ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<up8>:
  endbr64
  mov    %edi,%eax
  test   $0x7,%dil
  je     L2
  lea    0x8(%rdi),%eax
  mov    %edi,%edx
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%edi
  and    $0x7,%edi
  sub    %edx,%edi
  sub    %edi,%eax
L2:
  ret
```

assistant: ```c
int up8(int i) {
  int r;
  r = i % 8;
  if (r == 0)
    return i;
  return i + 8 - r;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<up8>:
  endbr64
  mov    %edi,%eax
  and    $0x7,%edi
  je     L0
  mov    %eax,%ecx
  lea    0x8(%rax),%esi
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rax,%rcx,1),%edx
  mov    %esi,%eax
  and    $0x7,%edx
  sub    %ecx,%edx
  sub    %edx,%eax
L0:
  ret
```

assistant: ```c
int up8(int i) {
  int r;
  r = i % 8;
  if (r == 0)
    return i;
  return i + 8 - r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<up8>:
  endbr64
  mov    %edi,%eax
  and    $0x7,%edi
  je     L4
  mov    %eax,%ecx
  lea    0x8(%rax),%esi
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rax,%rcx,1),%edx
  mov    %esi,%eax
  and    $0x7,%edx
  sub    %ecx,%edx
  sub    %edx,%eax
L4:
  ret
```

assistant: ```c
int up8(int i) {
  int r;
  r = i % 8;
  if (r == 0)
    return i;
  return i + 8 - r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_mixed_case>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L93
L63:
  addq   $0x1,-0x18(%rbp)
L93:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L81
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L63
L81:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L82
  mov    $0x0,%eax
  jmp    L29
L82:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <islower@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L34
L75:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L98
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <islower@plt>
  cmp    %eax,-0x4(%rbp)
  je     L98
  mov    $0x1,%eax
  jmp    L29
L98:
  addq   $0x1,-0x18(%rbp)
L34:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L75
  mov    $0x0,%eax
L29:
  leave
  ret
```

assistant: ```c
int str_mixed_case(const char *s) {
  int start_lower;
  for (; *s && !isalpha(*s); s++)
    ;
  if (!*s)
    return 0;
  start_lower = islower(*s);
  for (; *s; s++)
    if (isalpha(*s) && islower(*s) != start_lower)
      return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<str_mixed_case>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L51
L7:
  movsbl %dil,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  jne    L86
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L7
L51:
  mov    $0x0,%eax
L116:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L86:
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L71
  movsbl %dil,%edi
  call   <islower@plt>
  mov    %eax,%ebp
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L109
  mov    $0x0,%eax
  jmp    L116
L6:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L80
L109:
  movsbl %dil,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L6
  movsbl (%rbx),%edi
  call   <islower@plt>
  cmp    %ebp,%eax
  je     L6
  mov    $0x1,%eax
  jmp    L116
L80:
  mov    $0x0,%eax
  jmp    L116
L71:
  mov    $0x0,%eax
  jmp    L116
```

assistant: ```c
int str_mixed_case(const char *s) {
  int start_lower;
  for (; *s && !isalpha(*s); s++)
    ;
  if (!*s)
    return 0;
  start_lower = islower(*s);
  for (; *s; s++)
    if (isalpha(*s) && islower(*s) != start_lower)
      return 1;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<str_mixed_case>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L8
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L74:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L6
L8:
  call   <isalpha@plt>
  test   %rax,%rax
  je     L74
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L6
  call   <islower@plt>
  movsbl (%rbx),%edi
  mov    %eax,%ebp
  test   %dil,%dil
  jne    L45
  jmp    L6
L4:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L6
L45:
  call   <isalpha@plt>
  test   %rax,%rax
  je     L4
  movsbl (%rbx),%edi
  call   <islower@plt>
  cmp    %ebp,%eax
  je     L4
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nop
L6:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int str_mixed_case(const char *s) {
  int start_lower;
  for (; *s && !isalpha(*s); s++)
    ;
  if (!*s)
    return 0;
  start_lower = islower(*s);
  for (; *s; s++)
    if (isalpha(*s) && islower(*s) != start_lower)
      return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<str_mixed_case>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L9
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L57:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L36
L9:
  call   <isalpha@plt>
  test   %rax,%rax
  je     L57
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L36
  call   <islower@plt>
  movsbl (%rbx),%edi
  mov    %eax,%ebp
  test   %dil,%dil
  jne    L6
  jmp    L36
L24:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L36
L6:
  call   <isalpha@plt>
  test   %rax,%rax
  je     L24
  movsbl (%rbx),%edi
  call   <islower@plt>
  cmp    %ebp,%eax
  je     L24
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nop
L36:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int str_mixed_case(const char *s) {
  int start_lower;
  for (; *s && !isalpha(*s); s++)
    ;
  if (!*s)
    return 0;
  start_lower = islower(*s);
  for (; *s; s++)
    if (isalpha(*s) && islower(*s) != start_lower)
      return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<checkRomanN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0x24(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L93
L77:
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x49,%al
  je     L99
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x56,%al
  je     L99
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x58,%al
  je     L99
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x4c,%al
  je     L99
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x43,%al
  je     L99
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x43,%eax
  cmp    $0x15,%eax
  ja     L99
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D19(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D19(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L99
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L99
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L99
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L99
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  nop
L99:
  addl   $0x1,-0x28(%rbp)
L93:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x28(%rbp)
  jl     L77
  movl   $0x0,-0x28(%rbp)
  jmp    L117
L87:
  mov    -0x28(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    $0x3,%eax
  jle    L168
  mov    $0x0,%eax
  jmp    L9
L168:
  addl   $0x1,-0x28(%rbp)
L117:
  cmpl   $0x4,-0x28(%rbp)
  jle    L87
  mov    $0x1,%eax
L9:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L110
  call   <__stack_chk_fail@plt>
L110:
  leave
  ret
```

assistant: ```c
int checkRomanN(char romanNumber[20]) {
  int numerals[5] = {0}, correct = 1, i;
  for (i = 0; i < strlen(romanNumber); i++) {
    if (romanNumber[i] != 'I' && romanNumber[i] != 'V' && romanNumber[i] != 'X' && romanNumber[i] != 'L' && romanNumber[i] != 'C') {
      switch (romanNumber[i]) {
      case 'I':
        numerals[0]++;
        break;
      case 'V':
        numerals[1]++;
        break;
      case 'X':
        numerals[2]++;
        break;
      case 'L':
        numerals[3]++;
        break;
      case 'C':
        numerals[4]++;
        break;
      }
    }
  }
  for (i = 0; i < 5; i++) {
    if (numerals[i] > 3)
      return 0;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<checkRomanN>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movl   $0x0,0x10(%rsp)
  mov    $0x0,%ebx
  lea    D3(%rip),%r12
  jmp    L93
  addl   $0x1,(%rsp)
L120:
  add    $0x1,%rbx
L93:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L56
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    $0x49,%al
  je     L120
  cmp    $0x56,%al
  je     L120
  lea    -0x43(%rax),%edx
  cmp    $0x15,%dl
  ja     L58
  mov    $0x200201,%ecx
  bt     %rdx,%rcx
  jb     L120
L58:
  sub    $0x43,%eax
  cmp    $0x15,%al
  ja     L120
  movzbl %al,%eax
  movslq (%r12,%rax,4),%rax
  add    %r12,%rax
  notrack jmp *%rax
  addl   $0x1,0x4(%rsp)
  jmp    L120
  addl   $0x1,0x8(%rsp)
  jmp    L120
  addl   $0x1,0xc(%rsp)
  jmp    L120
  addl   $0x1,0x10(%rsp)
  jmp    L120
L56:
  mov    %rsp,%rax
  lea    0x14(%rsp),%rdx
L137:
  cmpl   $0x3,(%rax)
  jg     L22
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L137
  mov    $0x1,%eax
L0:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L79
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  mov    $0x0,%eax
  jmp    L0
L79:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int checkRomanN(char romanNumber[20]) {
  int numerals[5] = {0}, correct = 1, i;
  for (i = 0; i < strlen(romanNumber); i++) {
    if (romanNumber[i] != 'I' && romanNumber[i] != 'V' && romanNumber[i] != 'X' && romanNumber[i] != 'L' && romanNumber[i] != 'C') {
      switch (romanNumber[i]) {
      case 'I':
        numerals[0]++;
        break;
      case 'V':
        numerals[1]++;
        break;
      case 'X':
        numerals[2]++;
        break;
      case 'L':
        numerals[3]++;
        break;
      case 'C':
        numerals[4]++;
        break;
      }
    }
  }
  for (i = 0; i < 5; i++) {
    if (numerals[i] > 3)
      return 0;
  }
  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
<checkRomanN>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  mov    $0x200201,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  jmp    L76
  nopl   0x0(%rax)
L105:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    $0x49,%al
  je     L113
  cmp    $0x56,%al
  je     L113
  lea    -0x43(%rax),%edx
  cmp    $0x15,%dl
  ja     L46
  bt     %rdx,%r12
  jb     L113
L46:
  cmp    $0x58,%al
  je     L120
  cmp    $0x43,%al
  je     L108
  nopl   0x0(%rax,%rax,1)
L113:
  add    $0x1,%rbx
L76:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L105
  mov    %rsp,%rax
  lea    0x14(%rsp),%rdx
L17:
  add    $0x4,%rax
  cmp    %rdx,%rax
  je     L132
  cmpl   $0x3,(%rax)
  jle    L17
  xor    %eax,%eax
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L132:
  mov    $0x1,%eax
L32:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L53
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L108:
  addl   $0x1,0x10(%rsp)
  jmp    L113
  nopw   0x0(%rax,%rax,1)
L120:
  addl   $0x1,0x8(%rsp)
  jmp    L113
L53:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int checkRomanN(char romanNumber[20]) {
  int numerals[5] = {0}, correct = 1, i;
  for (i = 0; i < strlen(romanNumber); i++) {
    if (romanNumber[i] != 'I' && romanNumber[i] != 'V' && romanNumber[i] != 'X' && romanNumber[i] != 'L' && romanNumber[i] != 'C') {
      switch (romanNumber[i]) {
      case 'I':
        numerals[0]++;
        break;
      case 'V':
        numerals[1]++;
        break;
      case 'X':
        numerals[2]++;
        break;
      case 'L':
        numerals[3]++;
        break;
      case 'C':
        numerals[4]++;
        break;
      }
    }
  }
  for (i = 0; i < 5; i++) {
    if (numerals[i] > 3)
      return 0;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<checkRomanN>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  mov    $0x200201,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  jmp    L45
  nopl   0x0(%rax)
L96:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    $0x49,%al
  je     L133
  cmp    $0x56,%al
  je     L133
  lea    -0x43(%rax),%edx
  cmp    $0x15,%dl
  ja     L90
  bt     %rdx,%r12
  jb     L133
L90:
  cmp    $0x58,%al
  je     L36
  cmp    $0x43,%al
  je     L116
  nopl   0x0(%rax,%rax,1)
L133:
  add    $0x1,%rbx
L45:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L96
  cmpl   $0x3,0x4(%rsp)
  jg     L48
  cmpl   $0x3,0x8(%rsp)
  jg     L48
  xor    %eax,%eax
  cmpl   $0x3,0x10(%rsp)
  setle  %al
L6:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L36:
  addl   $0x1,0x8(%rsp)
  jmp    L133
  nopw   0x0(%rax,%rax,1)
L116:
  addl   $0x1,0x10(%rsp)
  jmp    L133
  nopw   0x0(%rax,%rax,1)
L48:
  xor    %eax,%eax
  jmp    L6
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int checkRomanN(char romanNumber[20]) {
  int numerals[5] = {0}, correct = 1, i;
  for (i = 0; i < strlen(romanNumber); i++) {
    if (romanNumber[i] != 'I' && romanNumber[i] != 'V' && romanNumber[i] != 'X' && romanNumber[i] != 'L' && romanNumber[i] != 'C') {
      switch (romanNumber[i]) {
      case 'I':
        numerals[0]++;
        break;
      case 'V':
        numerals[1]++;
        break;
      case 'X':
        numerals[2]++;
        break;
      case 'L':
        numerals[3]++;
        break;
      case 'C':
        numerals[4]++;
        break;
      }
    }
  }
  for (i = 0; i < 5; i++) {
    if (numerals[i] > 3)
      return 0;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <colour_distance@plt>
  mov    %eax,-0x8(%rbp)
L5:
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <colour_distance@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L0
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L5
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L15
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  jmp    L8
L15:
  mov    $0x0,%eax
L8:
  leave
  ret
```

assistant: ```c
int count(const int x, const int y, const int dir) {
  int dist = 0;
  int my_colour = colour_distance(x, y, dir, 0);
  int last_colour;
  do {
    ++dist;
  } while ((last_colour = colour_distance(x, y, dir, dist)) != 0 && last_colour != my_colour);
  if (last_colour == my_colour)
    return dist - 1;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<count>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %edx,%r14d
  mov    $0x0,%ecx
  call   <colour_distance@plt>
  mov    %eax,%ebp
  mov    $0x0,%ebx
L14:
  mov    %ebx,%r15d
  add    $0x1,%ebx
  mov    %ebx,%ecx
  mov    %r14d,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <colour_distance@plt>
  cmp    %eax,%ebp
  je     L0
  test   %eax,%eax
  jne    L14
L0:
  cmp    %eax,%ebp
  mov    $0x0,%eax
  cmovne %eax,%r15d
  mov    %r15d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int count(const int x, const int y, const int dir) {
  int dist = 0;
  int my_colour = colour_distance(x, y, dir, 0);
  int last_colour;
  do {
    ++dist;
  } while ((last_colour = colour_distance(x, y, dir, dist)) != 0 && last_colour != my_colour);
  if (last_colour == my_colour)
    return dist - 1;
  else
    return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<count>:
  endbr64
  push   %r15
  xor    %ecx,%ecx
  xor    %r15d,%r15d
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <colour_distance@plt>
  mov    %eax,%r13d
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L3:
  cmp    %eax,%r13d
  je     L11
L25:
  mov    %r15d,%r14d
  add    $0x1,%r15d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r15d,%ecx
  mov    %ebx,%edi
  call   <colour_distance@plt>
  test   %eax,%eax
  jne    L3
  test   %r13d,%r13d
  cmovne %eax,%r14d
L11:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int count(const int x, const int y, const int dir) {
  int dist = 0;
  int my_colour = colour_distance(x, y, dir, 0);
  int last_colour;
  do {
    ++dist;
  } while ((last_colour = colour_distance(x, y, dir, dist)) != 0 && last_colour != my_colour);
  if (last_colour == my_colour)
    return dist - 1;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count>:
  endbr64
  push   %r15
  xor    %ecx,%ecx
  xor    %r15d,%r15d
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <colour_distance@plt>
  mov    %eax,%r13d
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L13:
  cmp    %eax,%r13d
  je     L28
L3:
  mov    %r15d,%r14d
  add    $0x1,%r15d
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r15d,%ecx
  mov    %ebx,%edi
  call   <colour_distance@plt>
  test   %eax,%eax
  jne    L13
  test   %r13d,%r13d
  cmovne %eax,%r14d
L28:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int count(const int x, const int y, const int dir) {
  int dist = 0;
  int my_colour = colour_distance(x, y, dir, 0);
  int last_colour;
  do {
    ++dist;
  } while ((last_colour = colour_distance(x, y, dir, dist)) != 0 && last_colour != my_colour);
  if (last_colour == my_colour)
    return dist - 1;
  else
    return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rdx,%rax
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L88
L0:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L94
  subl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L19
L94:
  subl   $0x1,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
L19:
  subq   $0x4,-0x8(%rbp)
L88:
  cmpl   $0x0,-0x1c(%rbp)
  jle    L92
  cmpl   $0x0,-0x20(%rbp)
  jg     L0
L92:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L104
  jmp    L77
L26:
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%rax
  lea    -0x4(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    (%rcx),%edx
  mov    %edx,(%rax)
  subl   $0x1,-0x20(%rbp)
L77:
  cmpl   $0x0,-0x20(%rbp)
  jne    L26
L104:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L33
  jmp    L65
L32:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
L65:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L32
  subl   $0x1,-0x1c(%rbp)
L33:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void merge(int *nums1, int m, int *nums2, int n) {
  int *end = nums1 + n + m - 1;
  while (m > 0 && n > 0) {
    if (nums1[m - 1] > nums2[n - 1])
      *end = nums1[--m];
    else
      *end = nums2[--n];
    end--;
  }
  if (0 == m) {
    while (n != 0) {
      *end-- = nums2[n - 1];
      n--;
    }
  }
  if (0 == m) {
    while (m != 0)
      *end = nums1[m - 1];
    m--;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<merge>:
  endbr64
  mov    %rdi,%r9
  mov    %esi,%eax
  mov    %rdx,%rdi
  movslq %ecx,%rdx
  movslq %esi,%rsi
  add    %rsi,%rdx
  lea    -0x4(%r9,%rdx,4),%rsi
  test   %eax,%eax
  jle    L80
  test   %ecx,%ecx
  jg     L76
L80:
  test   %eax,%eax
  jne    L72
  test   %ecx,%ecx
  je     L72
  movslq %ecx,%rcx
  mov    %rcx,%rax
  neg    %rax
  lea    (%rsi,%rax,4),%rdx
L17:
  mov    -0x4(%rdi,%rcx,4),%eax
  mov    %eax,(%rdx,%rcx,4)
  sub    $0x1,%rcx
  test   %ecx,%ecx
  jne    L17
L72:
  ret
L14:
  sub    $0x1,%ecx
  movslq %ecx,%rdx
  mov    (%rdi,%rdx,4),%edx
L27:
  mov    %edx,(%rsi)
  sub    $0x4,%rsi
  test   %eax,%eax
  jle    L80
  test   %ecx,%ecx
  jle    L80
L76:
  movslq %eax,%rdx
  movslq %ecx,%r8
  mov    -0x4(%rdi,%r8,4),%r10d
  cmp    %r10d,-0x4(%r9,%rdx,4)
  jle    L14
  sub    $0x1,%eax
  movslq %eax,%rdx
  mov    (%r9,%rdx,4),%edx
  jmp    L27
```

assistant: ```c
void merge(int *nums1, int m, int *nums2, int n) {
  int *end = nums1 + n + m - 1;
  while (m > 0 && n > 0) {
    if (nums1[m - 1] > nums2[n - 1])
      *end = nums1[--m];
    else
      *end = nums2[--n];
    end--;
  }
  if (0 == m) {
    while (n != 0) {
      *end-- = nums2[n - 1];
      n--;
    }
  }
  if (0 == m) {
    while (m != 0)
      *end = nums1[m - 1];
    m--;
  }
}
```