ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  movdqa D23(%rip),%xmm0
  movslq %edi,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movdqa D28(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  mov    (%rsp,%rdi,4),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x3,%eax
  add    $0x38,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D23
data_type: byte[16]
name: extract_symbolic_data
arguments: data_tag: D28
data_type: byte[16]
name: extract_symbolic_data
arguments: data_tag: D4
data_type: string

tool: {"D23": [0x01,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D28": [0x09,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x0d,0x00,0x00,0x00,0x0f,0x00,0x00,0x00]}

tool: D4: %d


assistant: ```c
int main(int argc, char **argv) {
  int a[8];
  for (int i = 0; i < 8; i++)
    a[i] = 2 * i + 1;
  int *p = &a[0];
  __builtin_printf("%d\n", a[argc]);
  return *(++p);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_stat_success_150>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x656c694620303531,%rax
  movabs $0x2073757461747320,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x6261203b79616b6f,%rax
  movabs $0x6f206f742074756f,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x61746164206e6570,%rax
  movabs $0x7463656e6e6f6320,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x2e6e6f69,-0x10(%rbp)
  movw   $0xa0d,-0xc(%rbp)
  movb   $0x0,-0xa(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <_response@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void _stat_success_150(void) {
  const char msg[] = "150 File status okay; about to open data connection.\r\n";
  _response(msg);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_stat_success_150>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movabs $0x656c694620303531,%rax
  movabs $0x2073757461747320,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movabs $0x6261203b79616b6f,%rax
  movabs $0x6f206f742074756f,%rdx
  mov    %rax,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  movabs $0x61746164206e6570,%rax
  movabs $0x7463656e6e6f6320,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movl   $0x2e6e6f69,0x30(%rsp)
  movw   $0xa0d,0x34(%rsp)
  movb   $0x0,0x36(%rsp)
  mov    %rsp,%rdi
  call   <_response@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x48,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _stat_success_150(void) {
  const char msg[] = "150 File status okay; about to open data connection.\r\n";
  _response(msg);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_stat_success_150>:
  endbr64
  sub    $0x48,%rsp
  movdqa D26(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0xa0d,%eax
  movb   $0x0,0x36(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D12(%rip),%xmm0
  movl   $0x2e6e6f69,0x30(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D14(%rip),%xmm0
  mov    %ax,0x34(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <_response@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x48,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D26
data_type: byte[16]
name: unpack_data
arguments: label: D12
data_type: byte[16]
name: unpack_data
arguments: label: D14
data_type: byte[16]

tool: {"D26": [0x31,0x35,0x30,0x20,0x46,0x69,0x6c,0x65,0x20,0x73,0x74,0x61,0x74,0x75,0x73,0x20]}

tool: {"D12": [0x6f,0x6b,0x61,0x79,0x3b,0x20,0x61,0x62,0x6f,0x75,0x74,0x20,0x74,0x6f,0x20,0x6f]}

tool: {"D14": [0x70,0x65,0x6e,0x20,0x64,0x61,0x74,0x61,0x20,0x63,0x6f,0x6e,0x6e,0x65,0x63,0x74]}

assistant: ```c
void _stat_success_150(void) {
  const char msg[] = "150 File status okay; about to open data connection.\r\n";
  _response(msg);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_stat_success_150>:
  endbr64
  sub    $0x48,%rsp
  movdqa D20(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0xa0d,%eax
  movb   $0x0,0x36(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D26(%rip),%xmm0
  movl   $0x2e6e6f69,0x30(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D17(%rip),%xmm0
  mov    %ax,0x34(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <_response@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x48,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D20
datatype: byte[16]
name: disassemble_data
arguments: data_tag: D26
datatype: byte[16]
name: disassemble_data
arguments: data_tag: D17
datatype: byte[16]

tool: {"D20": [0x31,0x35,0x30,0x20,0x46,0x69,0x6c,0x65,0x20,0x73,0x74,0x61,0x74,0x75,0x73,0x20]}

tool: {"D26": [0x6f,0x6b,0x61,0x79,0x3b,0x20,0x61,0x62,0x6f,0x75,0x74,0x20,0x74,0x6f,0x20,0x6f]}

tool: {"D17": [0x70,0x65,0x6e,0x20,0x64,0x61,0x74,0x61,0x20,0x63,0x6f,0x6e,0x6e,0x65,0x63,0x74]}

assistant: ```c
void _stat_success_150(void) {
  const char msg[] = "150 File status okay; about to open data connection.\r\n";
  _response(msg);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fac>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L11
  mov    $0x1,%eax
  jmp    L5
L11:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fac>
  imul   -0x4(%rbp),%eax
L5:
  leave
  ret
```

assistant: ```c
int fac(int n) {
  if (n < 2)
    return 1;
  return n * fac(n - 1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fac>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L2
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <fac>
  imul   %ebx,%eax
  pop    %rbx
  ret
L2:
  ret
```

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

```asm
<fac>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L14
  xchg   %ax,%ax
L10:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L10
L14:
  ret
```

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

```asm
<fac>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L7
  xchg   %ax,%ax
L3:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L3
L7:
  ret
```

assistant: ```c
int fac(int n) {
  if (n < 2)
    return 1;
  return n * fac(n - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<is_tex_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x4,-0xc(%rbp)
  jg     L12
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L3
L12:
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L3:
  cmpl   $0x4,-0xc(%rbp)
  jle    L10
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L10
  movl   $0x1,-0x10(%rbp)
  jmp    L24
L10:
  movl   $0x0,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type_guessed: string

tool: D5: .tex

assistant: ```c
int is_tex_file(char *filename) {
  char *ptr;
  int length;
  int return_value;
  length = strlen(filename);
  if (length < 5)
    ptr = filename;
  else
    ptr = filename + length - 4;
  if (0 >= 10)
    (void)printf(".... is_tex_file .... \nfilename: <%s>\nptr     : <%s>\n", filename, ptr);
  if (length > 4 && !strcmp(ptr, ".tex"))
    return_value = 1;
  else
    return_value = 0;
  if (0 >= 10)
    (void)printf("is_tex_file: return_value is %d\n\n", return_value);
  return return_value;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_tex_file>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    $0x0,%edx
  cmp    $0x4,%eax
  jle    L9
  cltq
  lea    -0x4(%rbx,%rax,1),%rdi
  lea    D0(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %dl
  movzbl %dl,%edx
L9:
  mov    %edx,%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D0: .tex

assistant: ```c
int is_tex_file(char *filename) {
  char *ptr;
  int length;
  int return_value;
  length = strlen(filename);
  if (length < 5)
    ptr = filename;
  else
    ptr = filename + length - 4;
  if (0 >= 10)
    (void)printf(".... is_tex_file .... \nfilename: <%s>\nptr     : <%s>\n", filename, ptr);
  if (length > 4 && !strcmp(ptr, ".tex"))
    return_value = 1;
  else
    return_value = 0;
  if (0 >= 10)
    (void)printf("is_tex_file: return_value is %d\n\n", return_value);
  return return_value;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_tex_file>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    $0x4,%eax
  jle    L6
  cltq
  lea    D7(%rip),%rsi
  lea    -0x4(%rbx,%rax,1),%rdi
  call   <strcmp@plt>
  xor    %r8d,%r8d
  test   %eax,%eax
  sete   %r8b
L6:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
type_infered: string

tool: D7: .tex

assistant: ```c
int is_tex_file(char *filename) {
  char *ptr;
  int length;
  int return_value;
  length = strlen(filename);
  if (length < 5)
    ptr = filename;
  else
    ptr = filename + length - 4;
  if (0 >= 10)
    (void)printf(".... is_tex_file .... \nfilename: <%s>\nptr     : <%s>\n", filename, ptr);
  if (length > 4 && !strcmp(ptr, ".tex"))
    return_value = 1;
  else
    return_value = 0;
  if (0 >= 10)
    (void)printf("is_tex_file: return_value is %d\n\n", return_value);
  return return_value;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_tex_file>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    $0x4,%eax
  jle    L4
  cltq
  lea    D9(%rip),%rsi
  lea    -0x4(%rbx,%rax,1),%rdi
  call   <strcmp@plt>
  xor    %r8d,%r8d
  test   %eax,%eax
  sete   %r8b
L4:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type: string

tool: D9: .tex

assistant: ```c
int is_tex_file(char *filename) {
  char *ptr;
  int length;
  int return_value;
  length = strlen(filename);
  if (length < 5)
    ptr = filename;
  else
    ptr = filename + length - 4;
  if (0 >= 10)
    (void)printf(".... is_tex_file .... \nfilename: <%s>\nptr     : <%s>\n", filename, ptr);
  if (length > 4 && !strcmp(ptr, ".tex"))
    return_value = 1;
  else
    return_value = 0;
  if (0 >= 10)
    (void)printf("is_tex_file: return_value is %d\n\n", return_value);
  return return_value;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fitsBits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    $0x20,%eax
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  cmp    %eax,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fitsBits(int number, int word) {
  const int temp = sizeof(int) * 8 + 1 + ~word;
  int ans = !(((number << temp) >> temp) ^ number);
  return ans;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fitsBits>:
  endbr64
  mov    $0x20,%ecx
  sub    %esi,%ecx
  mov    %edi,%eax
  shl    %cl,%eax
  sar    %cl,%eax
  cmp    %edi,%eax
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int fitsBits(int number, int word) {
  const int temp = sizeof(int) * 8 + 1 + ~word;
  int ans = !(((number << temp) >> temp) ^ number);
  return ans;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fitsBits>:
  endbr64
  mov    $0x20,%ecx
  mov    %edi,%eax
  sub    %esi,%ecx
  shl    %cl,%eax
  sar    %cl,%eax
  cmp    %edi,%eax
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int fitsBits(int number, int word) {
  const int temp = sizeof(int) * 8 + 1 + ~word;
  int ans = !(((number << temp) >> temp) ^ number);
  return ans;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fitsBits>:
  endbr64
  mov    $0x20,%ecx
  mov    %edi,%eax
  sub    %esi,%ecx
  shl    %cl,%eax
  sar    %cl,%eax
  cmp    %edi,%eax
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int fitsBits(int number, int word) {
  const int temp = sizeof(int) * 8 + 1 + ~word;
  int ans = !(((number << temp) >> temp) ^ number);
  return ans;
}
```
user: What is the c source code of the assembly code below?

```asm
<f1_wh>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x4(%rbp)
  jmp    L15
L37:
  cmpw   $0x0,-0x4(%rbp)
  jg     L15
  mov    $0x64,%eax
  jmp    L34
L15:
  movzwl -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %ax,-0x4(%rbp)
  cmpw   $0x0,-0x4(%rbp)
  jne    L37
  mov    $0xc8,%eax
L34:
  pop    %rbp
  ret
```

assistant: ```c
short f1_wh(short x) {
  while (--x)
    if (x <= 0)
      return 100;
  return 200;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<f1_wh>:
  endbr64
L3:
  sub    $0x1,%di
  je     L18
  test   %di,%di
  jg     L3
  mov    $0x64,%eax
  ret
L18:
  mov    $0xc8,%eax
  ret
```

assistant: ```c
short f1_wh(short x) {
  while (--x)
    if (x <= 0)
      return 100;
  return 200;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f1_wh>:
  endbr64
  jmp    L12
  cs nopw 0x0(%rax,%rax,1)
L25:
  jle    L11
L12:
  sub    $0x1,%edi
  test   %di,%di
  jne    L25
  mov    $0xc8,%eax
  ret
L11:
  mov    $0x64,%eax
  ret
```

assistant: ```c
short f1_wh(short x) {
  while (--x)
    if (x <= 0)
      return 100;
  return 200;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<f1_wh>:
  endbr64
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L9:
  jle    L19
L24:
  sub    $0x1,%edi
  test   %di,%di
  jne    L9
  mov    $0xc8,%eax
  ret
L19:
  mov    $0x64,%eax
  ret
```

assistant: ```c
short f1_wh(short x) {
  while (--x)
    if (x <= 0)
      return 100;
  return 200;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movzbl -0x4(%rbp),%eax
  test   %al,%al
  jns    L5
  mov    $0x0,%eax
  call   <bar@plt>
L5:
  nop
  leave
  ret
```

assistant: ```c
void foo4(unsigned char c) {
  if (c & 0x80)
    bar();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<foo4>:
  endbr64
  test   %dil,%dil
  js     L6
  ret
L6:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <bar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void foo4(unsigned char c) {
  if (c & 0x80)
    bar();
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo4>:
  endbr64
  test   %dil,%dil
  js     L4
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  jmp    1050 <bar@plt>
```

assistant: ```c
void foo4(unsigned char c) {
  if (c & 0x80)
    bar();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo4>:
  endbr64
  test   %dil,%dil
  js     L1
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  jmp    1050 <bar@plt>
```

assistant: ```c
void foo4(unsigned char c) {
  if (c & 0x80)
    bar();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<quick_sort>:
  endbr64
  test   %rdi,%rdi
  je     L68
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r9d
  mov    %edx,%r12d
  cmp    %edx,%esi
  jge    L4
  movslq %esi,%rax
  lea    (%rdi,%rax,4),%r10
  mov    (%r10),%ecx
  lea    0x1(%rdx),%ebx
  cmp    %ebx,%esi
  jl     L88
  jmp    L61
L21:
  cmp    %esi,%ebx
  jle    L64
  movslq %esi,%rax
  lea    0x0(%rbp,%rax,4),%rax
  xor    (%rax),%edx
  mov    %edx,(%rax)
  xor    (%rdi),%edx
  mov    %edx,(%rdi)
  xor    %edx,(%rax)
L88:
  movslq %ebx,%rax
  lea    -0x4(%rbp,%rax,4),%rax
L1:
  mov    %ebx,%r8d
  sub    $0x1,%ebx
  mov    %rax,%rdi
  sub    $0x4,%rax
  mov    0x4(%rax),%edx
  cmp    %ecx,%edx
  jg     L1
  lea    0x1(%rsi),%eax
  cltq
L107:
  mov    %eax,%esi
  cmp    %eax,%ebx
  jl     L21
  add    $0x1,%rax
  cmp    %ecx,-0x4(%rbp,%rax,4)
  jle    L107
  jmp    L21
L105:
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,4),%rdx
  mov    (%r10),%eax
  xor    (%rdx),%eax
  mov    %eax,(%r10)
  xor    (%rdx),%eax
  mov    %eax,(%rdx)
  xor    %eax,(%r10)
  lea    -0x2(%r8),%edx
  mov    %r9d,%esi
  mov    %rbp,%rdi
  call   <quick_sort>
  jmp    L61
L4:
  mov    $0xffffffff,%eax
  jmp    L93
L64:
  cmp    %ebx,%r9d
  jl     L105
L61:
  lea    0x1(%rbx),%esi
  mov    %r12d,%edx
  mov    %rbp,%rdi
  call   <quick_sort>
  mov    $0x0,%eax
L93:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L68:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int quick_sort(int *arr, int left, int right) {
  int pivot;
  int i, j;
  if (NULL == arr || left >= right)
    return -1;
  pivot = arr[left];
  i = left;
  j = right + 1;
  while (i < j) {
    do {
      j--;
    } while (arr[j] > pivot);
    do {
      i++;
    } while (i <= j && arr[i] <= pivot);
    if (i < j) {
      do {
        arr[i] = arr[i] ^ arr[j];
        arr[j] = arr[i] ^ arr[j];
        arr[i] = arr[i] ^ arr[j];
      } while (0);
    }
  }
  if (left < j) {
    do {
      arr[left] = arr[left] ^ arr[j];
      arr[j] = arr[left] ^ arr[j];
      arr[left] = arr[left] ^ arr[j];
    } while (0);
    quick_sort(arr, left, j - 1);
  }
  quick_sort(arr, j + 1, right);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quick_sort>:
  endbr64
  test   %rdi,%rdi
  je     L109
  push   %r13
  mov    %esi,%r8d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %edx,%esi
  jge    L68
  movslq %esi,%rax
  mov    %rdi,%rbp
  lea    0x2(%rdx),%r13d
  mov    %esi,%ecx
  lea    (%rdi,%rax,4),%r9
  lea    0x1(%rdx),%eax
  mov    (%r9),%r11d
  cmp    %eax,%esi
  jge    L67
  nop
L62:
  movslq %eax,%rdx
  lea    -0x4(%rbp,%rdx,4),%rdx
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %rdx,%rdi
  mov    (%rdx),%esi
  mov    %eax,%r13d
  lea    -0x4(%rdx),%rdx
  sub    $0x1,%eax
  cmp    %r11d,%esi
  jg     L7
  movslq %ecx,%rdx
  lea    0x4(%rbp,%rdx,4),%rdx
  jmp    L28
  xchg   %ax,%ax
L10:
  mov    %rdx,%rbx
  mov    (%rdx),%r10d
  add    $0x4,%rdx
  cmp    %r11d,%r10d
  jg     L51
L28:
  add    $0x1,%ecx
  cmp    %ecx,%eax
  jge    L10
L87:
  cmp    %eax,%r8d
  jge    L67
  xor    (%r9),%esi
  lea    -0x2(%r13),%edx
  mov    %esi,(%r9)
  xor    (%rdi),%esi
  mov    %esi,(%rdi)
  mov    %rbp,%rdi
  xor    %esi,(%r9)
  mov    %r8d,%esi
  call   <quick_sort>
L67:
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <quick_sort>
  xor    %eax,%eax
L94:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L51:
  cmp    %ecx,%eax
  jle    L87
  xor    %r10d,%esi
  mov    %esi,(%rbx)
  xor    (%rdi),%esi
  mov    %esi,(%rdi)
  xor    %esi,(%rbx)
  jmp    L62
L68:
  mov    $0xffffffff,%eax
  jmp    L94
L109:
  or     $0xffffffff,%eax
  ret
```

assistant: ```c
int quick_sort(int *arr, int left, int right) {
  int pivot;
  int i, j;
  if (NULL == arr || left >= right)
    return -1;
  pivot = arr[left];
  i = left;
  j = right + 1;
  while (i < j) {
    do {
      j--;
    } while (arr[j] > pivot);
    do {
      i++;
    } while (i <= j && arr[i] <= pivot);
    if (i < j) {
      do {
        arr[i] = arr[i] ^ arr[j];
        arr[j] = arr[i] ^ arr[j];
        arr[i] = arr[i] ^ arr[j];
      } while (0);
    }
  }
  if (left < j) {
    do {
      arr[left] = arr[left] ^ arr[j];
      arr[j] = arr[left] ^ arr[j];
      arr[left] = arr[left] ^ arr[j];
    } while (0);
    quick_sort(arr, left, j - 1);
  }
  quick_sort(arr, j + 1, right);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<quick_sort>:
  endbr64
  test   %rdi,%rdi
  je     L58
  push   %r13
  mov    %esi,%r8d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %edx,%esi
  jge    L35
  movslq %esi,%rax
  mov    %rdi,%rbp
  lea    0x2(%rdx),%r13d
  mov    %esi,%ecx
  lea    (%rdi,%rax,4),%r9
  lea    0x1(%rdx),%eax
  mov    (%r9),%r11d
  cmp    %eax,%esi
  jge    L13
  nop
L50:
  movslq %eax,%rdx
  lea    -0x4(%rbp,%rdx,4),%rdx
  nopl   0x0(%rax,%rax,1)
L72:
  mov    %rdx,%rdi
  mov    (%rdx),%esi
  mov    %eax,%r13d
  lea    -0x4(%rdx),%rdx
  sub    $0x1,%eax
  cmp    %r11d,%esi
  jg     L72
  movslq %ecx,%rdx
  lea    0x4(%rbp,%rdx,4),%rdx
  jmp    L82
  xchg   %ax,%ax
L22:
  mov    %rdx,%rbx
  mov    (%rdx),%r10d
  add    $0x4,%rdx
  cmp    %r11d,%r10d
  jg     L60
L82:
  add    $0x1,%ecx
  cmp    %ecx,%eax
  jge    L22
L59:
  cmp    %eax,%r8d
  jge    L13
  xor    (%r9),%esi
  lea    -0x2(%r13),%edx
  mov    %esi,(%r9)
  xor    (%rdi),%esi
  mov    %esi,(%rdi)
  mov    %rbp,%rdi
  xor    %esi,(%r9)
  mov    %r8d,%esi
  call   <quick_sort>
L13:
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <quick_sort>
  xor    %eax,%eax
L105:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L60:
  cmp    %ecx,%eax
  jle    L59
  xor    %r10d,%esi
  mov    %esi,(%rbx)
  xor    (%rdi),%esi
  mov    %esi,(%rdi)
  xor    %esi,(%rbx)
  jmp    L50
L35:
  mov    $0xffffffff,%eax
  jmp    L105
L58:
  or     $0xffffffff,%eax
  ret
```

assistant: ```c
int quick_sort(int *arr, int left, int right) {
  int pivot;
  int i, j;
  if (NULL == arr || left >= right)
    return -1;
  pivot = arr[left];
  i = left;
  j = right + 1;
  while (i < j) {
    do {
      j--;
    } while (arr[j] > pivot);
    do {
      i++;
    } while (i <= j && arr[i] <= pivot);
    if (i < j) {
      do {
        arr[i] = arr[i] ^ arr[j];
        arr[j] = arr[i] ^ arr[j];
        arr[i] = arr[i] ^ arr[j];
      } while (0);
    }
  }
  if (left < j) {
    do {
      arr[left] = arr[left] ^ arr[j];
      arr[j] = arr[left] ^ arr[j];
      arr[left] = arr[left] ^ arr[j];
    } while (0);
    quick_sort(arr, left, j - 1);
  }
  quick_sort(arr, j + 1, right);
  return 0;
}
```