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
<any>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L26
L29:
  movl   $0x0,-0x8(%rbp)
  jmp    L16
L27:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L20
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L20
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L20:
  addl   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L27
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L29
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int any(char s[], char b[]) {
  int i, j;
  int charloc = -1;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = 0; b[j] != '\0'; j++) {
      if (s[i] == b[j] && charloc == -1) {
        charloc = i;
      }
    }
  }
  return charloc;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<any>:
  endbr64
  movzbl (%rdi),%r8d
  test   %r8b,%r8b
  je     L23
  movzbl (%rsi),%r11d
  mov    $0x0,%r10d
  mov    $0xffffffff,%ecx
  jmp    L43
L60:
  add    $0x1,%rax
  movzbl -0x1(%rax),%edx
  test   %dl,%dl
  je     L62
L75:
  cmp    $0xffffffff,%ecx
  jne    L60
  cmp    %r8b,%dl
  cmove  %r9d,%ecx
  jmp    L60
L62:
  add    $0x1,%r10
  movzbl (%rdi,%r10,1),%r8d
  test   %r8b,%r8b
  je     L27
L43:
  mov    %r10d,%r9d
  lea    0x1(%rsi),%rax
  mov    %r11d,%edx
  test   %r11b,%r11b
  jne    L75
  jmp    L62
L23:
  mov    $0xffffffff,%ecx
L27:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int any(char s[], char b[]) {
  int i, j;
  int charloc = -1;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = 0; b[j] != '\0'; j++) {
      if (s[i] == b[j] && charloc == -1) {
        charloc = i;
      }
    }
  }
  return charloc;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<any>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L25
  movzbl (%rsi),%r11d
  xor    %r10d,%r10d
  mov    $0xffffffff,%r8d
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r10d,%r9d
  lea    0x1(%rsi),%rax
  mov    %r11d,%edx
  test   %r11b,%r11b
  je     L35
  nop
L39:
  cmp    $0xffffffff,%r8d
  jne    L11
  cmp    %cl,%dl
  cmove  %r9d,%r8d
L11:
  movzbl (%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  jne    L39
L35:
  add    $0x1,%r10
  movzbl (%rdi,%r10,1),%ecx
  test   %cl,%cl
  jne    L19
  mov    %r8d,%eax
  ret
L25:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int any(char s[], char b[]) {
  int i, j;
  int charloc = -1;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = 0; b[j] != '\0'; j++) {
      if (s[i] == b[j] && charloc == -1) {
        charloc = i;
      }
    }
  }
  return charloc;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<any>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L9
  movzbl (%rsi),%r11d
  xor    %r10d,%r10d
  mov    $0xffffffff,%r8d
  nopl   0x0(%rax,%rax,1)
L47:
  mov    %r10d,%r9d
  lea    0x1(%rsi),%rdx
  mov    %r11d,%eax
  test   %r11b,%r11b
  je     L13
  nop
L21:
  cmp    %cl,%al
  jne    L6
  cmp    $0xffffffff,%r8d
  cmove  %r9d,%r8d
L6:
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  jne    L21
L13:
  add    $0x1,%r10
  movzbl (%rdi,%r10,1),%ecx
  test   %cl,%cl
  jne    L47
  mov    %r8d,%eax
  ret
L9:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int any(char s[], char b[]) {
  int i, j;
  int charloc = -1;
  for (i = 0; s[i] != '\0'; i++) {
    for (j = 0; b[j] != '\0'; j++) {
      if (s[i] == b[j] && charloc == -1) {
        charloc = i;
      }
    }
  }
  return charloc;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<OutputSymbol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L65
  cmpl   $0x3,-0x4(%rbp)
  jg     L20
  cmpl   $0x2,-0x4(%rbp)
  je     L5
  cmpl   $0x2,-0x4(%rbp)
  jg     L20
  cmpl   $0x0,-0x4(%rbp)
  je     L11
  cmpl   $0x1,-0x4(%rbp)
  je     L25
  jmp    L20
L11:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x41,%edi
  call   <putc@plt>
  jmp    L17
L25:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x43,%edi
  call   <putc@plt>
  jmp    L17
L5:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x47,%edi
  call   <putc@plt>
  jmp    L17
L65:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x54,%edi
  call   <putc@plt>
  jmp    L17
L20:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2d,%edi
  call   <putc@plt>
  nop
L17:
  nop
  leave
  ret
```

assistant: ```c
void OutputSymbol(int symbol, FILE *stream) {
  switch (symbol) {
  case 0:
    putc('A', stream);
    break;
  case 1:
    putc('C', stream);
    break;
  case 2:
    putc('G', stream);
    break;
  case 3:
    putc('T', stream);
    break;
  default:
    putc('-', stream);
    break;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<OutputSymbol>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L68
  jg     L66
  test   %edi,%edi
  je     L8
  cmp    $0x1,%edi
  jne    L73
  mov    $0x43,%edi
  call   <putc@plt>
  jmp    L7
L66:
  cmp    $0x3,%edi
  jne    L73
  mov    $0x54,%edi
  call   <putc@plt>
  jmp    L7
L8:
  mov    $0x41,%edi
  call   <putc@plt>
L7:
  add    $0x8,%rsp
  ret
L68:
  mov    $0x47,%edi
  call   <putc@plt>
  jmp    L7
L73:
  mov    $0x2d,%edi
  call   <putc@plt>
  jmp    L7
```

assistant: ```c
void OutputSymbol(int symbol, FILE *stream) {
  switch (symbol) {
  case 0:
    putc('A', stream);
    break;
  case 1:
    putc('C', stream);
    break;
  case 2:
    putc('G', stream);
    break;
  case 3:
    putc('T', stream);
    break;
  default:
    putc('-', stream);
    break;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<OutputSymbol>:
  endbr64
  cmp    $0x2,%edi
  je     L13
  jg     L37
  test   %edi,%edi
  je     L8
  cmp    $0x1,%edi
  jne    L6
  mov    $0x43,%edi
  jmp    1050 <putc@plt>
  xchg   %ax,%ax
L37:
  cmp    $0x3,%edi
  jne    L6
  mov    $0x54,%edi
  jmp    1050 <putc@plt>
  nop
L13:
  mov    $0x47,%edi
  jmp    1050 <putc@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  mov    $0x2d,%edi
  jmp    1050 <putc@plt>
  nopw   0x0(%rax,%rax,1)
L8:
  mov    $0x41,%edi
  jmp    1050 <putc@plt>
```

assistant: ```c
void OutputSymbol(int symbol, FILE *stream) {
  switch (symbol) {
  case 0:
    putc('A', stream);
    break;
  case 1:
    putc('C', stream);
    break;
  case 2:
    putc('G', stream);
    break;
  case 3:
    putc('T', stream);
    break;
  default:
    putc('-', stream);
    break;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<OutputSymbol>:
  endbr64
  cmp    $0x2,%edi
  je     L13
  jg     L22
  test   %edi,%edi
  je     L1
  cmp    $0x1,%edi
  jne    L33
  mov    $0x43,%edi
  jmp    1050 <putc@plt>
  xchg   %ax,%ax
L22:
  cmp    $0x3,%edi
  jne    L33
  mov    $0x54,%edi
  jmp    1050 <putc@plt>
  nop
L13:
  mov    $0x47,%edi
  jmp    1050 <putc@plt>
  nopw   0x0(%rax,%rax,1)
L33:
  mov    $0x2d,%edi
  jmp    1050 <putc@plt>
  nopw   0x0(%rax,%rax,1)
L1:
  mov    $0x41,%edi
  jmp    1050 <putc@plt>
```

assistant: ```c
void OutputSymbol(int symbol, FILE *stream) {
  switch (symbol) {
  case 0:
    putc('A', stream);
    break;
  case 1:
    putc('C', stream);
    break;
  case 2:
    putc('G', stream);
    break;
  case 3:
    putc('T', stream);
    break;
  default:
    putc('-', stream);
    break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<XfDrawCross>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    0x10(%rbp),%eax
  lea    (%rdx,%rax,1),%edi
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    0x10(%rbp),%eax
  lea    (%rdx,%rax,1),%r9d
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    0x10(%rbp),%eax
  mov    %eax,%r8d
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    0x10(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  sub    $0x8,%rsp
  push   %rdi
  mov    %rax,%rdi
  call   <XDrawLine@plt>
  add    $0x10,%rsp
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    0x10(%rbp),%eax
  lea    (%rdx,%rax,1),%edi
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    0x10(%rbp),%eax
  mov    %eax,%r9d
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    0x10(%rbp),%eax
  mov    %eax,%r8d
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    0x10(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  sub    $0x8,%rsp
  push   %rdi
  mov    %rax,%rdi
  call   <XDrawLine@plt>
  add    $0x10,%rsp
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L6
  nop
  leave
  ret
```

assistant: ```c
void XfDrawCross(Display *display, Window window, GC gc, int *xverts, int *yverts, int nvert, int size) {
  int i;
  for (i = 0; i < nvert; i++) {
    XDrawLine(display, window, gc, xverts[i] - size, yverts[i] - size, xverts[i] + size, yverts[i] + size);
    XDrawLine(display, window, gc, xverts[i] + size, yverts[i] - size, xverts[i] - size, yverts[i] + size);
  }
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<XfDrawCross>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%ebx
  test   %r9d,%r9d
  jle    L17
  mov    %rdi,%r13
  mov    %esi,%r14d
  mov    %edx,%r15d
  mov    %r8,%rbp
  mov    %rcx,%r12
  lea    -0x1(%r9),%eax
  lea    0x4(%r8,%rax,4),%rax
  mov    %rax,0x8(%rsp)
L12:
  mov    0x0(%rbp),%r8d
  mov    (%r12),%eax
  mov    %eax,%ecx
  sub    %ebx,%ecx
  sub    $0x8,%rsp
  lea    (%r8,%rbx,1),%edx
  push   %rdx
  lea    (%rax,%rbx,1),%r9d
  sub    %ebx,%r8d
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <XDrawLine@plt>
  mov    0x0(%rbp),%r8d
  mov    (%r12),%r9d
  lea    (%r9,%rbx,1),%ecx
  lea    (%r8,%rbx,1),%eax
  mov    %eax,(%rsp)
  sub    %ebx,%r9d
  sub    %ebx,%r8d
  mov    %r15d,%edx
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <XDrawLine@plt>
  add    $0x4,%rbp
  add    $0x4,%r12
  add    $0x10,%rsp
  cmp    0x8(%rsp),%rbp
  jne    L12
L17:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void XfDrawCross(Display *display, Window window, GC gc, int *xverts, int *yverts, int nvert, int size) {
  int i;
  for (i = 0; i < nvert; i++) {
    XDrawLine(display, window, gc, xverts[i] - size, yverts[i] - size, xverts[i] + size, yverts[i] + size);
    XDrawLine(display, window, gc, xverts[i] + size, yverts[i] - size, xverts[i] - size, yverts[i] + size);
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<XfDrawCross>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%ebx
  test   %r9d,%r9d
  jle    L3
  lea    -0x1(%r9),%eax
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%r14d
  lea    0x4(%r8,%rax,4),%rax
  mov    %r8,%rbp
  mov    %rcx,%r15
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax)
L2:
  mov    0x0(%rbp),%r8d
  sub    $0x8,%rsp
  mov    (%r15),%eax
  mov    %r13d,%esi
  mov    %r12,%rdi
  add    $0x4,%rbp
  add    $0x4,%r15
  lea    (%r8,%rbx,1),%edx
  mov    %eax,%ecx
  lea    (%rax,%rbx,1),%r9d
  sub    %ebx,%r8d
  push   %rdx
  sub    %ebx,%ecx
  mov    %r14d,%edx
  call   <XDrawLine@plt>
  mov    -0x4(%rbp),%r8d
  mov    -0x4(%r15),%r9d
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  lea    (%r8,%rbx,1),%eax
  lea    (%r9,%rbx,1),%ecx
  sub    %ebx,%r8d
  sub    %ebx,%r9d
  mov    %eax,(%rsp)
  call   <XDrawLine@plt>
  pop    %rax
  pop    %rdx
  cmp    0x8(%rsp),%rbp
  jne    L2
L3:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void XfDrawCross(Display *display, Window window, GC gc, int *xverts, int *yverts, int nvert, int size) {
  int i;
  for (i = 0; i < nvert; i++) {
    XDrawLine(display, window, gc, xverts[i] - size, yverts[i] - size, xverts[i] + size, yverts[i] + size);
    XDrawLine(display, window, gc, xverts[i] + size, yverts[i] - size, xverts[i] - size, yverts[i] + size);
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<XfDrawCross>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%ebx
  test   %r9d,%r9d
  jle    L1
  lea    -0x1(%r9),%eax
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%r14d
  lea    0x4(%r8,%rax,4),%rax
  mov    %r8,%rbp
  mov    %rcx,%r15
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax)
L3:
  mov    0x0(%rbp),%r8d
  sub    $0x8,%rsp
  mov    (%r15),%eax
  mov    %r13d,%esi
  mov    %r12,%rdi
  add    $0x4,%rbp
  add    $0x4,%r15
  lea    (%r8,%rbx,1),%edx
  mov    %eax,%ecx
  lea    (%rax,%rbx,1),%r9d
  sub    %ebx,%r8d
  push   %rdx
  sub    %ebx,%ecx
  mov    %r14d,%edx
  call   <XDrawLine@plt>
  mov    -0x4(%rbp),%r8d
  mov    -0x4(%r15),%r9d
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  lea    (%r8,%rbx,1),%eax
  lea    (%r9,%rbx,1),%ecx
  sub    %ebx,%r8d
  sub    %ebx,%r9d
  mov    %eax,(%rsp)
  call   <XDrawLine@plt>
  pop    %rax
  pop    %rdx
  cmp    0x8(%rsp),%rbp
  jne    L3
L1:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void XfDrawCross(Display *display, Window window, GC gc, int *xverts, int *yverts, int nvert, int size) {
  int i;
  for (i = 0; i < nvert; i++) {
    XDrawLine(display, window, gc, xverts[i] - size, yverts[i] - size, xverts[i] + size, yverts[i] + size);
    XDrawLine(display, window, gc, xverts[i] + size, yverts[i] - size, xverts[i] - size, yverts[i] + size);
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_test_various_MIN_biggerulong_smallerint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xffffffff,%esi
  mov    $0x0,%edi
  call   <_test_MIN_biggerulong_smallerint@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int _test_various_MIN_biggerulong_smallerint() {
  _test_MIN_biggerulong_smallerint(0, -1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_test_various_MIN_biggerulong_smallerint>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%esi
  mov    $0x0,%edi
  call   <_test_MIN_biggerulong_smallerint@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int _test_various_MIN_biggerulong_smallerint() {
  _test_MIN_biggerulong_smallerint(0, -1);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_test_various_MIN_biggerulong_smallerint>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%esi
  xor    %edi,%edi
  call   <_test_MIN_biggerulong_smallerint@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int _test_various_MIN_biggerulong_smallerint() {
  _test_MIN_biggerulong_smallerint(0, -1);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_test_various_MIN_biggerulong_smallerint>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%esi
  xor    %edi,%edi
  call   <_test_MIN_biggerulong_smallerint@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int _test_various_MIN_biggerulong_smallerint() {
  _test_MIN_biggerulong_smallerint(0, -1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<highest_bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    %eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x2,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x4,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x10,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  sub    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int highest_bit(unsigned int x) {
  x |= (x >> 1);
  x |= (x >> 2);
  x |= (x >> 4);
  x |= (x >> 8);
  x |= (x >> 16);
  return x - (x >> 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<highest_bit>:
  endbr64
  mov    %edi,%eax
  shr    %eax
  or     %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  or     %edx,%eax
  mov    %eax,%edx
  shr    $0x8,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  or     %edx,%eax
  mov    %eax,%edx
  shr    %edx
  sub    %edx,%eax
  ret
```

assistant: ```c
unsigned int highest_bit(unsigned int x) {
  x |= (x >> 1);
  x |= (x >> 2);
  x |= (x >> 4);
  x |= (x >> 8);
  x |= (x >> 16);
  return x - (x >> 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<highest_bit>:
  endbr64
  mov    %edi,%eax
  shr    %eax
  or     %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  or     %edx,%eax
  mov    %eax,%edx
  shr    $0x8,%edx
  or     %eax,%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  or     %edx,%eax
  mov    %eax,%edx
  shr    %edx
  sub    %edx,%eax
  ret
```

assistant: ```c
unsigned int highest_bit(unsigned int x) {
  x |= (x >> 1);
  x |= (x >> 2);
  x |= (x >> 4);
  x |= (x >> 8);
  x |= (x >> 16);
  return x - (x >> 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmp6>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x1,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L36
  mov    -0x14(%rbp),%eax
  jmp    L75
L36:
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x2,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L50
  mov    -0x14(%rbp),%eax
  jmp    L75
L50:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L85
  mov    -0x14(%rbp),%eax
  jmp    L75
L85:
  mov    -0x10(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x3,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L64
  mov    -0x14(%rbp),%eax
  jmp    L75
L64:
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x8(%rbp),%rdx
  add    $0x4,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L35
  mov    -0x14(%rbp),%eax
  jmp    L75
L35:
  mov    $0x0,%eax
L75:
  pop    %rbp
  ret
```

assistant: ```c
int cmp6(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cmp6>:
  endbr64
  movsbl 0x1(%rdi),%eax
  movsbl 0x1(%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl 0x2(%rdi),%eax
  movsbl 0x2(%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl 0x3(%rdi),%eax
  movsbl 0x3(%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl 0x4(%rdi),%eax
  movsbl 0x4(%rsi),%edx
  sub    %edx,%eax
L15:
  ret
```

assistant: ```c
int cmp6(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cmp6>:
  endbr64
  movsbl 0x1(%rdi),%eax
  movsbl 0x1(%rsi),%edx
  sub    %edx,%eax
  jne    L38
  movsbl 0x2(%rdi),%eax
  movsbl 0x2(%rsi),%edx
  sub    %edx,%eax
  jne    L38
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  jne    L38
  movsbl 0x3(%rdi),%eax
  movsbl 0x3(%rsi),%edx
  sub    %edx,%eax
  jne    L38
  movsbl 0x4(%rdi),%eax
  movsbl 0x4(%rsi),%edx
  sub    %edx,%eax
L38:
  ret
```

assistant: ```c
int cmp6(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cmp6>:
  endbr64
  movsbl 0x1(%rdi),%eax
  movsbl 0x1(%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl 0x2(%rdi),%eax
  movsbl 0x2(%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl (%rdi),%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl 0x3(%rdi),%eax
  movsbl 0x3(%rsi),%edx
  sub    %edx,%eax
  jne    L15
  movsbl 0x4(%rdi),%eax
  movsbl 0x4(%rsi),%edx
  sub    %edx,%eax
L15:
  ret
```

assistant: ```c
int cmp6(const void *a, const void *b) {
  const char *sa = a;
  const char *sb = b;
  int diff;
  if ((diff = sa[1] - sb[1]) != 0)
    return diff;
  if ((diff = sa[2] - sb[2]) != 0)
    return diff;
  if ((diff = sa[0] - sb[0]) != 0)
    return diff;
  if ((diff = sa[3] - sb[3]) != 0)
    return diff;
  if ((diff = sa[4] - sb[4]) != 0)
    return diff;
  return 0;
}
```