ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<clean>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,-0x4c(%rbp)
  test   %eax,%eax
  je     L75
  mov    -0x4c(%rbp),%ebx
  mov    %rsp,-0x58(%rbp)
  lea    0x0(,%rbx,8),%edi
  call   <malloc@plt>
  mov    %rsp,%rcx
  mov    %rax,%r12
  movslq %ebx,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L86
L34:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L34
L86:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L6
  orq    $0x0,-0x8(%rsp,%rdx,1)
L6:
  mov    -0x4c(%rbp),%eax
  mov    %rsp,%r14
  test   %eax,%eax
  jle    L16
  sub    $0x1,%eax
  mov    %r13,%r15
  xor    %ebx,%ebx
  lea    0x1(%r13,%rax,1),%rax
  mov    %rax,-0x48(%rbp)
L106:
  movsbl (%r15),%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L11
  movzbl (%r15),%ecx
  movslq %ebx,%rax
  add    $0x1,%ebx
  mov    %cl,(%r14,%rax,1)
L11:
  add    $0x1,%r15
  cmp    -0x48(%rbp),%r15
  jne    L106
  cmp    -0x4c(%rbp),%ebx
  je     L116
  movslq %ebx,%r13
  xor    %r15d,%r15d
  test   %ebx,%ebx
  jle    L16
  nopl   (%rax)
L120:
  movsbl (%r14,%r15,1),%edi
  call   <toupper@plt>
  mov    %al,(%r12,%r15,1)
  add    $0x1,%r15
  cmp    %r13,%r15
  jne    L120
L16:
  mov    -0x58(%rbp),%rsp
  jmp    L21
  nopl   (%rax)
L75:
  mov    $0x8,%edi
  call   <malloc@plt>
  movb   $0x0,(%rax)
  mov    %rax,%r12
L21:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  lea    -0x28(%rbp),%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L116:
  mov    %ebx,%ebx
  xor    %r14d,%r14d
  nopl   (%rax)
L130:
  movsbl 0x0(%r13,%r14,1),%edi
  call   <toupper@plt>
  mov    %al,(%r12,%r14,1)
  add    $0x1,%r14
  cmp    %r14,%rbx
  jne    L130
  mov    -0x58(%rbp),%rsp
  jmp    L21
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *clean(const char *str) {
  int len = strlen(str);
  char *res;
  if (len == 0) {
    res = malloc(sizeof(str));
    res[0] = '\0';
    return (res);
  } else {
    res = malloc(sizeof(str) * len);
    char chars[len];
    int cnt = 0, i;
    for (i = 0; i < len; i++) {
      if (isalpha(str[i])) {
        chars[cnt++] = str[i];
      }
    }
    if (cnt == len) {
      for (i = 0; i < len; i++) {
        res[i] = toupper(str[i]);
      }
    } else {
      for (i = 0; i < cnt; i++) {
        res[i] = toupper(chars[i]);
      }
    }
  }
  return (res);
}
```
user: What is the c source code of the assembly code below:

```asm
<clean>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,-0x4c(%rbp)
  test   %eax,%eax
  je     L55
  lea    0x0(,%rax,8),%edi
  mov    %eax,%ebx
  mov    %rsp,-0x58(%rbp)
  call   <malloc@plt>
  mov    %rsp,%rcx
  mov    %rax,%r12
  movslq %ebx,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L12
L92:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L92
L12:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L9
  orq    $0x0,-0x8(%rsp,%rdx,1)
L9:
  mov    -0x4c(%rbp),%eax
  mov    %rsp,%r14
  test   %eax,%eax
  jle    L103
  sub    $0x1,%eax
  mov    %r13,%r15
  xor    %ebx,%ebx
  lea    0x1(%r13,%rax,1),%rax
  mov    %rax,-0x48(%rbp)
  nop
L5:
  movsbl (%r15),%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L73
  movzbl (%r15),%ecx
  movslq %ebx,%rax
  add    $0x1,%ebx
  mov    %cl,(%r14,%rax,1)
L73:
  add    $0x1,%r15
  cmp    -0x48(%rbp),%r15
  jne    L5
  cmp    -0x4c(%rbp),%ebx
  je     L138
  movslq %ebx,%r13
  xor    %r15d,%r15d
  test   %ebx,%ebx
  jle    L103
  nopl   (%rax)
L60:
  movsbl (%r14,%r15,1),%edi
  call   <toupper@plt>
  mov    %al,(%r12,%r15,1)
  add    $0x1,%r15
  cmp    %r15,%r13
  jne    L60
L103:
  mov    -0x58(%rbp),%rsp
  jmp    L78
  nopl   (%rax)
L55:
  mov    $0x8,%edi
  call   <malloc@plt>
  movb   $0x0,(%rax)
  mov    %rax,%r12
L78:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  lea    -0x28(%rbp),%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L138:
  mov    %ebx,%ebx
  xor    %r14d,%r14d
  nopl   (%rax)
L95:
  movsbl 0x0(%r13,%r14,1),%edi
  call   <toupper@plt>
  mov    %al,(%r12,%r14,1)
  add    $0x1,%r14
  cmp    %r14,%rbx
  jne    L95
  mov    -0x58(%rbp),%rsp
  jmp    L78
L66:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *clean(const char *str) {
  int len = strlen(str);
  char *res;
  if (len == 0) {
    res = malloc(sizeof(str));
    res[0] = '\0';
    return (res);
  } else {
    res = malloc(sizeof(str) * len);
    char chars[len];
    int cnt = 0, i;
    for (i = 0; i < len; i++) {
      if (isalpha(str[i])) {
        chars[cnt++] = str[i];
      }
    }
    if (cnt == len) {
      for (i = 0; i < len; i++) {
        res[i] = toupper(str[i]);
      }
    } else {
      for (i = 0; i < cnt; i++) {
        res[i] = toupper(chars[i]);
      }
    }
  }
  return (res);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<XfwfDrawString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,-0x34(%rbp)
  mov    %r8d,-0x38(%rbp)
  mov    %r9,-0x40(%rbp)
  mov    $0x0,%r13d
  mov    %r13d,%r14d
  mov    -0x40(%rbp),%r12
  jmp    L18
L11:
  mov    0x10(%rbp),%edx
  mov    $0x9,%esi
  mov    %r12,%rdi
  call   <strnchr@plt>
  mov    %rax,%rbx
  test   %rbx,%rbx
  je     L53
  cmpq   $0x0,0x18(%rbp)
  je     L53
  mov    %rbx,%rax
  sub    %r12,%rax
  mov    %eax,%r8d
  mov    -0x34(%rbp),%eax
  lea    0x0(%r13,%rax,1),%ecx
  mov    -0x38(%rbp),%edi
  mov    -0x30(%rbp),%edx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  sub    $0x8,%rsp
  push   %r8
  mov    %r12,%r9
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <XDrawString@plt>
  add    $0x10,%rsp
  mov    %r14d,%eax
  lea    0x1(%rax),%r14d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%r13d
  mov    0x10(%rbp),%eax
  mov    %rbx,%rdx
  sub    %r12,%rdx
  sub    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,0x10(%rbp)
  lea    0x1(%rbx),%r12
  jmp    L18
L53:
  mov    -0x34(%rbp),%eax
  lea    0x0(%r13,%rax,1),%ecx
  mov    -0x38(%rbp),%r8d
  mov    -0x30(%rbp),%edx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  sub    $0x8,%rsp
  mov    0x10(%rbp),%edi
  push   %rdi
  mov    %r12,%r9
  mov    %rax,%rdi
  call   <XDrawString@plt>
  add    $0x10,%rsp
  jmp    L52
L18:
  cmpl   $0x0,0x10(%rbp)
  jne    L11
  nop
L52:
  nop
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
```

assistant: ```c
void XfwfDrawString(Display *display, Drawable drawable, GC gc, int x, int y, String string, int length, int *tabs) {
  register char *p, *ep;
  register int tx, tab;
  tab = tx = 0;
  for (p = string; length;) {
    ep = strnchr(p, '\t', length);
    if (ep && tabs) {
      XDrawString(display, drawable, gc, x + tx, y, p, ep - p);
      tx = tabs[tab++];
      length -= ep - p + 1;
      p = ep + 1;
    } else {
      XDrawString(display, drawable, gc, x + tx, y, p, length);
      break;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<XfwfDrawString>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,0x10(%rsp)
  mov    %edx,0x14(%rsp)
  mov    %ecx,0x18(%rsp)
  mov    %r8d,0x1c(%rsp)
  mov    0x60(%rsp),%ebx
  test   %ebx,%ebx
  je     L40
  mov    %r9,%r12
  mov    0x68(%rsp),%r14
  mov    $0x0,%r13d
L23:
  mov    %ebx,%edx
  mov    $0x9,%esi
  mov    %r12,%rdi
  call   <strnchr@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L13
  cmpq   $0x0,0x68(%rsp)
  je     L13
  mov    %rax,%r15
  sub    %r12,%r15
  mov    0x18(%rsp),%eax
  lea    (%rax,%r13,1),%ecx
  sub    $0x8,%rsp
  push   %r15
  mov    %r12,%r9
  mov    0x2c(%rsp),%r8d
  mov    0x24(%rsp),%edx
  mov    0x20(%rsp),%esi
  mov    0x18(%rsp),%rdi
  call   <XDrawString@plt>
  mov    (%r14),%r13d
  lea    -0x1(%rbx),%eax
  sub    %r15d,%eax
  mov    %eax,%ebx
  lea    0x1(%rbp),%r12
  add    $0x4,%r14
  add    $0x10,%rsp
  test   %eax,%eax
  jne    L23
  jmp    L40
L13:
  mov    0x18(%rsp),%ecx
  add    %r13d,%ecx
  sub    $0x8,%rsp
  push   %rbx
  mov    %r12,%r9
  mov    0x2c(%rsp),%r8d
  mov    0x24(%rsp),%edx
  mov    0x20(%rsp),%esi
  mov    0x18(%rsp),%rdi
  call   <XDrawString@plt>
  add    $0x10,%rsp
L40:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void XfwfDrawString(Display *display, Drawable drawable, GC gc, int x, int y, String string, int length, int *tabs) {
  register char *p, *ep;
  register int tx, tab;
  tab = tx = 0;
  for (p = string; length;) {
    ep = strnchr(p, '\t', length);
    if (ep && tabs) {
      XDrawString(display, drawable, gc, x + tx, y, p, ep - p);
      tx = tabs[tab++];
      length -= ep - p + 1;
      p = ep + 1;
    } else {
      XDrawString(display, drawable, gc, x + tx, y, p, length);
      break;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<XfwfDrawString>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    0x60(%rsp),%ebx
  mov    0x68(%rsp),%r12
  mov    %ecx,0xc(%rsp)
  mov    %r8d,0x18(%rsp)
  test   %ebx,%ebx
  je     L19
  mov    %rdi,0x10(%rsp)
  test   %r12,%r12
  mov    %esi,%r14d
  mov    %edx,%r15d
  movl   $0x0,0x8(%rsp)
  setne  0x1f(%rsp)
  nopw   0x0(%rax,%rax,1)
L22:
  mov    %r9,%rdi
  mov    %ebx,%edx
  mov    $0x9,%esi
  mov    %r9,(%rsp)
  call   <strnchr@plt>
  mov    0x8(%rsp),%ecx
  add    0xc(%rsp),%ecx
  test   %rax,%rax
  mov    (%rsp),%r9
  mov    %rax,%r13
  je     L27
  cmpb   $0x0,0x1f(%rsp)
  je     L27
  mov    %rax,%rbp
  sub    $0x8,%rsp
  mov    %r15d,%edx
  mov    %r14d,%esi
  sub    %r9,%rbp
  sub    $0x1,%ebx
  add    $0x4,%r12
  push   %rbp
  mov    0x28(%rsp),%r8d
  mov    0x20(%rsp),%rdi
  call   <XDrawString@plt>
  mov    -0x4(%r12),%eax
  lea    0x1(%r13),%r9
  mov    %eax,0x18(%rsp)
  pop    %rax
  pop    %rdx
  sub    %ebp,%ebx
  jne    L22
L19:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  mov    0x10(%rsp),%r13
  mov    %ebx,0x60(%rsp)
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    0x18(%rsp),%r8d
  add    $0x28,%rsp
  pop    %rbx
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <XDrawString@plt>
```

assistant: ```c
void XfwfDrawString(Display *display, Drawable drawable, GC gc, int x, int y, String string, int length, int *tabs) {
  register char *p, *ep;
  register int tx, tab;
  tab = tx = 0;
  for (p = string; length;) {
    ep = strnchr(p, '\t', length);
    if (ep && tabs) {
      XDrawString(display, drawable, gc, x + tx, y, p, ep - p);
      tx = tabs[tab++];
      length -= ep - p + 1;
      p = ep + 1;
    } else {
      XDrawString(display, drawable, gc, x + tx, y, p, length);
      break;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<XfwfDrawString>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    0x60(%rsp),%ebx
  mov    0x68(%rsp),%r12
  mov    %ecx,0xc(%rsp)
  mov    %r8d,0x18(%rsp)
  test   %ebx,%ebx
  je     L15
  mov    %rdi,0x10(%rsp)
  test   %r12,%r12
  mov    %esi,%r14d
  mov    %edx,%r15d
  movl   $0x0,0x8(%rsp)
  setne  0x1f(%rsp)
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %r9,%rdi
  mov    %ebx,%edx
  mov    $0x9,%esi
  mov    %r9,(%rsp)
  call   <strnchr@plt>
  mov    0x8(%rsp),%ecx
  add    0xc(%rsp),%ecx
  test   %rax,%rax
  mov    (%rsp),%r9
  mov    %rax,%r13
  je     L8
  cmpb   $0x0,0x1f(%rsp)
  je     L8
  mov    %rax,%rbp
  sub    $0x8,%rsp
  mov    %r15d,%edx
  mov    %r14d,%esi
  sub    %r9,%rbp
  sub    $0x1,%ebx
  add    $0x4,%r12
  push   %rbp
  mov    0x28(%rsp),%r8d
  mov    0x20(%rsp),%rdi
  call   <XDrawString@plt>
  mov    -0x4(%r12),%eax
  lea    0x1(%r13),%r9
  mov    %eax,0x18(%rsp)
  pop    %rax
  pop    %rdx
  sub    %ebp,%ebx
  jne    L9
L15:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  mov    0x10(%rsp),%r13
  mov    %ebx,0x60(%rsp)
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    0x18(%rsp),%r8d
  add    $0x28,%rsp
  pop    %rbx
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <XDrawString@plt>
```

assistant: ```c
void XfwfDrawString(Display *display, Drawable drawable, GC gc, int x, int y, String string, int length, int *tabs) {
  register char *p, *ep;
  register int tx, tab;
  tab = tx = 0;
  for (p = string; length;) {
    ep = strnchr(p, '\t', length);
    if (ep && tabs) {
      XDrawString(display, drawable, gc, x + tx, y, p, ep - p);
      tx = tabs[tab++];
      length -= ep - p + 1;
      p = ep + 1;
    } else {
      XDrawString(display, drawable, gc, x + tx, y, p, length);
      break;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverse_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L4
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L25
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse_string>
  jmp    L41
L58:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L41:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
  mov    -0x10(%rbp),%rax
  movzbl -0x11(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L36
L4:
  nop
  jmp    L36
L25:
  nop
L36:
  leave
  ret
```

assistant: ```c
void reverse_string(char *str) {
  char c;
  char *p, *q;
  p = str;
  if (!p)
    return;
  q = p + 1;
  if (*q == '\0')
    return;
  c = *p;
  reverse_string(q);
  while (*q != '\0') {
    *p = *q;
    p++;
    q++;
  }
  *p = c;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<reverse_string>:
  endbr64
  test   %rdi,%rdi
  je     L16
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  cmpb   $0x0,0x1(%rdi)
  jne    L39
L45:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L39:
  movzbl (%rdi),%ebp
  lea    0x1(%rdi),%rdi
  call   <reverse_string>
  movzbl 0x1(%rbx),%eax
  test   %al,%al
  je     L10
L32:
  mov    %al,(%rbx)
  add    $0x1,%rbx
  movzbl 0x1(%rbx),%eax
  test   %al,%al
  jne    L32
L10:
  mov    %bpl,(%rbx)
  jmp    L45
L16:
  ret
```

assistant: ```c
void reverse_string(char *str) {
  char c;
  char *p, *q;
  p = str;
  if (!p)
    return;
  q = p + 1;
  if (*q == '\0')
    return;
  c = *p;
  reverse_string(q);
  while (*q != '\0') {
    *p = *q;
    p++;
    q++;
  }
  *p = c;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse_string>:
  endbr64
  test   %rdi,%rdi
  je     L79
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movzbl 0x1(%rdi),%r12d
  test   %r12b,%r12b
  jne    L69
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L69:
  movzbl 0x2(%rdi),%r14d
  lea    0x1(%rdi),%rbp
  movzbl (%rdi),%edx
  mov    %rbp,%r13
  test   %r14b,%r14b
  je     L66
  jmp    L115
  xchg   %ax,%ax
L111:
  add    $0x1,%rbp
L66:
  mov    %r12b,(%rbx)
  movzbl 0x1(%rbp),%r12d
  mov    %rbp,%rbx
  test   %r12b,%r12b
  jne    L111
  mov    %dl,(%rbx)
L117:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L79:
  ret
  nopl   0x0(%rax)
L115:
  cmpb   $0x0,0x3(%rdi)
  lea    0x2(%rdi),%r15
  je     L112
  lea    0x3(%rdi),%rdi
  mov    %dl,0xf(%rsp)
  call   <reverse_string>
  movzbl 0x3(%rbx),%eax
  movzbl 0xf(%rsp),%edx
  test   %al,%al
  je     L25
  nopl   0x0(%rax)
L1:
  mov    %al,(%r15)
  movzbl 0x2(%r15),%eax
  add    $0x1,%r15
  test   %al,%al
  jne    L1
L25:
  mov    %r14b,(%r15)
  movzbl 0x2(%rbx),%r14d
  test   %r14b,%r14b
  je     L92
  nopl   (%rax)
L112:
  mov    %r14b,0x0(%r13)
  movzbl 0x2(%r13),%r14d
  add    $0x1,%r13
  test   %r14b,%r14b
  jne    L112
L92:
  mov    %r12b,0x0(%r13)
  movzbl 0x1(%rbx),%r12d
  test   %r12b,%r12b
  jne    L66
  mov    %dl,(%rbx)
  jmp    L117
```

assistant: ```c
void reverse_string(char *str) {
  char c;
  char *p, *q;
  p = str;
  if (!p)
    return;
  q = p + 1;
  if (*q == '\0')
    return;
  c = *p;
  reverse_string(q);
  while (*q != '\0') {
    *p = *q;
    p++;
    q++;
  }
  *p = c;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse_string>:
  endbr64
  test   %rdi,%rdi
  je     L24
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movzbl 0x1(%rdi),%r12d
  test   %r12b,%r12b
  jne    L42
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  movzbl 0x2(%rdi),%r14d
  lea    0x1(%rdi),%rbp
  movzbl (%rdi),%edx
  mov    %rbp,%r13
  test   %r14b,%r14b
  je     L79
  jmp    L18
  xchg   %ax,%ax
L93:
  add    $0x1,%rbp
L79:
  mov    %r12b,(%rbx)
  movzbl 0x1(%rbp),%r12d
  mov    %rbp,%rbx
  test   %r12b,%r12b
  jne    L93
  mov    %dl,(%rbx)
L57:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L24:
  ret
  nopl   0x0(%rax)
L18:
  cmpb   $0x0,0x3(%rdi)
  lea    0x2(%rdi),%r15
  je     L41
  lea    0x3(%rdi),%rdi
  mov    %dl,0xf(%rsp)
  call   <reverse_string>
  movzbl 0x3(%rbx),%eax
  movzbl 0xf(%rsp),%edx
  test   %al,%al
  je     L13
  nopl   0x0(%rax)
L28:
  mov    %al,(%r15)
  movzbl 0x2(%r15),%eax
  add    $0x1,%r15
  test   %al,%al
  jne    L28
L13:
  mov    %r14b,(%r15)
  movzbl 0x2(%rbx),%r14d
  test   %r14b,%r14b
  je     L22
  nopl   (%rax)
L41:
  mov    %r14b,0x0(%r13)
  movzbl 0x2(%r13),%r14d
  add    $0x1,%r13
  test   %r14b,%r14b
  jne    L41
L22:
  mov    %r12b,0x0(%r13)
  movzbl 0x1(%rbx),%r12d
  test   %r12b,%r12b
  jne    L79
  mov    %dl,(%rbx)
  jmp    L57
```

assistant: ```c
void reverse_string(char *str) {
  char c;
  char *p, *q;
  p = str;
  if (!p)
    return;
  q = p + 1;
  if (*q == '\0')
    return;
  c = *p;
  reverse_string(q);
  while (*q != '\0') {
    *p = *q;
    p++;
    q++;
  }
  *p = c;
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcomp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    %rsi,-0x18(%rbp)
  mov    %rdx,%rax
  mov    %rcx,-0x20(%rbp)
  cmp    -0x20(%rbp),%rax
  je     L33
  mov    $0xffffffff,%eax
  jmp    L17
L33:
  mov    $0x1,%ebx
  jmp    L26
L25:
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x18(%rbp)
  add    $0x1,%rbx
L26:
  cmp    %rax,%rbx
  jae    L1
  mov    -0x10(%rbp),%rdx
  movzbl (%rdx),%ecx
  mov    -0x18(%rbp),%rdx
  movzbl (%rdx),%edx
  cmp    %dl,%cl
  je     L25
L1:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L17:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int strcomp(const unsigned char *s1, const unsigned char *s2, register unsigned long ws1, unsigned long ws2) {
  if (ws1 != ws2) {
    return -1;
  }
  {
    register unsigned long iters;
    iters = 1;
    while (iters < ws1 && *s1 == *s2) {
      s1++;
      s2++;
      iters++;
    }
    return (*s1 - *s2);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcomp>:
  endbr64
  cmp    %rcx,%rdx
  jne    L35
  cmp    $0x1,%rdx
  jbe    L5
  lea    -0x1(%rdi,%rdx,1),%rax
L32:
  movzbl (%rsi),%ecx
  cmp    %cl,(%rdi)
  jne    L5
  add    $0x1,%rdi
  add    $0x1,%rsi
  cmp    %rax,%rdi
  jne    L32
  mov    %rax,%rdi
L5:
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  sub    %edx,%eax
  ret
L35:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strcomp(const unsigned char *s1, const unsigned char *s2, register unsigned long ws1, unsigned long ws2) {
  if (ws1 != ws2) {
    return -1;
  }
  {
    register unsigned long iters;
    iters = 1;
    while (iters < ws1 && *s1 == *s2) {
      s1++;
      s2++;
      iters++;
    }
    return (*s1 - *s2);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strcomp>:
  endbr64
  cmp    %rcx,%rdx
  jne    L16
  lea    -0x1(%rdi,%rdx,1),%rcx
  cmp    $0x1,%rdx
  ja     L54
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L26:
  lea    0x1(%rdi),%rax
  lea    0x1(%rsi),%rdx
  cmp    %rcx,%rax
  je     L19
  mov    %rdx,%rsi
  mov    %rax,%rdi
L54:
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  cmp    %dl,%al
  je     L26
L43:
  sub    %edx,%eax
  ret
L19:
  movzbl 0x1(%rdi),%eax
  movzbl 0x1(%rsi),%edx
  sub    %edx,%eax
  ret
L59:
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  jmp    L43
L16:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strcomp(const unsigned char *s1, const unsigned char *s2, register unsigned long ws1, unsigned long ws2) {
  if (ws1 != ws2) {
    return -1;
  }
  {
    register unsigned long iters;
    iters = 1;
    while (iters < ws1 && *s1 == *s2) {
      s1++;
      s2++;
      iters++;
    }
    return (*s1 - *s2);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcomp>:
  endbr64
  cmp    %rcx,%rdx
  jne    L3
  lea    -0x1(%rdi,%rdx,1),%rcx
  cmp    $0x1,%rdx
  ja     L13
  jmp    L32
  cs nopw 0x0(%rax,%rax,1)
L45:
  lea    0x1(%rdi),%rax
  lea    0x1(%rsi),%rdx
  cmp    %rcx,%rax
  je     L36
  mov    %rdx,%rsi
  mov    %rax,%rdi
L13:
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  cmp    %dl,%al
  je     L45
L50:
  sub    %edx,%eax
  ret
L36:
  movzbl 0x1(%rdi),%eax
  movzbl 0x1(%rsi),%edx
  sub    %edx,%eax
  ret
L32:
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  jmp    L50
L3:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int strcomp(const unsigned char *s1, const unsigned char *s2, register unsigned long ws1, unsigned long ws2) {
  if (ws1 != ws2) {
    return -1;
  }
  {
    register unsigned long iters;
    iters = 1;
    while (iters < ws1 && *s1 == *s2) {
      s1++;
      s2++;
      iters++;
    }
    return (*s1 - *s2);
  }
}
```