ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<delay>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  nop
L6:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jg     L6
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void delay(int i) {
  while (i-- > 0)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<delay>:
  endbr64
L0:
  mov    %edi,%eax
  sub    $0x1,%edi
  test   %eax,%eax
  jg     L0
  ret
```

assistant: ```c
void delay(int i) {
  while (i-- > 0)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<delay>:
  endbr64
  ret
```

assistant: ```c
void delay(int i) {
  while (i-- > 0)
    ;
}
```
user: What is the c source code of the assembly code below?

```asm
<delay>:
  endbr64
  ret
```

assistant: ```c
void delay(int i) {
  while (i-- > 0)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<match_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L14
L3:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  je     L19
  addl   $0x1,-0x4(%rbp)
L19:
  addl   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int match_size(int *matching, int nleft) {
  int i, nmatch;
  nmatch = 0;
  for (i = 0; i < nleft; i++) {
    if (matching[i] != -1) {
      ++nmatch;
    }
  }
  return (nmatch);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<match_size>:
  endbr64
  test   %esi,%esi
  jle    L8
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rsi
  mov    $0x0,%edx
L14:
  cmpl   $0xffffffff,(%rax)
  setne  %cl
  movzbl %cl,%ecx
  add    %ecx,%edx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L14
L29:
  mov    %edx,%eax
  ret
L8:
  mov    $0x0,%edx
  jmp    L29
```

assistant: ```c
int match_size(int *matching, int nleft) {
  int i, nmatch;
  nmatch = 0;
  for (i = 0; i < nleft; i++) {
    if (matching[i] != -1) {
      ++nmatch;
    }
  }
  return (nmatch);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<match_size>:
  endbr64
  test   %esi,%esi
  jle    L16
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L14:
  xor    %edx,%edx
  cmpl   $0xffffffff,(%rdi)
  setne  %dl
  add    $0x4,%rdi
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L14
  ret
  nopl   0x0(%rax)
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int match_size(int *matching, int nleft) {
  int i, nmatch;
  nmatch = 0;
  for (i = 0; i < nleft; i++) {
    if (matching[i] != -1) {
      ++nmatch;
    }
  }
  return (nmatch);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<match_size>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L58
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L43
  mov    %esi,%ecx
  pxor   %xmm1,%xmm1
  pcmpeqd %xmm3,%xmm3
  mov    %rdi,%rax
  shr    $0x2,%ecx
  movdqa %xmm1,%xmm2
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L24:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  pcmpeqd %xmm3,%xmm0
  pcmpeqd %xmm2,%xmm0
  psubd  %xmm0,%xmm1
  cmp    %rax,%rcx
  jne    L24
  movdqa %xmm1,%xmm0
  mov    %esi,%ecx
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%ecx
  paddd  %xmm0,%xmm1
  movdqa %xmm1,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm1
  movd   %xmm1,%eax
  test   $0x3,%sil
  je     L99
L36:
  movslq %ecx,%rdi
  cmpl   $0xffffffff,(%rdx,%rdi,4)
  lea    0x0(,%rdi,4),%r8
  je     L95
  add    $0x1,%eax
L95:
  lea    0x1(%rcx),%edi
  cmp    %esi,%edi
  jge    L90
  cmpl   $0xffffffff,0x4(%rdx,%r8,1)
  je     L25
  add    $0x1,%eax
L25:
  add    $0x2,%ecx
  cmp    %ecx,%esi
  jle    L90
  cmpl   $0xffffffff,0x8(%rdx,%r8,1)
  je     L90
  add    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L58:
  xor    %eax,%eax
L90:
  ret
  nopl   0x0(%rax,%rax,1)
L99:
  ret
L43:
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L36
```

assistant: ```c
int match_size(int *matching, int nleft) {
  int i, nmatch;
  nmatch = 0;
  for (i = 0; i < nleft; i++) {
    if (matching[i] != -1) {
      ++nmatch;
    }
  }
  return (nmatch);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <pthread_exit@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  pthread_exit(NULL);
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <pthread_exit@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  pthread_exit(NULL);
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <pthread_exit@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  pthread_exit(NULL);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <pthread_exit@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  pthread_exit(NULL);
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bn_cmp_words>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L82
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L9
  mov    $0x1,%eax
  jmp    L67
L9:
  mov    $0xffffffff,%eax
  jmp    L67
L82:
  mov    -0x34(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L91
L61:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L37
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L18
  mov    $0x1,%eax
  jmp    L67
L18:
  mov    $0xffffffff,%eax
  jmp    L67
L37:
  subl   $0x1,-0x14(%rbp)
L91:
  cmpl   $0x0,-0x14(%rbp)
  jns    L61
  mov    $0x0,%eax
L67:
  pop    %rbp
  ret
```

assistant: ```c
int bn_cmp_words(unsigned long *a, unsigned long *b, int n) {
  int i;
  unsigned long aa, bb;
  aa = a[n - 1];
  bb = b[n - 1];
  if (aa != bb)
    return ((aa > bb) ? 1 : -1);
  for (i = n - 2; i >= 0; i--) {
    aa = a[i];
    bb = b[i];
    if (aa != bb)
      return ((aa > bb) ? 1 : -1);
  }
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bn_cmp_words>:
  endbr64
  movslq %edx,%rax
  shl    $0x3,%rax
  mov    -0x8(%rdi,%rax,1),%rcx
  mov    -0x8(%rsi,%rax,1),%rax
  cmp    %rax,%rcx
  je     L34
  cmp    %rcx,%rax
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  ret
L34:
  sub    $0x2,%edx
  js     L20
  movslq %edx,%rax
L22:
  mov    (%rdi,%rax,8),%rdx
  mov    (%rsi,%rax,8),%rcx
  cmp    %rcx,%rdx
  jne    L35
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L22
  mov    $0x0,%eax
  ret
L35:
  cmp    %rdx,%rcx
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  ret
L20:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int bn_cmp_words(unsigned long *a, unsigned long *b, int n) {
  int i;
  unsigned long aa, bb;
  aa = a[n - 1];
  bb = b[n - 1];
  if (aa != bb)
    return ((aa > bb) ? 1 : -1);
  for (i = n - 2; i >= 0; i--) {
    aa = a[i];
    bb = b[i];
    if (aa != bb)
      return ((aa > bb) ? 1 : -1);
  }
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bn_cmp_words>:
  endbr64
  movslq %edx,%rax
  shl    $0x3,%rax
  mov    -0x8(%rdi,%rax,1),%rcx
  mov    -0x8(%rsi,%rax,1),%rax
  cmp    %rax,%rcx
  je     L9
  cmp    %rcx,%rax
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  sub    $0x2,%edx
  js     L20
  movslq %edx,%rax
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L24:
  sub    $0x1,%rax
  test   %eax,%eax
  js     L20
L15:
  mov    (%rdi,%rax,8),%rdx
  mov    (%rsi,%rax,8),%rcx
  cmp    %rcx,%rdx
  je     L24
  cmp    %rdx,%rcx
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  ret
  nopl   0x0(%rax)
L20:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bn_cmp_words(unsigned long *a, unsigned long *b, int n) {
  int i;
  unsigned long aa, bb;
  aa = a[n - 1];
  bb = b[n - 1];
  if (aa != bb)
    return ((aa > bb) ? 1 : -1);
  for (i = n - 2; i >= 0; i--) {
    aa = a[i];
    bb = b[i];
    if (aa != bb)
      return ((aa > bb) ? 1 : -1);
  }
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bn_cmp_words>:
  endbr64
  movslq %edx,%rax
  shl    $0x3,%rax
  mov    -0x8(%rdi,%rax,1),%rcx
  mov    -0x8(%rsi,%rax,1),%rax
  cmp    %rax,%rcx
  je     L14
  cmp    %rcx,%rax
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  sub    $0x2,%edx
  js     L48
  movslq %edx,%rax
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L0:
  sub    $0x1,%rax
  test   %eax,%eax
  js     L48
L21:
  mov    (%rdi,%rax,8),%rdx
  mov    (%rsi,%rax,8),%rcx
  cmp    %rcx,%rdx
  je     L0
  cmp    %rdx,%rcx
  sbb    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  ret
  nopl   0x0(%rax)
L48:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bn_cmp_words(unsigned long *a, unsigned long *b, int n) {
  int i;
  unsigned long aa, bb;
  aa = a[n - 1];
  bb = b[n - 1];
  if (aa != bb)
    return ((aa > bb) ? 1 : -1);
  for (i = n - 2; i >= 0; i--) {
    aa = a[i];
    bb = b[i];
    if (aa != bb)
      return ((aa > bb) ? 1 : -1);
  }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<__assert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    $0x0,%eax
  call   <abort@plt>
  nop
  leave
  ret
```

assistant: ```c
void __assert(const char *file, int line, const char *func, const char *failedexpr) {
  (void)(file);
  (void)(line);
  (void)(func);
  (void)(failedexpr);
  abort();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__assert>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <abort@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __assert(const char *file, int line, const char *func, const char *failedexpr) {
  (void)(file);
  (void)(line);
  (void)(func);
  (void)(failedexpr);
  abort();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__assert>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <abort@plt>
```

assistant: ```c
void __assert(const char *file, int line, const char *func, const char *failedexpr) {
  (void)(file);
  (void)(line);
  (void)(func);
  (void)(failedexpr);
  abort();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<__assert>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <abort@plt>
```

assistant: ```c
void __assert(const char *file, int line, const char *func, const char *failedexpr) {
  (void)(file);
  (void)(line);
  (void)(func);
  (void)(failedexpr);
  abort();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mbsinit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int mbsinit(const mbstate_t *ps) {
  ps = ps;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mbsinit>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int mbsinit(const mbstate_t *ps) {
  ps = ps;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mbsinit>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int mbsinit(const mbstate_t *ps) {
  ps = ps;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<mbsinit>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int mbsinit(const mbstate_t *ps) {
  ps = ps;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<USBHostRequestStatus>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x40050000,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  add    $0x102,%eax
  cltq
  movl   $0x60,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void USBHostRequestStatus(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000102))) = 0x00000020 | 0x00000040;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<USBHostRequestStatus>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x40050000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  add    $0x102,%ebx
  movslq %ebx,%rbx
  movl   $0x60,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void USBHostRequestStatus(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000102))) = 0x00000020 | 0x00000040;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<USBHostRequestStatus>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x40050000,%ebx
  sete   %dil
  add    $0x102,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  movl   $0x60,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void USBHostRequestStatus(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000102))) = 0x00000020 | 0x00000040;
}
```
user: What is the c source code of the assembly code below?

```asm
<USBHostRequestStatus>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x40050000,%ebx
  sete   %dil
  add    $0x102,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  movl   $0x60,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void USBHostRequestStatus(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000102))) = 0x00000020 | 0x00000040;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strnprcbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x150,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %rsi,-0x140(%rbp)
  mov    %rdx,-0x148(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x138(%rbp),%rax
  mov    %rax,-0x120(%rbp)
  movq   $0x0,-0x118(%rbp)
  lea    -0x110(%rbp),%rax
  mov    $0x100,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L24
L27:
  movsbq -0x121(%rbp),%rax
  movb   $0x1,-0x110(%rbp,%rax,1)
L24:
  mov    -0x140(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x140(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x121(%rbp)
  cmpb   $0x0,-0x121(%rbp)
  jne    L27
  jmp    L56
L7:
  movsbq -0x121(%rbp),%rax
  movzbl -0x110(%rbp,%rax,1),%eax
  test   %al,%al
  je     L56
  mov    -0x120(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x118(%rbp)
L56:
  mov    -0x148(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x148(%rbp)
  test   %rax,%rax
  je     L8
  mov    -0x120(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x120(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x121(%rbp)
  cmpb   $0x0,-0x121(%rbp)
  jne    L7
L8:
  mov    -0x118(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
char *(strnprcbrk)(const char *string, const char *skipset, size_t maxlen) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256 * sizeof(char));
  while ((c = *skipset++))
    set[(size_t)c] = 1;
  while (maxlen-- && ((c = *s++)))
    if (set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: What is the c source code of the assembly code below:

```asm
<strnprcbrk>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x100,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  lea    0x1(%r12),%rdx
  movzbl (%r12),%eax
  test   %al,%al
  je     L53
L23:
  movsbq %al,%rax
  movb   $0x1,(%rsp,%rax,1)
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L23
L53:
  test   %rbp,%rbp
  je     L26
  lea    (%rbx,%rbp,1),%rsi
  mov    $0x0,%edx
L25:
  mov    %rbx,%rcx
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  je     L69
  movsbq %al,%rax
  cmpb   $0x0,(%rsp,%rax,1)
  cmovne %rcx,%rdx
  cmp    %rbx,%rsi
  jne    L25
L69:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    %rdx,%rax
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  mov    $0x0,%edx
  jmp    L69
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *(strnprcbrk)(const char *string, const char *skipset, size_t maxlen) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256 * sizeof(char));
  while ((c = *skipset++))
    set[(size_t)c] = 1;
  while (maxlen-- && ((c = *s++)))
    if (set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strnprcbrk>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x100,%edx
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memset@plt>
  movsbq (%r12),%rax
  test   %al,%al
  je     L26
  lea    0x1(%r12),%rcx
  nopl   0x0(%rax,%rax,1)
L70:
  add    $0x1,%rcx
  movb   $0x1,(%rsp,%rax,1)
  movsbq -0x1(%rcx),%rax
  test   %al,%al
  jne    L70
L26:
  test   %rbp,%rbp
  je     L58
  lea    (%rbx,%rbp,1),%rdx
  xor    %r8d,%r8d
  jmp    L67
  nop
L64:
  cmpb   $0x0,(%rsp,%rax,1)
  cmovne %rcx,%r8
  cmp    %rdx,%rbx
  je     L52
L67:
  mov    %rbx,%rcx
  movsbq (%rbx),%rax
  add    $0x1,%rbx
  test   %al,%al
  jne    L64
L52:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x110,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L58:
  xor    %r8d,%r8d
  jmp    L52
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *(strnprcbrk)(const char *string, const char *skipset, size_t maxlen) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256 * sizeof(char));
  while ((c = *skipset++))
    set[(size_t)c] = 1;
  while (maxlen-- && ((c = *s++)))
    if (set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: What is the c source code of the assembly code below:

```asm
<strnprcbrk>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x100,%edx
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memset@plt>
  movsbq (%r12),%rax
  test   %al,%al
  je     L78
  lea    0x1(%r12),%rcx
  nopl   0x0(%rax,%rax,1)
L66:
  add    $0x1,%rcx
  movb   $0x1,(%rsp,%rax,1)
  movsbq -0x1(%rcx),%rax
  test   %al,%al
  jne    L66
L78:
  test   %rbp,%rbp
  je     L43
  lea    (%rbx,%rbp,1),%rdx
  xor    %r8d,%r8d
  jmp    L79
  nop
L13:
  cmpb   $0x0,(%rsp,%rax,1)
  cmovne %rcx,%r8
  cmp    %rdx,%rbx
  je     L38
L79:
  mov    %rbx,%rcx
  movsbq (%rbx),%rax
  add    $0x1,%rbx
  test   %al,%al
  jne    L13
L38:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x110,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L43:
  xor    %r8d,%r8d
  jmp    L38
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *(strnprcbrk)(const char *string, const char *skipset, size_t maxlen) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256 * sizeof(char));
  while ((c = *skipset++))
    set[(size_t)c] = 1;
  while (maxlen-- && ((c = *s++)))
    if (set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<GetReverseComplimentAnyCase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L13
L7:
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x30(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <GetReverseComplimentAnyCaseBase@plt>
  mov    %al,(%rbx)
  subl   $0x1,-0x14(%rbp)
L13:
  cmpl   $0x0,-0x14(%rbp)
  jns    L7
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void GetReverseComplimentAnyCase(char *s, char *r, int length) {
  int i;
  for (i = length - 1; i >= 0; i--) {
    r[i] = GetReverseComplimentAnyCaseBase(s[length - 1 - i]);
  }
  r[length] = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<GetReverseComplimentAnyCase>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r14
  mov    %edx,%r13d
  mov    %edx,%ebp
  sub    $0x1,%ebp
  js     L8
  mov    %rdi,%rbx
  movslq %ebp,%rbp
  add    %rsi,%rbp
  lea    -0x1(%rdx),%eax
  lea    0x1(%rdi,%rax,1),%r12
L17:
  movsbl (%rbx),%edi
  call   <GetReverseComplimentAnyCaseBase@plt>
  mov    %al,0x0(%rbp)
  add    $0x1,%rbx
  sub    $0x1,%rbp
  cmp    %r12,%rbx
  jne    L17
L8:
  movslq %r13d,%r13
  movb   $0x0,(%r14,%r13,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void GetReverseComplimentAnyCase(char *s, char *r, int length) {
  int i;
  for (i = length - 1; i >= 0; i--) {
    r[i] = GetReverseComplimentAnyCaseBase(s[length - 1 - i]);
  }
  r[length] = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<GetReverseComplimentAnyCase>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  movslq %edx,%r12
  mov    %r12d,%eax
  push   %rbp
  push   %rbx
  sub    $0x1,%eax
  js     L6
  movslq %eax,%rbp
  mov    %eax,%eax
  mov    %rdi,%rbx
  add    %rsi,%rbp
  lea    0x1(%rdi,%rax,1),%r14
  nopw   0x0(%rax,%rax,1)
L17:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  sub    $0x1,%rbp
  call   <GetReverseComplimentAnyCaseBase@plt>
  mov    %al,0x1(%rbp)
  cmp    %r14,%rbx
  jne    L17
L6:
  pop    %rbx
  pop    %rbp
  movb   $0x0,0x0(%r13,%r12,1)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void GetReverseComplimentAnyCase(char *s, char *r, int length) {
  int i;
  for (i = length - 1; i >= 0; i--) {
    r[i] = GetReverseComplimentAnyCaseBase(s[length - 1 - i]);
  }
  r[length] = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<GetReverseComplimentAnyCase>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  movslq %edx,%r12
  mov    %r12d,%eax
  push   %rbp
  push   %rbx
  sub    $0x1,%eax
  js     L2
  movslq %eax,%rbp
  mov    %eax,%eax
  mov    %rdi,%rbx
  add    %rsi,%rbp
  lea    0x1(%rdi,%rax,1),%r14
  nopw   0x0(%rax,%rax,1)
L3:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  sub    $0x1,%rbp
  call   <GetReverseComplimentAnyCaseBase@plt>
  mov    %al,0x1(%rbp)
  cmp    %r14,%rbx
  jne    L3
L2:
  pop    %rbx
  pop    %rbp
  movb   $0x0,0x0(%r13,%r12,1)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void GetReverseComplimentAnyCase(char *s, char *r, int length) {
  int i;
  for (i = length - 1; i >= 0; i--) {
    r[i] = GetReverseComplimentAnyCaseBase(s[length - 1 - i]);
  }
  r[length] = '\0';
}
```