ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strlcat>:
  endbr64
  push   %rbx
  test   %rdx,%rdx
  je     L61
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
L58:
  cmpb   $0x0,(%rax)
  je     L78
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L58
  mov    %rcx,%rax
L78:
  mov    %rax,%rbx
  sub    %r8,%rbx
  sub    %rbx,%rdx
  je     L85
  movzbl (%rsi),%r8d
  mov    %rsi,%rcx
  test   %r8b,%r8b
  jne    L11
L2:
  movb   $0x0,(%rax)
  sub    %rsi,%rcx
  lea    (%rcx,%rbx,1),%rax
L26:
  pop    %rbx
  ret
L61:
  mov    %rdx,%rbx
L85:
  mov    %rsi,%rdi
  call   <ft_strlen@plt>
  add    %rbx,%rax
  jmp    L26
L1:
  add    $0x1,%rcx
  movzbl (%rcx),%r8d
  test   %r8b,%r8b
  je     L2
L11:
  cmp    $0x1,%rdx
  je     L1
  mov    %r8b,(%rax)
  sub    $0x1,%rdx
  lea    0x1(%rax),%rax
  jmp    L1
```

assistant: ```c
size_t ft_strlcat(char *dst, const char *src, size_t size) {
  const char *s;
  char *d;
  size_t n;
  size_t dst_len;
  s = src;
  d = dst;
  n = size;
  while (n-- != 0 && *d != '\0')
    d++;
  dst_len = d - dst;
  n = size - dst_len;
  if (n == 0)
    return (dst_len + ft_strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dst_len + (s - src));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_strlcat>:
  endbr64
  push   %r12
  test   %rdx,%rdx
  je     L26
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L95
L25:
  cmpb   $0x0,(%rax)
  jne    L0
L95:
  mov    %rax,%r12
  sub    %r8,%r12
  sub    %r12,%rdx
  je     L59
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  test   %cl,%cl
  jne    L67
  jmp    L15
  nopl   (%rax)
L62:
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  mov    %r8,%rcx
  add    $0x1,%rax
L48:
  lea    0x1(%rcx),%r8
  movzbl 0x1(%rcx),%ecx
  test   %cl,%cl
  je     L75
L67:
  cmp    $0x1,%rdx
  jne    L62
  cmpb   $0x0,0x1(%r8)
  lea    0x1(%r8),%rcx
  jne    L48
  sub    %rsi,%rcx
  add    %rcx,%r12
L15:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L75:
  sub    %rsi,%r8
  add    %r8,%r12
  jmp    L15
L26:
  xor    %r12d,%r12d
L59:
  mov    %rsi,%rdi
  call   <ft_strlen@plt>
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
size_t ft_strlcat(char *dst, const char *src, size_t size) {
  const char *s;
  char *d;
  size_t n;
  size_t dst_len;
  s = src;
  d = dst;
  n = size;
  while (n-- != 0 && *d != '\0')
    d++;
  dst_len = d - dst;
  n = size - dst_len;
  if (n == 0)
    return (dst_len + ft_strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dst_len + (s - src));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_strlcat>:
  endbr64
  push   %r12
  test   %rdx,%rdx
  je     L55
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
  jmp    L77
  nopl   0x0(%rax,%rax,1)
L32:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L38
L77:
  cmpb   $0x0,(%rax)
  jne    L32
L38:
  mov    %rax,%r12
  sub    %r8,%r12
  sub    %r12,%rdx
  je     L87
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  test   %cl,%cl
  jne    L41
  jmp    L76
  nopl   (%rax)
L66:
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  mov    %r8,%rcx
  add    $0x1,%rax
L42:
  lea    0x1(%rcx),%r8
  movzbl 0x1(%rcx),%ecx
  test   %cl,%cl
  je     L70
L41:
  cmp    $0x1,%rdx
  jne    L66
  cmpb   $0x0,0x1(%r8)
  lea    0x1(%r8),%rcx
  jne    L42
  sub    %rsi,%rcx
  add    %rcx,%r12
L76:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L70:
  sub    %rsi,%r8
  add    %r8,%r12
  jmp    L76
L55:
  xor    %r12d,%r12d
L87:
  mov    %rsi,%rdi
  call   <ft_strlen@plt>
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
size_t ft_strlcat(char *dst, const char *src, size_t size) {
  const char *s;
  char *d;
  size_t n;
  size_t dst_len;
  s = src;
  d = dst;
  n = size;
  while (n-- != 0 && *d != '\0')
    d++;
  dst_len = d - dst;
  n = size - dst_len;
  if (n == 0)
    return (dst_len + ft_strlen(s));
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return (dst_len + (s - src));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<srec_skip_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
L8:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L29
  mov    $0x1,%eax
  jmp    L26
L29:
  cmpl   $0xa,-0x4(%rbp)
  je     L4
  cmpl   $0xd,-0x4(%rbp)
  jne    L8
L4:
  mov    $0x0,%eax
L26:
  leave
  ret
```

assistant: ```c
int srec_skip_line(FILE *fp) {
  int c;
  while (1) {
    c = fgetc(fp);
    if (c == EOF) {
      return (1);
    }
    if ((c == 0x0a) || (c == 0x0d)) {
      return (0);
    }
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<srec_skip_line>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
L8:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  mov    D1(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L27
  cmp    $0xa,%eax
  je     L13
  cmp    $0xd,%eax
  jne    L8
L13:
  mov    $0x0,%eax
  jmp    L14
L27:
  mov    $0x1,%eax
L14:
  pop    %rbx
  ret
```

assistant: ```c
int srec_skip_line(FILE *fp) {
  int c;
  while (1) {
    c = fgetc(fp);
    if (c == EOF) {
      return (1);
    }
    if ((c == 0x0a) || (c == 0x0d)) {
      return (0);
    }
  }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<srec_skip_line>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    D3(%rip),%rbp
  jmp    L2
  cs nopw 0x0(%rax,%rax,1)
L13:
  cmp    $0xa,%eax
  je     L37
  cmp    $0xd,%eax
  je     L37
L2:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    %eax,0x0(%rbp)
  jne    L13
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L37:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int srec_skip_line(FILE *fp) {
  int c;
  while (1) {
    c = fgetc(fp);
    if (c == EOF) {
      return (1);
    }
    if ((c == 0x0a) || (c == 0x0d)) {
      return (0);
    }
  }
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<srec_skip_line>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    D6(%rip),%rbp
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L35:
  cmp    $0xa,%eax
  je     L9
  cmp    $0xd,%eax
  je     L9
L6:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    %eax,0x0(%rbp)
  jne    L35
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int srec_skip_line(FILE *fp) {
  int c;
  while (1) {
    c = fgetc(fp);
    if (c == EOF) {
      return (1);
    }
    if ((c == 0x0a) || (c == 0x0d)) {
      return (0);
    }
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L16:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <screen_control@plt>
  mov    %eax,-0x4(%rbp)
L18:
  cmpl   $0x0,-0x4(%rbp)
  jns    L16
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int game_state = 0;
  while (game_state >= 0) {
    game_state = screen_control(game_state);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
L9:
  call   <screen_control@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jns    L9
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int game_state = 0;
  while (game_state >= 0) {
    game_state = screen_control(game_state);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L7:
  call   <screen_control@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jns    L7
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int game_state = 0;
  while (game_state >= 0) {
    game_state = screen_control(game_state);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L1:
  call   <screen_control@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jns    L1
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int game_state = 0;
  while (game_state >= 0) {
    game_state = screen_control(game_state);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<kpnum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %edx,-0x4c(%rbp)
  mov    %ecx,%eax
  mov    %al,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x30(%rbp)
  movl   $0xa,-0x2c(%rbp)
  movl   $0x64,-0x28(%rbp)
  movl   $0x3e8,-0x24(%rbp)
  movl   $0x2710,-0x20(%rbp)
  movabs $0x3736353433323130,%rax
  mov    %rax,-0x13(%rbp)
  movw   $0x3938,-0xb(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x40(%rbp)
  movl   $0x1,-0x38(%rbp)
  cmpl   $0x0,-0x48(%rbp)
  je     L41
  cmpl   $0x0,-0x40(%rbp)
  jns    L41
  mov    $0x2d,%edi
  call   <kputc@plt>
  andl   $0x7fff,-0x40(%rbp)
  mov    -0x40(%rbp),%edx
  mov    $0x8000,%eax
  sub    %edx,%eax
  mov    %eax,-0x40(%rbp)
L41:
  mov    -0x4c(%rbp),%eax
  mov    %eax,-0x3c(%rbp)
  jmp    L115
L22:
  movsbl -0x50(%rbp),%eax
  mov    %eax,%edi
  call   <kputc@plt>
  movl   $0x0,-0x38(%rbp)
  subl   $0x1,-0x3c(%rbp)
L115:
  cmpl   $0x5,-0x3c(%rbp)
  jg     L22
  movl   $0x4,-0x3c(%rbp)
  jmp    L50
L95:
  mov    -0x40(%rbp),%eax
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x30(%rbp,%rdx,4),%esi
  mov    $0x0,%edx
  div    %esi
  mov    %eax,-0x34(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  je     L93
  movl   $0x0,-0x38(%rbp)
L93:
  cmpl   $0x0,-0x38(%rbp)
  je     L101
  cmpl   $0x0,-0x3c(%rbp)
  je     L101
  mov    -0x3c(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jbe    L38
  movsbl -0x50(%rbp),%eax
  mov    %eax,%edi
  call   <kputc@plt>
  jmp    L38
L101:
  mov    -0x34(%rbp),%eax
  movzbl -0x13(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <kputc@plt>
L38:
  mov    -0x40(%rbp),%eax
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x30(%rbp,%rdx,4),%ecx
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%eax
  mov    %eax,-0x40(%rbp)
  subl   $0x1,-0x3c(%rbp)
L50:
  cmpl   $0x0,-0x3c(%rbp)
  jns    L95
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L61
  call   <__stack_chk_fail@plt>
L61:
  leave
  ret
```

assistant: ```c
void kpnum(int n, int is_signed, unsigned int len, char pad) {
  unsigned int divisors[] = {1, 10, 100, 1000, 10000};
  char digits[] = "0123456789";
  unsigned int digit;
  int num = n;
  int d;
  int lz = 1;
  if (is_signed && num < 0) {
    kputc('-');
    num = num & 0x7fffU;
    num = 0x8000U - num;
  }
  for (d = len; d > 5; d--) {
    kputc(pad);
    lz = 0;
  }
  for (d = 4; d >= 0; d--) {
    digit = num / divisors[d];
    if (digit)
      lz = 0;
    if (lz && d) {
      if (d < len)
        kputc(pad);
    } else {
      kputc(digits[digit]);
    }
    num = num % divisors[d];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<kpnum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,%r13d
  mov    %edx,%r14d
  mov    %cl,0xf(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movl   $0xa,0x14(%rsp)
  movl   $0x64,0x18(%rsp)
  movl   $0x3e8,0x1c(%rsp)
  movl   $0x2710,0x20(%rsp)
  movabs $0x3736353433323130,%rax
  mov    %rax,0x2d(%rsp)
  movw   $0x3938,0x35(%rsp)
  movb   $0x0,0x37(%rsp)
  test   %esi,%esi
  je     L94
  test   %edi,%edi
  js     L71
L94:
  mov    %r14d,%ebx
  cmp    $0x5,%r14d
  jle    L67
  movsbl 0xf(%rsp),%ebp
L131:
  mov    %ebp,%edi
  call   <kputc@plt>
  sub    $0x1,%ebx
  cmp    $0x5,%ebx
  jne    L131
  mov    $0x0,%r15d
L38:
  mov    $0x4,%ebx
  jmp    L66
L71:
  mov    $0x2d,%edi
  call   <kputc@plt>
  mov    %r13d,%eax
  and    $0x7fff,%eax
  mov    $0x8000,%r13d
  sub    %eax,%r13d
  jmp    L94
L67:
  mov    $0x1,%r15d
  jmp    L38
L127:
  mov    $0x0,%r15d
L78:
  mov    %ebp,%eax
  mov    $0x0,%edx
  div    %r12d
  movsbl 0x2d(%rsp,%rax,1),%edi
  call   <kputc@plt>
L35:
  mov    %ebp,%eax
  mov    $0x0,%edx
  div    %r12d
  mov    %edx,%r13d
  sub    $0x1,%rbx
  cmp    $0xffffffffffffffff,%rbx
  je     L51
L66:
  mov    %r13d,%ebp
  mov    0x10(%rsp,%rbx,4),%r12d
  cmp    %r12d,%r13d
  jae    L127
  test   %r15d,%r15d
  je     L78
  test   %ebx,%ebx
  je     L78
  cmp    %ebx,%r14d
  jbe    L35
  movsbl 0xf(%rsp),%edi
  call   <kputc@plt>
  jmp    L35
L51:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L116
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L116:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void kpnum(int n, int is_signed, unsigned int len, char pad) {
  unsigned int divisors[] = {1, 10, 100, 1000, 10000};
  char digits[] = "0123456789";
  unsigned int digit;
  int num = n;
  int d;
  int lz = 1;
  if (is_signed && num < 0) {
    kputc('-');
    num = num & 0x7fffU;
    num = 0x8000U - num;
  }
  for (d = len; d > 5; d--) {
    kputc(pad);
    lz = 0;
  }
  for (d = 4; d >= 0; d--) {
    digit = num / divisors[d];
    if (digit)
      lz = 0;
    if (lz && d) {
      if (d < len)
        kputc(pad);
    } else {
      kputc(digits[digit]);
    }
    num = num % divisors[d];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<kpnum>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0xa00000001,%rax
  movb   $0x0,0x27(%rsp)
  mov    %rax,(%rsp)
  movabs $0x3e800000064,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x3736353433323130,%rax
  mov    %rax,0x1d(%rsp)
  mov    $0x3938,%eax
  mov    %ax,0x25(%rsp)
  test   %esi,%esi
  je     L74
  test   %edi,%edi
  js     L16
L74:
  mov    %r13d,%ebp
  cmp    $0x5,%r13d
  jle    L125
L128:
  movsbl %r12b,%r12d
  nopl   0x0(%rax,%rax,1)
L122:
  mov    %r12d,%edi
  sub    $0x1,%ebp
  call   <kputc@plt>
  cmp    $0x5,%ebp
  jne    L122
  xor    %r15d,%r15d
L90:
  mov    $0x2710,%r12d
  mov    $0x4,%ebp
  cmp    %r12d,%ebx
  jb     L119
L127:
  xor    %r15d,%r15d
L23:
  mov    %ebx,%eax
  xor    %edx,%edx
  div    %r12d
  movsbl 0x1d(%rsp,%rax,1),%edi
  call   <kputc@plt>
L17:
  mov    %ebx,%eax
  xor    %edx,%edx
  div    %r12d
  mov    %edx,%ebx
  sub    $0x1,%rbp
  jb     L37
L76:
  mov    (%rsp,%rbp,4),%r12d
  cmp    %r12d,%ebx
  jae    L127
L119:
  test   %ebp,%ebp
  je     L23
  test   %r15d,%r15d
  je     L23
  cmp    %ebp,%r13d
  jbe    L17
  movsbl %r14b,%edi
  call   <kputc@plt>
  mov    %ebx,%eax
  xor    %edx,%edx
  div    %r12d
  mov    %edx,%ebx
  sub    $0x1,%rbp
  jae    L76
L37:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L16:
  mov    $0x2d,%edi
  mov    %r13d,%ebp
  call   <kputc@plt>
  mov    %ebx,%eax
  mov    $0x8000,%ebx
  and    $0x7fff,%eax
  sub    %eax,%ebx
  cmp    $0x5,%r13d
  jg     L128
L125:
  mov    $0x1,%r15d
  jmp    L90
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void kpnum(int n, int is_signed, unsigned int len, char pad) {
  unsigned int divisors[] = {1, 10, 100, 1000, 10000};
  char digits[] = "0123456789";
  unsigned int digit;
  int num = n;
  int d;
  int lz = 1;
  if (is_signed && num < 0) {
    kputc('-');
    num = num & 0x7fffU;
    num = 0x8000U - num;
  }
  for (d = len; d > 5; d--) {
    kputc(pad);
    lz = 0;
  }
  for (d = 4; d >= 0; d--) {
    digit = num / divisors[d];
    if (digit)
      lz = 0;
    if (lz && d) {
      if (d < len)
        kputc(pad);
    } else {
      kputc(digits[digit]);
    }
    num = num % divisors[d];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<kpnum>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x48,%rsp
  movdqa D9(%rip),%xmm0
  mov    %cl,0xf(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x2710,0x20(%rsp)
  movabs $0x3736353433323130,%rax
  mov    %rax,0x2d(%rsp)
  mov    $0x3938,%eax
  mov    %ax,0x35(%rsp)
  movb   $0x0,0x37(%rsp)
  movaps %xmm0,0x10(%rsp)
  test   %esi,%esi
  je     L62
  test   %edi,%edi
  js     L18
L62:
  mov    %r13d,%ebp
  cmp    $0x5,%r13d
  jle    L73
L100:
  movsbl %r12b,%r12d
  xchg   %ax,%ax
L117:
  mov    %r12d,%edi
  sub    $0x1,%ebp
  call   <kputc@plt>
  cmp    $0x5,%ebp
  jne    L117
  xor    %r15d,%r15d
L125:
  mov    $0x4,%ebp
  lea    0x10(%rsp),%r14
L3:
  mov    (%r14,%rbp,4),%r12d
  cmp    %r12d,%ebx
  jb     L41
  xor    %r15d,%r15d
L36:
  mov    %ebx,%eax
  xor    %edx,%edx
  div    %r12d
  movsbl 0x2d(%rsp,%rax,1),%edi
  call   <kputc@plt>
L120:
  mov    %ebx,%eax
  xor    %edx,%edx
  div    %r12d
  mov    %edx,%ebx
  sub    $0x1,%rbp
  jae    L3
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  test   %ebp,%ebp
  je     L36
  test   %r15d,%r15d
  je     L36
  cmp    %ebp,%r13d
  jbe    L120
  movsbl 0xf(%rsp),%edi
  call   <kputc@plt>
  jmp    L120
  nopw   0x0(%rax,%rax,1)
L18:
  mov    $0x2d,%edi
  mov    %r13d,%ebp
  call   <kputc@plt>
  mov    %ebx,%eax
  mov    $0x8000,%ebx
  and    $0x7fff,%eax
  sub    %eax,%ebx
  cmp    $0x5,%r13d
  jg     L100
L73:
  mov    $0x1,%r15d
  jmp    L125
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type_guessed: byte[16]

tool: {"D9": [0x01,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x64,0x00,0x00,0x00,0xe8,0x03,0x00,0x00]}

assistant: ```c
void kpnum(int n, int is_signed, unsigned int len, char pad) {
  unsigned int divisors[] = {1, 10, 100, 1000, 10000};
  char digits[] = "0123456789";
  unsigned int digit;
  int num = n;
  int d;
  int lz = 1;
  if (is_signed && num < 0) {
    kputc('-');
    num = num & 0x7fffU;
    num = 0x8000U - num;
  }
  for (d = len; d > 5; d--) {
    kputc(pad);
    lz = 0;
  }
  for (d = 4; d >= 0; d--) {
    digit = num / divisors[d];
    if (digit)
      lz = 0;
    if (lz && d) {
      if (d < len)
        kputc(pad);
    } else {
      kputc(digits[digit]);
    }
    num = num % divisors[d];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L32
L37:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  jne    L26
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    $0x30,%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L6
L26:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <get_char_key@plt>
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L6:
  addq   $0x1,-0x8(%rbp)
L32:
  cmpq   $0x6,-0x8(%rbp)
  jbe    L37
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void print_number(int places[], int number[]) {
  size_t i;
  for (i = 0; i < 7; i++) {
    if (places[i] == -1) {
      putchar(number[i] + '0');
    } else {
      putchar(get_char_key(number[i], places[i]));
    }
  }
  putchar('\n');
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_number>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%ebx
  jmp    L8
L12:
  movzbl (%r12,%rbx,4),%eax
  lea    0x30(%rax),%edi
  movsbl %dil,%edi
  call   <putchar@plt>
L23:
  add    $0x1,%rbx
  cmp    $0x7,%rbx
  je     L6
L8:
  mov    0x0(%rbp,%rbx,4),%esi
  cmp    $0xffffffff,%esi
  je     L12
  mov    (%r12,%rbx,4),%edi
  call   <get_char_key@plt>
  movsbl %al,%edi
  call   <putchar@plt>
  jmp    L23
L6:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void print_number(int places[], int number[]) {
  size_t i;
  for (i = 0; i < 7; i++) {
    if (places[i] == -1) {
      putchar(number[i] + '0');
    } else {
      putchar(get_char_key(number[i], places[i]));
    }
  }
  putchar('\n');
}
```
user: What is the c source code of the assembly code below:

```asm
<print_number>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L36:
  call   <get_char_key@plt>
  add    $0x1,%rbx
  movsbl %al,%edi
  call   <putchar@plt>
  cmp    $0x7,%rbx
  je     L27
L30:
  mov    (%r12,%rbx,4),%esi
  mov    0x0(%rbp,%rbx,4),%edi
  cmp    $0xffffffff,%esi
  jne    L36
  add    $0x30,%edi
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <putchar@plt>
  cmp    $0x7,%rbx
  jne    L30
L27:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
```

assistant: ```c
void print_number(int places[], int number[]) {
  size_t i;
  for (i = 0; i < 7; i++) {
    if (places[i] == -1) {
      putchar(number[i] + '0');
    } else {
      putchar(get_char_key(number[i], places[i]));
    }
  }
  putchar('\n');
}
```
user: What is the c source code of the assembly code below:

```asm
<print_number>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L22:
  call   <get_char_key@plt>
  add    $0x1,%rbx
  movsbl %al,%edi
  call   <putchar@plt>
  cmp    $0x7,%rbx
  je     L12
L6:
  mov    (%r12,%rbx,4),%esi
  mov    0x0(%rbp,%rbx,4),%edi
  cmp    $0xffffffff,%esi
  jne    L22
  add    $0x30,%edi
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <putchar@plt>
  cmp    $0x7,%rbx
  jne    L6
L12:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
```

assistant: ```c
void print_number(int places[], int number[]) {
  size_t i;
  for (i = 0; i < 7; i++) {
    if (places[i] == -1) {
      putchar(number[i] + '0');
    } else {
      putchar(get_char_key(number[i], places[i]));
    }
  }
  putchar('\n');
}
```