)
        printf("%5d %3c", mat[i][j], ' ');
      else
        printf("    -   ");
    }
    printf("\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  lea    D18(%rip),%rdi
  xor    %r15d,%r15d
  push   %r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  lea    D45(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  movq   $0x0,(%rsp)
  mov    %rax,0x8(%rsp)
L24:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%rcx
  add    $0x1,%r15d
  xor    %r14d,%r14d
  mov    %rax,%rbp
  lea    (%rax,%rcx,4),%rbx
L59:
  add    $0x1,%r14d
  mov    %r15d,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  mov    %r14d,%edx
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    $0x3,%r14d
  jne    L59
  addq   $0x3,(%rsp)
  cmp    $0x3,%r15d
  jne    L24
  lea    D22(%rip),%rdi
  xor    %r12d,%r12d
  lea    D56(%rip),%r13
  call   <puts@plt>
  lea    D0(%rip),%r14
L27:
  xor    %ebx,%ebx
L53:
  cmp    %ebx,%r12d
  je     L22
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L35:
  add    $0x1,%rbx
  cmp    $0x3,%rbx
  jne    L53
  mov    $0xa,%edi
  add    $0x1,%r12d
  add    $0xc,%rbp
  call   <putchar@plt>
  cmp    $0x3,%r12d
  jne    L27
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L22:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    $0x20,%edx
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L35
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D18
type_guessed: string
name: disassemble_data
arguments: data_label: D1
type_guessed: string
name: disassemble_data
arguments: data_label: D45
type_guessed: string
name: disassemble_data
arguments: data_label: D22
type_guessed: string
name: disassemble_data
arguments: data_label: D56
type_guessed: string
name: disassemble_data
arguments: data_label: D0
type_guessed: string

tool: D18: Matriz Quadrada 3x3:


tool: D1: mat[%d][%d] = 

tool: D45: %i

tool: D22: Mostrando a Diagonal Prncipal:


tool: D56:     -   

tool: D0: %5d %3c

assistant: ```c
int main() {
  int mat[3][3], i, j;
  printf("Matriz Quadrada 3x3:\n\n");
  for (i = 0; i < 3; i++)
    for (j = 0; j < 3; j++) {
      printf("mat[%d][%d] = ", i + 1, j + 1);
      scanf("%i", &mat[i][j]);
    }
  printf("Mostrando a Diagonal Prncipal:\n\n");
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++) {
      if (i == j)
        printf("%5d %3c", mat[i][j], ' ');
      else
        printf("    -   ");
    }
    printf("\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<permute_args>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    -0x38(%rbp),%eax
  sub    -0x34(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x3c(%rbp),%eax
  sub    -0x38(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <gcd@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x3c(%rbp),%eax
  sub    -0x34(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L37
L7:
  mov    -0x38(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L10
L24:
  mov    -0x20(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L14
  mov    -0x1c(%rbp),%eax
  sub    %eax,-0x20(%rbp)
  jmp    L39
L14:
  mov    -0x18(%rbp),%eax
  add    %eax,-0x20(%rbp)
L39:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x48(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%rax
  mov    %rax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
  addl   $0x1,-0x24(%rbp)
L10:
  mov    -0x24(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L24
  addl   $0x1,-0x28(%rbp)
L37:
  mov    -0x28(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void permute_args(int nonopt_start, int nonopt_end, int opt_end, char *const *nargv) {
  int cstart, cyclelen, i, j, ncycle, nnonopts, nopts, pos;
  char *swap;
  nnonopts = nonopt_end - nonopt_start;
  nopts = opt_end - nonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - nonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    cstart = nonopt_end + i;
    pos = cstart;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= nonopt_end)
        pos -= nnonopts;
      else
        pos += nopts;
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<permute_args>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r15d
  mov    %esi,%ebx
  mov    %edx,%r14d
  mov    %rcx,%rbp
  mov    %esi,%r12d
  sub    %edi,%r12d
  mov    %edx,%r13d
  sub    %esi,%r13d
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <gcd@plt>
  mov    %eax,%r11d
  mov    %r14d,%eax
  sub    %r15d,%eax
  cltd
  idiv   %r11d
  test   %r11d,%r11d
  jle    L13
  movslq %ebx,%rdx
  lea    0x0(%rbp,%rdx,8),%rdi
  mov    %ebx,%r10d
  add    %ebx,%r11d
L28:
  mov    %r10d,%edx
  test   %eax,%eax
  jle    L27
  mov    $0x0,%ecx
L21:
  mov    %edx,%esi
  sub    %r12d,%esi
  lea    (%rdx,%r13,1),%r8d
  cmp    %edx,%ebx
  mov    %esi,%edx
  cmovg  %r8d,%edx
  movslq %edx,%rsi
  lea    0x0(%rbp,%rsi,8),%rsi
  mov    (%rsi),%r8
  mov    (%rdi),%r9
  mov    %r9,(%rsi)
  mov    %r8,(%rdi)
  add    $0x1,%ecx
  cmp    %ecx,%eax
  jne    L21
L27:
  add    $0x8,%rdi
  add    $0x1,%r10d
  cmp    %r11d,%r10d
  jne    L28
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void permute_args(int nonopt_start, int nonopt_end, int opt_end, char *const *nargv) {
  int cstart, cyclelen, i, j, ncycle, nnonopts, nopts, pos;
  char *swap;
  nnonopts = nonopt_end - nonopt_start;
  nopts = opt_end - nonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - nonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    cstart = nonopt_end + i;
    pos = cstart;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= nonopt_end)
        pos -= nnonopts;
      else
        pos += nopts;
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<permute_args>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  sub    %esi,%r13d
  push   %rbp
  sub    %edi,%r12d
  mov    %rcx,%rbp
  push   %rbx
  mov    %r12d,%edi
  mov    %esi,%ebx
  mov    %r13d,%esi
  sub    $0x8,%rsp
  call   <gcd@plt>
  mov    %eax,%r11d
  mov    %r14d,%eax
  sub    %r15d,%eax
  cltd
  idiv   %r11d
  test   %r11d,%r11d
  jle    L30
  movslq %ebx,%rdx
  mov    %ebx,%r10d
  add    %ebx,%r11d
  lea    0x0(%rbp,%rdx,8),%r9
  nopl   0x0(%rax)
L22:
  mov    %r10d,%edx
  test   %eax,%eax
  jle    L29
  mov    (%r9),%r8
  xor    %ecx,%ecx
  jmp    L25
  cs nopw 0x0(%rax,%rax,1)
L33:
  mov    %rdi,%r8
L25:
  mov    %edx,%esi
  lea    (%rdx,%r13,1),%edi
  sub    %r12d,%esi
  cmp    %edx,%ebx
  mov    %esi,%edx
  cmovg  %edi,%edx
  add    $0x1,%ecx
  movslq %edx,%rsi
  lea    0x0(%rbp,%rsi,8),%rsi
  mov    (%rsi),%rdi
  mov    %r8,(%rsi)
  mov    %rdi,(%r9)
  cmp    %ecx,%eax
  jne    L33
L29:
  add    $0x1,%r10d
  add    $0x8,%r9
  cmp    %r11d,%r10d
  jne    L22
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void permute_args(int nonopt_start, int nonopt_end, int opt_end, char *const *nargv) {
  int cstart, cyclelen, i, j, ncycle, nnonopts, nopts, pos;
  char *swap;
  nnonopts = nonopt_end - nonopt_start;
  nopts = opt_end - nonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - nonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    cstart = nonopt_end + i;
    pos = cstart;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= nonopt_end)
        pos -= nnonopts;
      else
        pos += nopts;
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<permute_args>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  sub    %esi,%r13d
  push   %rbp
  sub    %edi,%r12d
  mov    %rcx,%rbp
  push   %rbx
  mov    %r12d,%edi
  mov    %esi,%ebx
  mov    %r13d,%esi
  sub    $0x8,%rsp
  call   <gcd@plt>
  mov    %eax,%r11d
  mov    %r14d,%eax
  sub    %r15d,%eax
  cltd
  idiv   %r11d
  test   %r11d,%r11d
  jle    L47
  test   %eax,%eax
  jle    L47
  movslq %ebx,%rdx
  mov    %ebx,%r10d
  add    %ebx,%r11d
  lea    0x0(%rbp,%rdx,8),%r9
  nopl   0x0(%rax,%rax,1)
L11:
  mov    (%r9),%r8
  mov    %r10d,%edx
  xor    %ecx,%ecx
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %rdi,%r8
L28:
  lea    (%rdx,%r13,1),%esi
  mov    %edx,%edi
  sub    %r12d,%edi
  cmp    %edx,%ebx
  mov    %esi,%edx
  cmovle %edi,%edx
  add    $0x1,%ecx
  movslq %edx,%rsi
  lea    0x0(%rbp,%rsi,8),%rsi
  mov    (%rsi),%rdi
  mov    %r8,(%rsi)
  mov    %rdi,(%r9)
  cmp    %ecx,%eax
  jne    L39
  add    $0x1,%r10d
  add    $0x8,%r9
  cmp    %r10d,%r11d
  jne    L11
L47:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void permute_args(int nonopt_start, int nonopt_end, int opt_end, char *const *nargv) {
  int cstart, cyclelen, i, j, ncycle, nnonopts, nopts, pos;
  char *swap;
  nnonopts = nonopt_end - nonopt_start;
  nopts = opt_end - nonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - nonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    cstart = nonopt_end + i;
    pos = cstart;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= nonopt_end)
        pos -= nnonopts;
      else
        pos += nopts;
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```