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
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D72(%rip),%rdi
  call   <system@plt>
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D21(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%eax
  cmp    %eax,%edx
  jle    L35
  mov    %edx,(%rsp)
  mov    %eax,0x4(%rsp)
L35:
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  je     L18
  mov    (%rsp),%esi
  mov    %esi,%eax
  cltd
  idiv   %edi
  mov    %eax,%ecx
  mov    %edi,%edx
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  mov    0x4(%rsp),%edi
  mov    %esi,%eax
  cltd
  idiv   %edi
  mov    %edx,%ecx
  mov    %edi,%edx
  lea    D66(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdl (%rsp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdl 0x4(%rsp),%xmm1
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  lea    D38(%rip),%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L13:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  ret
L18:
  lea    D78(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L13
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D72
datatype: string
name: disassemble_data
arguments: tag: D9
datatype: string
name: disassemble_data
arguments: tag: D21
datatype: string
name: disassemble_data
arguments: tag: D58
datatype: string
name: disassemble_data
arguments: tag: D2
datatype: string
name: disassemble_data
arguments: tag: D66
datatype: string
name: disassemble_data
arguments: tag: D38
datatype: string
name: disassemble_data
arguments: tag: D78
datatype: string

tool: D72: clear

tool: D9: Please enter an integer: 

tool: D21: %d

tool: D58: Please enter a second integer: 

tool: D2: 
%d / %d = %d


tool: D66: %d %% %d = %d


tool: D38: %lf / %lf = %lf



tool: D78: 
You cannot divide by zero, program ending...


assistant: ```c
int main(int argc, char **argv) {
  int userint1;
  int userint2;
  int swap;
  double double1;
  double double2;
  system("clear");
  printf("Please enter an integer: ");
  scanf("%d", &userint1);
  printf("Please enter a second integer: ");
  scanf("%d", &userint2);
  if (userint2 > userint1) {
    swap = userint1;
    userint1 = userint2;
    userint2 = swap;
  }
  if (userint2 == 0) {
    printf("\nYou cannot divide by zero, program ending...\n\n");
    return (1);
  }
  printf("\n%d / %d = %d\n", userint1, userint2, (userint1 / userint2));
  printf("%d %% %d = %d\n", userint1, userint2, (userint1 % userint2));
  double1 = userint1;
  double2 = userint2;
  printf("%lf / %lf = %lf\n\n", double1, double2, (double1 / double2));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D39(%rip),%rdi
  lea    D59(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <system@plt>
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D67(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%eax
  cmp    %eax,%r8d
  jle    L6
  mov    %eax,0x4(%rsp)
  mov    %r8d,(%rsp)
  mov    %eax,%r8d
L6:
  test   %r8d,%r8d
  je     L27
  mov    (%rsp),%esi
  lea    D53(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r8d
  mov    %r8d,%edx
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  mov    0x4(%rsp),%r8d
  lea    D41(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r8d
  xor    %eax,%eax
  mov    %edx,%ecx
  mov    %r8d,%edx
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  lea    D16(%rip),%rdi
  cvtsi2sdl (%rsp),%xmm0
  mov    $0x3,%eax
  cvtsi2sdl 0x4(%rsp),%xmm1
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  call   <printf@plt>
  xor    %eax,%eax
L3:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x10,%rsp
  pop    %rbp
  ret
L27:
  lea    D74(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L3
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D39
datatype: string
name: read_data
arguments: data_tag: D59
datatype: string
name: read_data
arguments: data_tag: D27
datatype: string
name: read_data
arguments: data_tag: D67
datatype: string
name: read_data
arguments: data_tag: D53
datatype: string
name: read_data
arguments: data_tag: D41
datatype: string
name: read_data
arguments: data_tag: D16
datatype: string
name: read_data
arguments: data_tag: D74
datatype: string

tool: D39: clear

tool: D59: %d

tool: D27: Please enter an integer: 

tool: D67: Please enter a second integer: 

tool: D53: 
%d / %d = %d


tool: D41: %d %% %d = %d


tool: D16: %lf / %lf = %lf



tool: D74: 
You cannot divide by zero, program ending...


assistant: ```c
int main(int argc, char **argv) {
  int userint1;
  int userint2;
  int swap;
  double double1;
  double double2;
  system("clear");
  printf("Please enter an integer: ");
  scanf("%d", &userint1);
  printf("Please enter a second integer: ");
  scanf("%d", &userint2);
  if (userint2 > userint1) {
    swap = userint1;
    userint1 = userint2;
    userint2 = swap;
  }
  if (userint2 == 0) {
    printf("\nYou cannot divide by zero, program ending...\n\n");
    return (1);
  }
  printf("\n%d / %d = %d\n", userint1, userint2, (userint1 / userint2));
  printf("%d %% %d = %d\n", userint1, userint2, (userint1 % userint2));
  double1 = userint1;
  double2 = userint2;
  printf("%lf / %lf = %lf\n\n", double1, double2, (double1 / double2));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D29(%rip),%rdi
  lea    D23(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <system@plt>
  lea    D36(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%eax
  cmp    %eax,%r8d
  jle    L24
  mov    %eax,0x4(%rsp)
  mov    %r8d,(%rsp)
  mov    %eax,%r8d
L24:
  test   %r8d,%r8d
  je     L19
  mov    (%rsp),%esi
  lea    D62(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r8d
  mov    %r8d,%edx
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%esi
  mov    0x4(%rsp),%r8d
  lea    D13(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r8d
  xor    %eax,%eax
  mov    %edx,%ecx
  mov    %r8d,%edx
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  lea    D59(%rip),%rdi
  cvtsi2sdl (%rsp),%xmm0
  mov    $0x3,%eax
  cvtsi2sdl 0x4(%rsp),%xmm1
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  call   <printf@plt>
  xor    %eax,%eax
L20:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x10,%rsp
  pop    %rbp
  ret
L19:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L20
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D29
type: string
name: parse_data
arguments: data_label: D23
type: string
name: parse_data
arguments: data_label: D36
type: string
name: parse_data
arguments: data_label: D10
type: string
name: parse_data
arguments: data_label: D62
type: string
name: parse_data
arguments: data_label: D13
type: string
name: parse_data
arguments: data_label: D59
type: string
name: parse_data
arguments: data_label: D35
type: string

tool: D29: clear

tool: D23: %d

tool: D36: Please enter an integer: 

tool: D10: Please enter a second integer: 

tool: D62: 
%d / %d = %d


tool: D13: %d %% %d = %d


tool: D59: %lf / %lf = %lf



tool: D35: 
You cannot divide by zero, program ending...


assistant: ```c
int main(int argc, char **argv) {
  int userint1;
  int userint2;
  int swap;
  double double1;
  double double2;
  system("clear");
  printf("Please enter an integer: ");
  scanf("%d", &userint1);
  printf("Please enter a second integer: ");
  scanf("%d", &userint2);
  if (userint2 > userint1) {
    swap = userint1;
    userint1 = userint2;
    userint2 = swap;
  }
  if (userint2 == 0) {
    printf("\nYou cannot divide by zero, program ending...\n\n");
    return (1);
  }
  printf("\n%d / %d = %d\n", userint1, userint2, (userint1 / userint2));
  printf("%d %% %d = %d\n", userint1, userint2, (userint1 % userint2));
  double1 = userint1;
  double2 = userint2;
  printf("%lf / %lf = %lf\n\n", double1, double2, (double1 / double2));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<remove_quote>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L1
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L14
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L14:
  addq   $0x1,-0x8(%rbp)
L1:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *remove_quote(char *p) {
  if (p[0] == '"') {
    if (p[strlen(p) - 1] == '"')
      p[strlen(p) - 1] = '\0';
    p++;
  }
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<remove_quote>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  cmpb   $0x22,(%rdi)
  je     L6
L0:
  mov    %rbx,%rax
  pop    %rbx
  ret
L6:
  call   <strlen@plt>
  cltq
  cmpb   $0x22,-0x1(%rbx,%rax,1)
  je     L20
L28:
  add    $0x1,%rbx
  jmp    L0
L20:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbx,%rax,1)
  jmp    L28
```

assistant: ```c
char *remove_quote(char *p) {
  if (p[0] == '"') {
    if (p[strlen(p) - 1] == '"')
      p[strlen(p) - 1] = '\0';
    p++;
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<remove_quote>:
  endbr64
  push   %r12
  cmpb   $0x22,(%rdi)
  mov    %rdi,%r12
  je     L25
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L25:
  call   <strlen@plt>
  cltq
  cmpb   $0x22,-0x1(%r12,%rax,1)
  je     L11
L4:
  add    $0x1,%r12
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L11:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  jmp    L4
```

assistant: ```c
char *remove_quote(char *p) {
  if (p[0] == '"') {
    if (p[strlen(p) - 1] == '"')
      p[strlen(p) - 1] = '\0';
    p++;
  }
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<remove_quote>:
  endbr64
  push   %r12
  cmpb   $0x22,(%rdi)
  mov    %rdi,%r12
  je     L4
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L4:
  call   <strlen@plt>
  cltq
  cmpb   $0x22,-0x1(%r12,%rax,1)
  je     L16
L8:
  add    $0x1,%r12
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L16:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%r12,%rax,1)
  jmp    L8
```

assistant: ```c
char *remove_quote(char *p) {
  if (p[0] == '"') {
    if (p[strlen(p) - 1] == '"')
      p[strlen(p) - 1] = '\0';
    p++;
  }
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L3:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D9: Out of memory!


assistant: ```c
void *my_malloc(long int size) {
  void *ptr;
  ptr = (void *)malloc(size);
  if (!ptr) {
    perror("Out of memory!\n");
    exit(1);
  }
  return (ptr);
}
```
user: What is the c source code of the assembly code below:

```asm
<my_malloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L11
L0:
  mov    %rbx,%rax
  pop    %rbx
  ret
L11:
  lea    D8(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L0
```

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

tool: D8: Out of memory!


assistant: ```c
void *my_malloc(long int size) {
  void *ptr;
  ptr = (void *)malloc(size);
  if (!ptr) {
    perror("Out of memory!\n");
    exit(1);
  }
  return (ptr);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<my_malloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  lea    D7(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D7: Out of memory!


assistant: ```c
void *my_malloc(long int size) {
  void *ptr;
  ptr = (void *)malloc(size);
  if (!ptr) {
    perror("Out of memory!\n");
    exit(1);
  }
  return (ptr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_malloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L7
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  lea    D0(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_infered: string

tool: D0: Out of memory!


assistant: ```c
void *my_malloc(long int size) {
  void *ptr;
  ptr = (void *)malloc(size);
  if (!ptr) {
    perror("Out of memory!\n");
    exit(1);
  }
  return (ptr);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<checksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0x24(%rbp)
  jle    L16
  movq   $0x0,-0x18(%rbp)
  movl   $0x1,-0x1c(%rbp)
  jmp    L10
L48:
  movb   $0x0,-0xa(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0xa(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L10
  movzbl -0xa(%rbp),%eax
  movzbl %al,%eax
  add    %rax,-0x18(%rbp)
L10:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L48
  mov    -0x18(%rbp),%rax
  jmp    L27
L16:
  mov    $0x0,%eax
L27:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

assistant: ```c
unsigned long long checksum(int file) {
  if (file > 0) {
    unsigned long long sum = 0;
    unsigned char buffer[2];
    int c = 1;
    while (c != 0) {
      buffer[0] = 0;
      buffer[1] = 0;
      c = read(file, buffer, 1);
      if (c != 0)
        sum += (unsigned long long)buffer[0];
    }
    return sum;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<checksum>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  test   %edi,%edi
  jg     L29
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    %rbp,%rax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  mov    %edi,%ebx
  lea    0x6(%rsp),%r12
L33:
  movb   $0x0,0x6(%rsp)
  movb   $0x0,0x7(%rsp)
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <read@plt>
  test   %eax,%eax
  je     L14
  movzbl 0x6(%rsp),%eax
  add    %rax,%rbp
  jmp    L33
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long checksum(int file) {
  if (file > 0) {
    unsigned long long sum = 0;
    unsigned char buffer[2];
    int c = 1;
    while (c != 0) {
      buffer[0] = 0;
      buffer[1] = 0;
      c = read(file, buffer, 1);
      if (c != 0)
        sum += (unsigned long long)buffer[0];
    }
    return sum;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<checksum>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jg     L20
L30:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L20:
  mov    %edi,%ebx
  lea    0x6(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  mov    %ax,0x6(%rsp)
  call   <read@plt>
  test   %eax,%eax
  je     L30
  movzbl 0x6(%rsp),%eax
  add    %rax,%r12
  jmp    L7
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long checksum(int file) {
  if (file > 0) {
    unsigned long long sum = 0;
    unsigned char buffer[2];
    int c = 1;
    while (c != 0) {
      buffer[0] = 0;
      buffer[1] = 0;
      c = read(file, buffer, 1);
      if (c != 0)
        sum += (unsigned long long)buffer[0];
    }
    return sum;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<checksum>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jg     L23
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L23:
  mov    %edi,%ebx
  lea    0x6(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L37:
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  mov    %ax,0x6(%rsp)
  call   <read@plt>
  test   %eax,%eax
  je     L4
  movzbl 0x6(%rsp),%eax
  add    %rax,%r12
  jmp    L37
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long long checksum(int file) {
  if (file > 0) {
    unsigned long long sum = 0;
    unsigned char buffer[2];
    int c = 1;
    while (c != 0) {
      buffer[0] = 0;
      buffer[1] = 0;
      c = read(file, buffer, 1);
      if (c != 0)
        sum += (unsigned long long)buffer[0];
    }
    return sum;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<audio_out_format_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4000,-0x4(%rbp)
  je     L123
  cmpl   $0x4000,-0x4(%rbp)
  jg     L41
  cmpl   $0x2000,-0x4(%rbp)
  je     L49
  cmpl   $0x2000,-0x4(%rbp)
  jg     L41
  cmpl   $0x1000,-0x4(%rbp)
  je     L144
  cmpl   $0x1000,-0x4(%rbp)
  jg     L41
  cmpl   $0x100,-0x4(%rbp)
  je     L96
  cmpl   $0x100,-0x4(%rbp)
  jg     L41
  cmpl   $0x80,-0x4(%rbp)
  je     L96
  cmpl   $0x80,-0x4(%rbp)
  jg     L41
  cmpl   $0x10,-0x4(%rbp)
  je     L96
  cmpl   $0x20,-0x4(%rbp)
  jne    L41
L96:
  mov    $0x10,%eax
  jmp    L134
L144:
  mov    $0x20,%eax
  jmp    L134
L49:
  mov    $0x20,%eax
  jmp    L134
L123:
  mov    $0x20,%eax
  jmp    L134
L41:
  mov    $0x8,%eax
L134:
  pop    %rbp
  ret
```

assistant: ```c
int audio_out_format_bits(int format) {
  switch (format) {
  case 0x00000010:
  case 0x00000020:
  case 0x00000080:
  case 0x00000100:
    return 16;
  case 0x00001000:
    return 32;
  case 0x00002000:
    return 32;
  case 0x00004000:
    return 32;
  case 0x00000001:
  case 0x00000002:
  case 0x00000004:
  case 0x00000040:
  case 0x00000008:
  case 0x00000200:
  case 0x00000400:
  default:
    return 8;
  }
  return 8;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<audio_out_format_bits>:
  endbr64
  cmp    $0x1000,%edi
  je     L39
  jle    L2
  sub    $0x2000,%edi
  and    $0xffffdfff,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x18,%eax
  add    $0x8,%eax
  ret
L2:
  cmp    $0x80,%edi
  je     L23
  jle    L16
  cmp    $0x100,%edi
  sete   %al
  movzbl %al,%eax
  lea    0x8(,%rax,8),%eax
  ret
L16:
  sub    $0x10,%edi
  and    $0xffffffef,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x8,%eax
  add    $0x8,%eax
  ret
L39:
  mov    $0x20,%eax
  ret
L23:
  mov    $0x10,%eax
  ret
```

assistant: ```c
int audio_out_format_bits(int format) {
  switch (format) {
  case 0x00000010:
  case 0x00000020:
  case 0x00000080:
  case 0x00000100:
    return 16;
  case 0x00001000:
    return 32;
  case 0x00002000:
    return 32;
  case 0x00004000:
    return 32;
  case 0x00000001:
  case 0x00000002:
  case 0x00000004:
  case 0x00000040:
  case 0x00000008:
  case 0x00000200:
  case 0x00000400:
  default:
    return 8;
  }
  return 8;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<audio_out_format_bits>:
  endbr64
  cmp    $0x1000,%edi
  je     L39
  jle    L14
  sub    $0x2000,%edi
  and    $0xffffdfff,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x18,%eax
  add    $0x8,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  cmp    $0x80,%edi
  je     L28
  jle    L12
  xor    %eax,%eax
  cmp    $0x100,%edi
  sete   %al
  lea    0x8(,%rax,8),%eax
  ret
  nopl   (%rax)
L12:
  sub    $0x10,%edi
  and    $0xffffffef,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x8,%eax
  add    $0x8,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  mov    $0x20,%eax
  ret
  xchg   %ax,%ax
L28:
  mov    $0x10,%eax
  ret
```

assistant: ```c
int audio_out_format_bits(int format) {
  switch (format) {
  case 0x00000010:
  case 0x00000020:
  case 0x00000080:
  case 0x00000100:
    return 16;
  case 0x00001000:
    return 32;
  case 0x00002000:
    return 32;
  case 0x00004000:
    return 32;
  case 0x00000001:
  case 0x00000002:
  case 0x00000004:
  case 0x00000040:
  case 0x00000008:
  case 0x00000200:
  case 0x00000400:
  default:
    return 8;
  }
  return 8;
}
```
user: What is the c source code of the assembly code below:

```asm
<audio_out_format_bits>:
  endbr64
  cmp    $0x1000,%edi
  je     L17
  jle    L37
  sub    $0x2000,%edi
  and    $0xffffdfff,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x18,%eax
  add    $0x8,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L37:
  cmp    $0x80,%edi
  je     L0
  jle    L36
  xor    %eax,%eax
  cmp    $0x100,%edi
  sete   %al
  lea    0x8(,%rax,8),%eax
  ret
  nopl   (%rax)
L36:
  sub    $0x10,%edi
  and    $0xffffffef,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x8,%eax
  add    $0x8,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    $0x20,%eax
  ret
  xchg   %ax,%ax
L0:
  mov    $0x10,%eax
  ret
```

assistant: ```c
int audio_out_format_bits(int format) {
  switch (format) {
  case 0x00000010:
  case 0x00000020:
  case 0x00000080:
  case 0x00000100:
    return 16;
  case 0x00001000:
    return 32;
  case 0x00002000:
    return 32;
  case 0x00004000:
    return 32;
  case 0x00000001:
  case 0x00000002:
  case 0x00000004:
  case 0x00000040:
  case 0x00000008:
  case 0x00000200:
  case 0x00000400:
  default:
    return 8;
  }
  return 8;
}
```