ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_mean>:
  endbr64
  test   %esi,%esi
  jle    L16
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  lea    0x8(%rdi,%rax,8),%rax
  nopl   0x0(%rax)
L17:
  addsd  (%rdi),%xmm0
  add    $0x8,%rdi
  cmp    %rax,%rdi
  jne    L17
  pxor   %xmm1,%xmm1
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double get_mean(double *data_array, int size) {
  double sum = 0;
  double mean;
  int i;
  for (i = 0; i < size; i++) {
    sum += data_array[i];
  }
  mean = sum / size;
  return (mean);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_mean>:
  endbr64
  test   %esi,%esi
  jle    L47
  cmp    $0x1,%esi
  je     L7
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L28:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L28
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%sil
  je     L21
L46:
  cltq
  addsd  (%rdi,%rax,8),%xmm0
L21:
  pxor   %xmm1,%xmm1
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L47:
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm1
  divsd  %xmm1,%xmm0
  ret
L7:
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  jmp    L46
```

assistant: ```c
double get_mean(double *data_array, int size) {
  double sum = 0;
  double mean;
  int i;
  for (i = 0; i < size; i++) {
    sum += data_array[i];
  }
  mean = sum / size;
  return (mean);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L7:
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L7
```

tool_calls: name: read_data
arguments: data_label: D6
data_type: string

tool: D6: hello world


assistant: ```c
int main() {
  while (1) {
    printf("hello world\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   %rbx
  lea    D5(%rip),%rbx
L5:
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L5
```

tool_calls: name: read_data
arguments: tag: D5
datatype: string

tool: D5: hello world


assistant: ```c
int main() {
  while (1) {
    printf("hello world\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D8(%rip),%rbx
  nopl   0x0(%rax)
L9:
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L9
```

tool_calls: name: unpack_data
arguments: label: D8
type_guessed: string

tool: D8: hello world


assistant: ```c
int main() {
  while (1) {
    printf("hello world\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   %rbx
  lea    D3(%rip),%rbx
  nopl   0x0(%rax)
L9:
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L9
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_infered: string

tool: D3: hello world


assistant: ```c
int main() {
  while (1) {
    printf("hello world\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rop3_136_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x1c(%rbp),%al
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_136_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D & S;
  *D = stk1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rop3_136_0_0>:
  endbr64
  and    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_136_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D & S;
  *D = stk1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rop3_136_0_0>:
  endbr64
  and    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_136_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D & S;
  *D = stk1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rop3_136_0_0>:
  endbr64
  and    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_136_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D & S;
  *D = stk1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dirname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L16
  mov    $0x0,%edi
  call   <errori@plt>
L16:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x10(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L19
L3:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
L19:
  leave
  ret
```

assistant: ```c
char *dirname(const char *path) {
  char *dir;
  char *sep;
  dir = malloc(strlen(path) + 1);
  if (!dir)
    errori(NULL);
  strcpy(dir, path);
  sep = strrchr(dir, '/');
  if (!sep)
    return NULL;
  *sep = '\0';
  return dir;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<dirname>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L22
L2:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L15
  movb   $0x0,(%rax)
L10:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L22:
  mov    $0x0,%edi
  call   <errori@plt>
  jmp    L2
L15:
  mov    %rax,%rbx
  jmp    L10
```

assistant: ```c
char *dirname(const char *path) {
  char *dir;
  char *sep;
  dir = malloc(strlen(path) + 1);
  if (!dir)
    errori(NULL);
  strcpy(dir, path);
  sep = strrchr(dir, '/');
  if (!sep)
    return NULL;
  *sep = '\0';
  return dir;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dirname>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L6
L7:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    $0x2f,%esi
  mov    %r12,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L34
  movb   $0x0,(%rax)
L2:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nop
L34:
  xor    %r12d,%r12d
  jmp    L2
L6:
  xor    %edi,%edi
  call   <errori@plt>
  jmp    L7
```

assistant: ```c
char *dirname(const char *path) {
  char *dir;
  char *sep;
  dir = malloc(strlen(path) + 1);
  if (!dir)
    errori(NULL);
  strcpy(dir, path);
  sep = strrchr(dir, '/');
  if (!sep)
    return NULL;
  *sep = '\0';
  return dir;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dirname>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
L11:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    $0x2f,%esi
  mov    %r12,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L36
  movb   $0x0,(%rax)
L32:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nop
L36:
  xor    %r12d,%r12d
  jmp    L32
L3:
  xor    %edi,%edi
  call   <errori@plt>
  jmp    L11
```

assistant: ```c
char *dirname(const char *path) {
  char *dir;
  char *sep;
  dir = malloc(strlen(path) + 1);
  if (!dir)
    errori(NULL);
  strcpy(dir, path);
  sep = strrchr(dir, '/');
  if (!sep)
    return NULL;
  *sep = '\0';
  return dir;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x6,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int k = (3) + (3);
  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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int k = (3) + (3);
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int k = (3) + (3);
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int k = (3) + (3);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<norm_quat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  divsd  -0x8(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void norm_quat(double *quat) {
  double mag = quat[0] * quat[0] + quat[1] * quat[1] + quat[2] * quat[2] + quat[3] * quat[3];
  mag = sqrt(mag);
  quat[0] /= mag;
  quat[1] /= mag;
  quat[2] /= mag;
  quat[3] /= mag;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<norm_quat>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%r14
  mov    0x8(%rdi),%r13
  mov    0x10(%rdi),%r12
  mov    0x18(%rdi),%rbp
  movq   %r14,%xmm0
  mulsd  %xmm0,%xmm0
  movq   %r13,%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  movq   %r12,%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  movq   %rbp,%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L0
  sqrtsd %xmm0,%xmm0
L11:
  movq   %r14,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rbx)
  movq   %r13,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rbx)
  movq   %r12,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rbx)
  movq   %rbp,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x18(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <sqrt@plt>
  jmp    L11
```

assistant: ```c
void norm_quat(double *quat) {
  double mag = quat[0] * quat[0] + quat[1] * quat[1] + quat[2] * quat[2] + quat[3] * quat[3];
  mag = sqrt(mag);
  quat[0] /= mag;
  quat[1] /= mag;
  quat[2] /= mag;
  quat[3] /= mag;
}
```
user: What is the c source code of the assembly code below?

```asm
<norm_quat>:
  endbr64
  sub    $0x38,%rsp
  movsd  (%rdi),%xmm4
  movsd  0x8(%rdi),%xmm3
  movsd  0x10(%rdi),%xmm2
  movsd  0x18(%rdi),%xmm1
  movapd %xmm4,%xmm0
  movapd %xmm3,%xmm5
  mulsd  %xmm3,%xmm5
  mulsd  %xmm4,%xmm0
  addsd  %xmm5,%xmm0
  movapd %xmm2,%xmm5
  mulsd  %xmm2,%xmm5
  addsd  %xmm5,%xmm0
  movapd %xmm1,%xmm5
  mulsd  %xmm1,%xmm5
  addsd  %xmm5,%xmm0
  pxor   %xmm5,%xmm5
  ucomisd %xmm0,%xmm5
  ja     L10
  sqrtsd %xmm0,%xmm0
L15:
  divsd  %xmm0,%xmm4
  divsd  %xmm0,%xmm3
  movsd  %xmm4,(%rdi)
  divsd  %xmm0,%xmm2
  movsd  %xmm3,0x8(%rdi)
  divsd  %xmm0,%xmm1
  movsd  %xmm2,0x10(%rdi)
  movsd  %xmm1,0x18(%rdi)
  add    $0x38,%rsp
  ret
L10:
  mov    %rdi,0x28(%rsp)
  movsd  %xmm1,0x20(%rsp)
  movsd  %xmm2,0x18(%rsp)
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm4,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x28(%rsp),%rdi
  movsd  0x20(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm3
  movsd  0x8(%rsp),%xmm4
  jmp    L15
```

assistant: ```c
void norm_quat(double *quat) {
  double mag = quat[0] * quat[0] + quat[1] * quat[1] + quat[2] * quat[2] + quat[3] * quat[3];
  mag = sqrt(mag);
  quat[0] /= mag;
  quat[1] /= mag;
  quat[2] /= mag;
  quat[3] /= mag;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<norm_quat>:
  endbr64
  sub    $0x38,%rsp
  movsd  (%rdi),%xmm0
  movsd  0x8(%rdi),%xmm4
  movupd 0x10(%rdi),%xmm1
  movsd  0x10(%rdi),%xmm3
  mulsd  %xmm0,%xmm0
  movupd (%rdi),%xmm5
  mulsd  %xmm4,%xmm4
  movapd %xmm1,%xmm6
  mulsd  %xmm3,%xmm3
  unpckhpd %xmm6,%xmm6
  movapd %xmm6,%xmm2
  mulsd  %xmm6,%xmm2
  addsd  %xmm4,%xmm0
  addsd  %xmm3,%xmm0
  addsd  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  ucomisd %xmm0,%xmm2
  ja     L19
  sqrtsd %xmm0,%xmm0
L11:
  unpcklpd %xmm0,%xmm0
  divpd  %xmm0,%xmm5
  divpd  %xmm0,%xmm1
  movups %xmm5,(%rdi)
  movups %xmm1,0x10(%rdi)
  add    $0x38,%rsp
  ret
L19:
  mov    %rdi,0x28(%rsp)
  movaps %xmm1,0x10(%rsp)
  movaps %xmm5,(%rsp)
  call   <sqrt@plt>
  mov    0x28(%rsp),%rdi
  movapd 0x10(%rsp),%xmm1
  movapd (%rsp),%xmm5
  jmp    L11
```

assistant: ```c
void norm_quat(double *quat) {
  double mag = quat[0] * quat[0] + quat[1] * quat[1] + quat[2] * quat[2] + quat[3] * quat[3];
  mag = sqrt(mag);
  quat[0] /= mag;
  quat[1] /= mag;
  quat[2] /= mag;
  quat[3] /= mag;
}
```
user: Converte this assembly snippet into its equivalent C code:

```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    L15
L16:
  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    L48
L44:
  mov    -0x20(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L35
  mov    -0x1c(%rbp),%eax
  sub    %eax,-0x20(%rbp)
  jmp    L28
L35:
  mov    -0x18(%rbp),%eax
  add    %eax,-0x20(%rbp)
L28:
  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)
L48:
  mov    -0x24(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L44
  addl   $0x1,-0x28(%rbp)
L15:
  mov    -0x28(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L16
  nop
  nop
  leave
  ret
```

assistant: ```c
void permute_args(int panonopt_start, int panonopt_end, int opt_end, char *const *nargv) {
  int cyclelen, i, j, ncycle, nnonopts, nopts;
  ;
  nnonopts = panonopt_end - panonopt_start;
  nopts = opt_end - panonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - panonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    int cstart = panonopt_end + i;
    int pos = cstart;
    char *swap;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= panonopt_end) {
        pos -= nnonopts;
      } else {
        pos += nopts;
      }
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```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    L31
  movslq %ebx,%rdx
  lea    0x0(%rbp,%rdx,8),%rdi
  mov    %ebx,%r10d
  add    %ebx,%r11d
L11:
  mov    %r10d,%edx
  test   %eax,%eax
  jle    L21
  mov    $0x0,%ecx
L27:
  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    L27
L21:
  add    $0x8,%rdi
  add    $0x1,%r10d
  cmp    %r11d,%r10d
  jne    L11
L31:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void permute_args(int panonopt_start, int panonopt_end, int opt_end, char *const *nargv) {
  int cyclelen, i, j, ncycle, nnonopts, nopts;
  ;
  nnonopts = panonopt_end - panonopt_start;
  nopts = opt_end - panonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - panonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    int cstart = panonopt_end + i;
    int pos = cstart;
    char *swap;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= panonopt_end) {
        pos -= nnonopts;
      } else {
        pos += nopts;
      }
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```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    L20
  movslq %ebx,%rdx
  mov    %ebx,%r10d
  add    %ebx,%r11d
  lea    0x0(%rbp,%rdx,8),%r9
  nopl   0x0(%rax)
L33:
  mov    %r10d,%edx
  test   %eax,%eax
  jle    L28
  mov    (%r9),%r8
  xor    %ecx,%ecx
  jmp    L43
  cs nopw 0x0(%rax,%rax,1)
L35:
  mov    %rdi,%r8
L43:
  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    L35
L28:
  add    $0x1,%r10d
  add    $0x8,%r9
  cmp    %r11d,%r10d
  jne    L33
L20:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void permute_args(int panonopt_start, int panonopt_end, int opt_end, char *const *nargv) {
  int cyclelen, i, j, ncycle, nnonopts, nopts;
  ;
  nnonopts = panonopt_end - panonopt_start;
  nopts = opt_end - panonopt_end;
  ncycle = gcd(nnonopts, nopts);
  cyclelen = (opt_end - panonopt_start) / ncycle;
  for (i = 0; i < ncycle; i++) {
    int cstart = panonopt_end + i;
    int pos = cstart;
    char *swap;
    for (j = 0; j < cyclelen; j++) {
      if (pos >= panonopt_end) {
        pos -= nnonopts;
      } else {
        pos += nopts;
      }
      swap = nargv[pos];
      ((char **)nargv)[pos] = nargv[cstart];
      ((char **)nargv)[cstart] = swap;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```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    L4
  test   %eax,%eax
  jle    L4
  movslq %ebx,%rdx
  mov    %ebx,%r10d
  add    %ebx,%r11d
  lea    0x0(%rbp,%rdx,8),%r9
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%r9),%r8
  mov    %r10d,%edx
  xor    %ecx,%ecx
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %rdi,%r8
L21:
  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    L13
  add    $0x1,%r10d
  add    $0x8,%r9
  cmp    %r10d,%r11d
  jne    L14
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

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