ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sortColors>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%esi
  jbe    L109
  movq   $0x0,0x8(%rsp)
  mov    %rdi,%r8
  movl   $0x0,0x10(%rsp)
  test   %esi,%esi
  jle    L98
  lea    -0x1(%rsi),%edx
  mov    %rdi,%rax
  lea    0x4(%rdi,%rdx,4),%rcx
  nopl   0x0(%rax)
L31:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,0x8(%rsp,%rdx,4)
  cmp    %rcx,%rax
  jne    L31
L98:
  xor    %edi,%edi
  xor    %r9d,%r9d
  lea    0x8(%rsp),%r11
L108:
  mov    (%r11,%rdi,4),%esi
  mov    %edi,%r10d
  test   %esi,%esi
  jle    L114
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L128
  movslq %r9d,%rax
  movd   %r10d,%xmm1
  lea    (%r8,%rax,4),%rdx
  mov    %esi,%eax
  pshufd $0x0,%xmm1,%xmm0
  shr    $0x2,%eax
  lea    -0x1(%rax),%ecx
  lea    0x10(%rdx),%rax
  shl    $0x4,%rcx
  add    %rax,%rcx
  jmp    L26
L4:
  add    $0x10,%rax
L26:
  movups %xmm0,(%rdx)
  mov    %rax,%rdx
  cmp    %rax,%rcx
  jne    L4
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  lea    (%r9,%rax,1),%edx
  cmp    %eax,%esi
  je     L112
L80:
  movslq %edx,%rdx
  mov    %r10d,(%r8,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%esi
  jle    L112
  add    $0x2,%eax
  mov    %r10d,0x4(%r8,%rcx,1)
  cmp    %eax,%esi
  jle    L112
  mov    %r10d,0x8(%r8,%rcx,1)
L112:
  add    %esi,%r9d
L114:
  add    $0x1,%rdi
  cmp    $0x3,%rdi
  jne    L108
L109:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L56
  add    $0x28,%rsp
  ret
L128:
  mov    %r9d,%edx
  xor    %eax,%eax
  jmp    L80
L56:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortColors(int *nums, int numsSize) {
  if (!numsSize || numsSize == 1)
    return;
  int color[3] = {0, 0, 0};
  int i, j, p;
  for (i = 0; i < numsSize; i++)
    color[nums[i]]++;
  for (p = i = 0; i < 3; i++) {
    for (j = 0; j < color[i]; j++)
      nums[p++] = i;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<chop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L17
L45:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L44
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  subq   $0x1,-0x20(%rbp)
L17:
  cmpq   $0x0,-0x20(%rbp)
  jne    L45
  jmp    L64
L44:
  nop
  jmp    L64
L16:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xd,%al
  jne    L10
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  subq   $0x1,-0x20(%rbp)
L64:
  cmpq   $0x0,-0x20(%rbp)
  jne    L16
  jmp    L51
L10:
  nop
L51:
  mov    -0x20(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t chop(char *buffer, size_t len) {
  char *ptr;
  ptr = buffer + len - 1;
  while (len) {
    if (*ptr != '\n')
      break;
    *ptr-- = 0;
    len--;
  }
  while (len) {
    if (*ptr != '\r')
      break;
    *ptr-- = 0;
    len--;
  }
  return len;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<chop>:
  endbr64
  lea    -0x1(%rsi,%rdi,1),%rax
  test   %rsi,%rsi
  je     L12
L7:
  cmpb   $0xa,(%rax)
  jne    L68
  sub    $0x1,%rax
  movb   $0x0,0x1(%rax)
  sub    $0x1,%rsi
  jne    L7
L12:
  mov    $0x0,%edx
  jmp    L41
L68:
  test   %rsi,%rsi
  je     L63
  mov    %rsi,%rdx
  sub    %rsi,%rax
L24:
  cmpb   $0xd,(%rax,%rdx,1)
  jne    L41
  movb   $0x0,(%rax,%rdx,1)
  sub    $0x1,%rdx
  jne    L24
L41:
  mov    %rdx,%rax
  ret
L63:
  mov    %rsi,%rdx
  jmp    L41
```

assistant: ```c
size_t chop(char *buffer, size_t len) {
  char *ptr;
  ptr = buffer + len - 1;
  while (len) {
    if (*ptr != '\n')
      break;
    *ptr-- = 0;
    len--;
  }
  while (len) {
    if (*ptr != '\r')
      break;
    *ptr-- = 0;
    len--;
  }
  return len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<chop>:
  endbr64
  mov    %rsi,%rax
  lea    -0x1(%rsi,%rdi,1),%rdx
  test   %rsi,%rsi
  jne    L13
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L48:
  sub    $0x1,%rdx
  movb   $0x0,0x1(%rdx)
  sub    $0x1,%rax
  je     L69
L13:
  movzbl (%rdx),%ecx
  cmp    $0xa,%cl
  je     L48
  sub    %rax,%rdx
  jmp    L49
  nopl   0x0(%rax,%rax,1)
L61:
  movb   $0x0,(%rdx,%rax,1)
  sub    $0x1,%rax
  je     L69
  movzbl (%rdx,%rax,1),%ecx
L49:
  cmp    $0xd,%cl
  je     L61
  ret
  nopl   0x0(%rax)
L69:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t chop(char *buffer, size_t len) {
  char *ptr;
  ptr = buffer + len - 1;
  while (len) {
    if (*ptr != '\n')
      break;
    *ptr-- = 0;
    len--;
  }
  while (len) {
    if (*ptr != '\r')
      break;
    *ptr-- = 0;
    len--;
  }
  return len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<chop>:
  endbr64
  mov    %rsi,%rax
  lea    -0x1(%rsi,%rdi,1),%rdx
  test   %rsi,%rsi
  jne    L17
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L0:
  sub    $0x1,%rdx
  movb   $0x0,0x1(%rdx)
  sub    $0x1,%rax
  je     L31
L17:
  cmpb   $0xa,(%rdx)
  je     L0
  sub    %rax,%rdx
  jmp    L42
L5:
  movb   $0x0,(%rdx,%rax,1)
  sub    $0x1,%rax
  je     L31
L42:
  cmpb   $0xd,(%rdx,%rax,1)
  je     L5
  ret
  nopl   0x0(%rax)
L31:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t chop(char *buffer, size_t len) {
  char *ptr;
  ptr = buffer + len - 1;
  while (len) {
    if (*ptr != '\n')
      break;
    *ptr-- = 0;
    len--;
  }
  while (len) {
    if (*ptr != '\r')
      break;
    *ptr-- = 0;
    len--;
  }
  return len;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L10:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L22
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
L22:
  addl   $0x1,-0xc(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L10
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strip(const char *str) {
  char *stripped = malloc(strlen(str)) + 1;
  int i = 0;
  while (i < strlen(str)) {
    if (str[i] != '\n') {
      stripped[i] = str[i];
    }
    i++;
  }
  stripped[i] = '\0';
  return stripped;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strip>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  lea    0x1(%rax),%r13
  mov    $0x0,%ebx
  jmp    L34
L16:
  add    $0x1,%rbx
L34:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L20
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    $0xa,%al
  je     L16
  mov    %al,0x1(%r12,%rbx,1)
  jmp    L16
L20:
  movslq %ebx,%r14
  movb   $0x0,0x1(%r12,%r14,1)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *strip(const char *str) {
  char *stripped = malloc(strlen(str)) + 1;
  int i = 0;
  while (i < strlen(str)) {
    if (str[i] != '\n') {
      stripped[i] = str[i];
    }
    i++;
  }
  stripped[i] = '\0';
  return stripped;
}
```
user: What is the c source code of the assembly code below?

```asm
<strip>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  lea    0x1(%rax),%r13
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L18:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    $0xa,%al
  je     L22
  mov    %al,0x1(%r12,%rbx,1)
L22:
  add    $0x1,%rbx
L2:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L18
  movb   $0x0,0x1(%r12,%rbx,1)
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strip(const char *str) {
  char *stripped = malloc(strlen(str)) + 1;
  int i = 0;
  while (i < strlen(str)) {
    if (str[i] != '\n') {
      stripped[i] = str[i];
    }
    i++;
  }
  stripped[i] = '\0';
  return stripped;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strip>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  lea    0x1(%rax),%r13
  jmp    L12
  nopl   0x0(%rax,%rax,1)
L7:
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    $0xa,%al
  je     L20
  mov    %al,0x1(%r12,%rbx,1)
L20:
  add    $0x1,%rbx
L12:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L7
  movb   $0x0,0x1(%r12,%rbx,1)
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strip(const char *str) {
  char *stripped = malloc(strlen(str)) + 1;
  int i = 0;
  while (i < strlen(str)) {
    if (str[i] != '\n') {
      stripped[i] = str[i];
    }
    i++;
  }
  stripped[i] = '\0';
  return stripped;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<udiv64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    $0x20,%edx
  mov    -0x28(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  test   %eax,%eax
  jne    L53
  movl   $0x0,-0x10(%rbp)
  mov    $0x20,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <divl@plt>
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  imul   -0x10(%rbp),%eax
  add    %ecx,%eax
  jmp    L21
L53:
  mov    -0x24(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jge    L40
  mov    $0x0,%eax
  jmp    L21
L40:
  mov    $0x20,%edx
  mov    -0x28(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <nlz@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    -0x28(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    $0x20,%edx
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  sar    %eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <divl@plt>
  mov    %eax,%edx
  mov    $0x1f,%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  sub    %edx,%eax
  cmp    %eax,-0x28(%rbp)
  jle    L33
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  jmp    L21
L33:
  mov    -0x14(%rbp),%eax
L21:
  leave
  ret
```

assistant: ```c
uint64_t udiv64(uint64_t n, uint64_t d) {
  if ((d >> 32) == 0) {
    uint64_t b = 1ULL << 32;
    uint32_t n1 = n >> 32;
    uint32_t n0 = n;
    uint32_t d0 = d;
    return divl(b * (n1 % d0) + n0, d0) + b * (n1 / d0);
  } else {
    if (n < d) {
      return 0;
    } else {
      uint32_t d1 = d >> 32;
      int s = nlz(d1);
      uint64_t q = divl(n >> 1, (d << s) >> 32) >> (31 - s);
      return n - (q - 1) * d < d ? q - 1 : q;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<udiv64>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %esi,%edi
  mov    $0x20,%ecx
  sar    %cl,%edi
  test   %edi,%edi
  je     L23
  mov    $0x0,%eax
  cmp    %ebp,%esi
  jle    L15
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  mov    %ebp,%edi
  call   <divl@plt>
  jmp    L39
L15:
  call   <nlz@plt>
  mov    %eax,%r12d
  mov    %ebx,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    $0x20,%ecx
  sar    %cl,%esi
  mov    %ebp,%edi
  sar    %edi
  call   <divl@plt>
  mov    %eax,%esi
  mov    $0x1f,%ecx
  sub    %r12d,%ecx
  sar    %cl,%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  imul   %ebx,%edi
  mov    %ebp,%edx
  sub    %edi,%edx
  sar    %cl,%esi
  cmp    %ebx,%edx
  cmovge %esi,%eax
  jmp    L39
```

assistant: ```c
uint64_t udiv64(uint64_t n, uint64_t d) {
  if ((d >> 32) == 0) {
    uint64_t b = 1ULL << 32;
    uint32_t n1 = n >> 32;
    uint32_t n0 = n;
    uint32_t d0 = d;
    return divl(b * (n1 % d0) + n0, d0) + b * (n1 / d0);
  } else {
    if (n < d) {
      return 0;
    } else {
      uint32_t d1 = d >> 32;
      int s = nlz(d1);
      uint64_t q = divl(n >> 1, (d << s) >> 32) >> (31 - s);
      return n - (q - 1) * d < d ? q - 1 : q;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<udiv64>:
  endbr64
  push   %r12
  mov    $0x20,%ecx
  mov    %edi,%r12d
  mov    %esi,%edi
  sar    %cl,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  test   %edi,%edi
  je     L0
  xor    %eax,%eax
  cmp    %r12d,%esi
  jle    L11
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L11:
  call   <nlz@plt>
  mov    %ebp,%esi
  mov    %r12d,%edi
  mov    %eax,%ecx
  sar    %edi
  mov    %eax,%ebx
  shl    %cl,%esi
  mov    $0x20,%ecx
  sar    %cl,%esi
  call   <divl@plt>
  mov    $0x1f,%ecx
  mov    %r12d,%edx
  sub    %ebx,%ecx
  pop    %rbx
  sar    %cl,%eax
  lea    -0x1(%rax),%ecx
  mov    %ecx,%esi
  imul   %ebp,%esi
  sub    %esi,%edx
  cmp    %ebp,%edx
  pop    %rbp
  pop    %r12
  cmovl  %ecx,%eax
  ret
  xchg   %ax,%ax
L0:
  pop    %rbx
  mov    %r12d,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <divl@plt>
```

assistant: ```c
uint64_t udiv64(uint64_t n, uint64_t d) {
  if ((d >> 32) == 0) {
    uint64_t b = 1ULL << 32;
    uint32_t n1 = n >> 32;
    uint32_t n0 = n;
    uint32_t d0 = d;
    return divl(b * (n1 % d0) + n0, d0) + b * (n1 / d0);
  } else {
    if (n < d) {
      return 0;
    } else {
      uint32_t d1 = d >> 32;
      int s = nlz(d1);
      uint64_t q = divl(n >> 1, (d << s) >> 32) >> (31 - s);
      return n - (q - 1) * d < d ? q - 1 : q;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<udiv64>:
  endbr64
  push   %r12
  mov    $0x20,%ecx
  mov    %edi,%r12d
  mov    %esi,%edi
  sar    %cl,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  test   %edi,%edi
  je     L19
  xor    %eax,%eax
  cmp    %r12d,%esi
  jle    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L10:
  call   <nlz@plt>
  mov    %ebp,%esi
  mov    %r12d,%edi
  mov    %eax,%ecx
  sar    %edi
  mov    %eax,%ebx
  shl    %cl,%esi
  mov    $0x20,%ecx
  sar    %cl,%esi
  call   <divl@plt>
  mov    $0x1f,%ecx
  mov    %r12d,%edx
  sub    %ebx,%ecx
  pop    %rbx
  sar    %cl,%eax
  lea    -0x1(%rax),%ecx
  mov    %ecx,%esi
  imul   %ebp,%esi
  sub    %esi,%edx
  cmp    %ebp,%edx
  pop    %rbp
  pop    %r12
  cmovl  %ecx,%eax
  ret
  xchg   %ax,%ax
L19:
  pop    %rbx
  mov    %r12d,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <divl@plt>
```

assistant: ```c
uint64_t udiv64(uint64_t n, uint64_t d) {
  if ((d >> 32) == 0) {
    uint64_t b = 1ULL << 32;
    uint32_t n1 = n >> 32;
    uint32_t n0 = n;
    uint32_t d0 = d;
    return divl(b * (n1 % d0) + n0, d0) + b * (n1 / d0);
  } else {
    if (n < d) {
      return 0;
    } else {
      uint32_t d1 = d >> 32;
      int s = nlz(d1);
      uint64_t q = divl(n >> 1, (d << s) >> 32) >> (31 - s);
      return n - (q - 1) * d < d ? q - 1 : q;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ReadCFG@plt>
  mov    $0x0,%eax
  call   <ReadDATfile@plt>
  mov    $0x0,%eax
  call   <OpenOurFiles@plt>
  mov    $0x0,%eax
  call   <Looper@plt>
  mov    $0x0,%eax
  call   <fix_dat@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  ReadCFG();
  ReadDATfile();
  OpenOurFiles();
  Looper();
  fix_dat();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ReadCFG@plt>
  mov    $0x0,%eax
  call   <ReadDATfile@plt>
  mov    $0x0,%eax
  call   <OpenOurFiles@plt>
  mov    $0x0,%eax
  call   <Looper@plt>
  mov    $0x0,%eax
  call   <fix_dat@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  ReadCFG();
  ReadDATfile();
  OpenOurFiles();
  Looper();
  fix_dat();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ReadCFG@plt>
  xor    %eax,%eax
  call   <ReadDATfile@plt>
  xor    %eax,%eax
  call   <OpenOurFiles@plt>
  xor    %eax,%eax
  call   <Looper@plt>
  xor    %eax,%eax
  call   <fix_dat@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  ReadCFG();
  ReadDATfile();
  OpenOurFiles();
  Looper();
  fix_dat();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ReadCFG@plt>
  xor    %eax,%eax
  call   <ReadDATfile@plt>
  xor    %eax,%eax
  call   <OpenOurFiles@plt>
  xor    %eax,%eax
  call   <Looper@plt>
  xor    %eax,%eax
  call   <fix_dat@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  ReadCFG();
  ReadDATfile();
  OpenOurFiles();
  Looper();
  fix_dat();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ultoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %edx,-0x74(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x70(%rbp),%rax
  mov    %rax,-0x58(%rbp)
  cmpl   $0x1,-0x74(%rbp)
  jle    L35
  cmpl   $0x24,-0x74(%rbp)
  jg     L35
  movl   $0x0,-0x60(%rbp)
L8:
  mov    -0x74(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x68(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  mov    %eax,-0x5c(%rbp)
  cmpl   $0x9,-0x5c(%rbp)
  jg     L32
  mov    -0x5c(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%ecx
  jmp    L66
L32:
  mov    -0x5c(%rbp),%eax
  add    $0x57,%eax
  mov    %eax,%ecx
L66:
  mov    -0x60(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x60(%rbp)
  cltq
  mov    %cl,-0x50(%rbp,%rax,1)
  mov    -0x74(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x68(%rbp),%rax
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,-0x68(%rbp)
  cmpq   $0x0,-0x68(%rbp)
  jne    L8
L41:
  subl   $0x1,-0x60(%rbp)
  mov    -0x58(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x58(%rbp)
  mov    -0x60(%rbp),%edx
  movslq %edx,%rdx
  movzbl -0x50(%rbp,%rdx,1),%edx
  mov    %dl,(%rax)
  cmpl   $0x0,-0x60(%rbp)
  jg     L41
L35:
  mov    -0x58(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x70(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
char *ultoa(unsigned long val, char *s, int radix) {
  char *_s = s;
  if (radix >= 2 && radix <= 36) {
    char buf[sizeof(val) * 8];
    int i = 0;
    do {
      int mod = val % radix;
      buf[i++] = (mod < 10) ? '0' + mod : 'a' + mod - 10;
      val /= radix;
    } while (val > 0);
    do {
      *_s++ = buf[--i];
    } while (i > 0);
  }
  *_s = '\0';
  return s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ultoa>:
  endbr64
  sub    $0x58,%rsp
  mov    %rsi,%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    -0x2(%rdx),%esi
  mov    %r9,%rax
  cmp    $0x22,%esi
  ja     L5
  mov    %rdi,%rcx
  mov    $0x1,%edi
  movslq %edx,%rsi
  lea    -0x1(%rsp),%r10
  jmp    L1
L10:
  mov    %rax,%rcx
L1:
  mov    %rcx,%rax
  mov    $0x0,%edx
  div    %rsi
  lea    0x30(%rdx),%r8d
  lea    0x57(%rdx),%eax
  cmp    $0x9,%edx
  cmovle %r8d,%eax
  mov    %edi,%r8d
  mov    %al,(%r10,%rdi,1)
  mov    %rcx,%rax
  mov    $0x0,%edx
  div    %rsi
  add    $0x1,%rdi
  cmp    %rcx,%rsi
  jbe    L10
  movslq %r8d,%rax
  mov    %r9,%rdx
  lea    -0x1(%rsp),%rsi
L29:
  add    $0x1,%rdx
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,-0x1(%rdx)
  sub    $0x1,%rax
  test   %eax,%eax
  jg     L29
  lea    -0x1(%r8),%eax
  add    $0x1,%rax
  test   %r8d,%r8d
  mov    $0x1,%edx
  cmovle %rdx,%rax
  add    %r9,%rax
L5:
  movb   $0x0,(%rax)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    %r9,%rax
  add    $0x58,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ultoa(unsigned long val, char *s, int radix) {
  char *_s = s;
  if (radix >= 2 && radix <= 36) {
    char buf[sizeof(val) * 8];
    int i = 0;
    do {
      int mod = val % radix;
      buf[i++] = (mod < 10) ? '0' + mod : 'a' + mod - 10;
      val /= radix;
    } while (val > 0);
    do {
      *_s++ = buf[--i];
    } while (i > 0);
  }
  *_s = '\0';
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ultoa>:
  endbr64
  sub    $0x58,%rsp
  lea    -0x2(%rdx),%ecx
  mov    %rsi,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsi,%rax
  cmp    $0x22,%ecx
  ja     L18
  movslq %edx,%r8
  mov    $0x1,%esi
  lea    -0x1(%rsp),%r11
  jmp    L47
  nopl   (%rax)
L8:
  mov    %rax,%rdi
L47:
  xor    %edx,%edx
  mov    %rdi,%rax
  div    %r8
  lea    0x30(%rdx),%r9d
  cmp    $0x9,%rdx
  lea    0x57(%rdx),%ecx
  mov    %r9d,%edx
  mov    %esi,%r9d
  cmova  %ecx,%edx
  mov    %dl,(%r11,%rsi,1)
  add    $0x1,%rsi
  cmp    %r8,%rdi
  jae    L8
  movslq %r9d,%rdi
  lea    -0x1(%r9),%esi
  mov    %r10,%rax
  lea    (%rsp,%rdi,1),%rcx
  add    %r10,%rsi
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L59:
  movzbl -0x1(%rcx),%edx
  add    $0x1,%rax
L39:
  mov    %dl,(%rax)
  sub    $0x1,%rcx
  cmp    %rsi,%rax
  jne    L59
  lea    (%r10,%rdi,1),%rax
L18:
  movb   $0x0,(%rax)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    %r10,%rax
  add    $0x58,%rsp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ultoa(unsigned long val, char *s, int radix) {
  char *_s = s;
  if (radix >= 2 && radix <= 36) {
    char buf[sizeof(val) * 8];
    int i = 0;
    do {
      int mod = val % radix;
      buf[i++] = (mod < 10) ? '0' + mod : 'a' + mod - 10;
      val /= radix;
    } while (val > 0);
    do {
      *_s++ = buf[--i];
    } while (i > 0);
  }
  *_s = '\0';
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ultoa>:
  endbr64
  sub    $0x58,%rsp
  mov    %rsi,%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    -0x2(%rdx),%eax
  cmp    $0x22,%eax
  ja     L21
  mov    %rdi,%rcx
  mov    $0x1,%esi
  lea    -0x1(%rsp),%r10
  movslq %edx,%rdi
  jmp    L35
  nopl   (%rax)
L14:
  add    $0x30,%edx
  mov    %dl,(%r10,%rsi,1)
  add    $0x1,%rsi
  cmp    %rdi,%rcx
  jb     L37
L66:
  mov    %rax,%rcx
L35:
  mov    %rcx,%rax
  xor    %edx,%edx
  mov    %esi,%r8d
  div    %rdi
  cmp    $0x9,%rdx
  jbe    L14
  add    $0x57,%edx
  mov    %dl,(%r10,%rsi,1)
  add    $0x1,%rsi
  cmp    %rdi,%rcx
  jae    L66
L37:
  movslq %r8d,%rsi
  mov    %r9,%rax
  lea    (%rsp,%rsi,1),%rdx
  add    %r9,%rsi
  nopl   0x0(%rax)
L7:
  movzbl -0x1(%rdx),%ecx
  add    $0x1,%rax
  sub    $0x1,%rdx
  mov    %cl,-0x1(%rax)
  cmp    %rax,%rsi
  jne    L7
L21:
  movb   $0x0,(%rsi)
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    %r9,%rax
  add    $0x58,%rsp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ultoa(unsigned long val, char *s, int radix) {
  char *_s = s;
  if (radix >= 2 && radix <= 36) {
    char buf[sizeof(val) * 8];
    int i = 0;
    do {
      int mod = val % radix;
      buf[i++] = (mod < 10) ? '0' + mod : 'a' + mod - 10;
      val /= radix;
    } while (val > 0);
    do {
      *_s++ = buf[--i];
    } while (i > 0);
  }
  *_s = '\0';
  return s;
}
```