ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<cmd_showbind>:
  endbr64
  push   %rbp
  mov    $0x20,%esi
  mov    %rdi,%rbp
  call   <strchr@plt>
  test   %rax,%rax
  je     L12
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rsi
  movzbl 0x1(%rax),%eax
  cmp    $0x20,%al
  jne    L15
  cs nopw 0x0(%rax,%rax,1)
L36:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  cmp    $0x20,%al
  je     L36
L15:
  test   %al,%al
  jne    L6
L12:
  lea    D4(%rip),%rdi
  pop    %rbp
  jmp    1070 <error_msg@plt>
  nopl   (%rax)
L6:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1080 <show_binding@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type: string

tool: D4: expecting 2 arguments (context and key)


assistant: ```c
void cmd_showbind(char *arg) {
  char *key;
  key = strchr(arg, ' ');
  if (key == NULL)
    goto err;
  *key++ = 0;
  while (*key == ' ')
    key++;
  if (*key == 0)
    goto err;
  show_binding(arg, key);
  return;
err:
  error_msg("expecting 2 arguments (context and key)\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L10:
  mov    -0x4(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cltq
  test   %rax,%rax
  je     L1
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
L1:
  addl   $0x1,-0x4(%rbp)
L8:
  cmpl   $0x9,-0x4(%rbp)
  jle    L10
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int res = 0;
  for (int i = 0; i < 10; i++) {
    if (__builtin_expect(i % 2, 0)) {
      res += i;
    }
  }
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  mov    $0x19,%eax
  ret
```

assistant: ```c
int main() {
  int res = 0;
  for (int i = 0; i < 10; i++) {
    if (__builtin_expect(i % 2, 0)) {
      res += i;
    }
  }
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  mov    $0x19,%eax
  ret
```

assistant: ```c
int main() {
  int res = 0;
  for (int i = 0; i < 10; i++) {
    if (__builtin_expect(i % 2, 0)) {
      res += i;
    }
  }
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x19,%eax
  ret
```

assistant: ```c
int main() {
  int res = 0;
  for (int i = 0; i < 10; i++) {
    if (__builtin_expect(i % 2, 0)) {
      res += i;
    }
  }
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<ListFirstPrimes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movq   $0x2,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movq   $0x3,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movq   $0x5,(%rax)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movq   $0x7,(%rax)
  movq   $0x8,-0x10(%rbp)
  jmp    L51
L18:
  movl   $0x0,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L40
L57:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rcx
  mov    -0x10(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L0
  movl   $0x1,-0x1c(%rbp)
  jmp    L33
L0:
  addq   $0x1,-0x8(%rbp)
L40:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  imul   %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  jge    L57
L33:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L31
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,(%rdx)
L31:
  addq   $0x1,-0x10(%rbp)
L51:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jl     L18
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long int ListFirstPrimes(long int mpr, long int *prlist) {
  long int prnum = 0;
  int composed = 0;
  long int nmb = 0, j = 0;
  prlist[prnum++] = 2;
  prlist[prnum++] = 3;
  prlist[prnum++] = 5;
  prlist[prnum++] = 7;
  for (nmb = 8; nmb < mpr; nmb++) {
    composed = 0;
    for (j = 0; prlist[j] * prlist[j] <= nmb; j++) {
      if (nmb - prlist[j] * ((long int)(nmb / prlist[j])) == 0) {
        composed = 1;
        break;
      }
    }
    if (composed == 0)
      prlist[prnum++] = nmb;
  }
  return prnum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ListFirstPrimes>:
  endbr64
  movq   $0x2,(%rsi)
  movq   $0x3,0x8(%rsi)
  movq   $0x5,0x10(%rsi)
  movq   $0x7,0x18(%rsi)
  mov    $0x8,%r8d
  mov    $0x4,%r10d
  cmp    $0x8,%rdi
  jg     L33
L57:
  mov    %r10,%rax
  ret
L12:
  mov    %r8,(%rsi,%r10,8)
  lea    0x1(%r10),%r10
L44:
  add    $0x1,%r8
  cmp    %r8,%rdi
  je     L57
L33:
  mov    (%rsi),%rcx
  mov    %rcx,%rax
  imul   %rcx,%rax
  lea    0x8(%rsi),%r9
  cmp    %r8,%rax
  jg     L12
L25:
  mov    %r8,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  je     L44
  mov    (%r9),%rcx
  add    $0x8,%r9
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %r8,%rax
  jle    L25
  jmp    L12
```

assistant: ```c
long int ListFirstPrimes(long int mpr, long int *prlist) {
  long int prnum = 0;
  int composed = 0;
  long int nmb = 0, j = 0;
  prlist[prnum++] = 2;
  prlist[prnum++] = 3;
  prlist[prnum++] = 5;
  prlist[prnum++] = 7;
  for (nmb = 8; nmb < mpr; nmb++) {
    composed = 0;
    for (j = 0; prlist[j] * prlist[j] <= nmb; j++) {
      if (nmb - prlist[j] * ((long int)(nmb / prlist[j])) == 0) {
        composed = 1;
        break;
      }
    }
    if (composed == 0)
      prlist[prnum++] = nmb;
  }
  return prnum;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ListFirstPrimes>:
  endbr64
  movq   $0x2,(%rsi)
  mov    $0x2,%ecx
  mov    $0x8,%r8d
  mov    $0x4,%r10d
  movq   $0x3,0x8(%rsi)
  movq   $0x5,0x10(%rsi)
  movq   $0x7,0x18(%rsi)
  cmp    $0x8,%rdi
  jle    L79
  mov    %rcx,%rax
  lea    0x8(%rsi),%r9
  imul   %rcx,%rax
  cmp    %r8,%rax
  jle    L17
  jmp    L36
  nopl   0x0(%rax)
L27:
  mov    (%r9),%rcx
  add    $0x8,%r9
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %r8,%rax
  jg     L36
L17:
  mov    %r8,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L27
  add    $0x1,%r8
  cmp    %r8,%rdi
  je     L79
L25:
  mov    (%rsi),%rcx
  lea    0x8(%rsi),%r9
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %r8,%rax
  jle    L17
  nopl   0x0(%rax)
L36:
  mov    %r8,(%rsi,%r10,8)
  add    $0x1,%r8
  add    $0x1,%r10
  cmp    %r8,%rdi
  jne    L25
L79:
  mov    %r10,%rax
  ret
```

assistant: ```c
long int ListFirstPrimes(long int mpr, long int *prlist) {
  long int prnum = 0;
  int composed = 0;
  long int nmb = 0, j = 0;
  prlist[prnum++] = 2;
  prlist[prnum++] = 3;
  prlist[prnum++] = 5;
  prlist[prnum++] = 7;
  for (nmb = 8; nmb < mpr; nmb++) {
    composed = 0;
    for (j = 0; prlist[j] * prlist[j] <= nmb; j++) {
      if (nmb - prlist[j] * ((long int)(nmb / prlist[j])) == 0) {
        composed = 1;
        break;
      }
    }
    if (composed == 0)
      prlist[prnum++] = nmb;
  }
  return prnum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ListFirstPrimes>:
  endbr64
  movdqa D19(%rip),%xmm0
  mov    $0x2,%ecx
  mov    $0x8,%r8d
  mov    $0x4,%r10d
  movups %xmm0,(%rsi)
  movdqa D0(%rip),%xmm0
  movups %xmm0,0x10(%rsi)
  cmp    $0x8,%rdi
  jle    L0
  mov    %rcx,%rax
  lea    0x8(%rsi),%r9
  imul   %rcx,%rax
  cmp    %r8,%rax
  jle    L66
  jmp    L21
  nopl   0x0(%rax)
L54:
  mov    (%r9),%rcx
  add    $0x8,%r9
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %r8,%rax
  jg     L21
L66:
  mov    %r8,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L54
  add    $0x1,%r8
  cmp    %r8,%rdi
  je     L0
L23:
  mov    (%rsi),%rcx
  lea    0x8(%rsi),%r9
  mov    %rcx,%rax
  imul   %rcx,%rax
  cmp    %r8,%rax
  jle    L66
  nopl   0x0(%rax)
L21:
  mov    %r8,(%rsi,%r10,8)
  add    $0x1,%r8
  add    $0x1,%r10
  cmp    %r8,%rdi
  jne    L23
L0:
  mov    %r10,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D19
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D0
type_guessed: byte[16]

tool: {"D19": [0x02,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D0": [0x05,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
long int ListFirstPrimes(long int mpr, long int *prlist) {
  long int prnum = 0;
  int composed = 0;
  long int nmb = 0, j = 0;
  prlist[prnum++] = 2;
  prlist[prnum++] = 3;
  prlist[prnum++] = 5;
  prlist[prnum++] = 7;
  for (nmb = 8; nmb < mpr; nmb++) {
    composed = 0;
    for (j = 0; prlist[j] * prlist[j] <= nmb; j++) {
      if (nmb - prlist[j] * ((long int)(nmb / prlist[j])) == 0) {
        composed = 1;
        break;
      }
    }
    if (composed == 0)
      prlist[prnum++] = nmb;
  }
  return prnum;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ft_putchar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x14(%rbp),%eax
  cmp    $0x7f,%eax
  jg     L73
  lea    -0x14(%rbp),%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L58
L73:
  mov    -0x14(%rbp),%eax
  cmp    $0x7ff,%eax
  jg     L60
  mov    -0x14(%rbp),%eax
  sar    $0x6,%eax
  or     $0xc0,%al
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  shl    $0x8,%eax
  and    $0x3f00,%eax
  or     $0x80,%ah
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    $0x2,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L58
L60:
  mov    -0x14(%rbp),%eax
  cmp    $0xffff,%eax
  jg     L0
  mov    -0x14(%rbp),%eax
  sar    $0xc,%eax
  or     $0xe0,%al
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x6,%eax
  shl    $0x8,%eax
  and    $0x3f00,%eax
  or     $0x80,%ah
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0x10,%eax
  and    $0x3f0000,%eax
  or     $0x800000,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    $0x3,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L58
L0:
  mov    -0x14(%rbp),%eax
  cmp    $0x10ffff,%eax
  jg     L58
  mov    -0x14(%rbp),%eax
  sar    $0x12,%eax
  or     $0xf0,%al
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0xc,%eax
  shl    $0x8,%eax
  and    $0x3f00,%eax
  or     $0x80,%ah
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x6,%eax
  shl    $0x10,%eax
  and    $0x3f0000,%eax
  or     $0x800000,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0x18,%eax
  and    $0x3f000000,%eax
  or     $0x80000000,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  nop
L58:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L57
  call   <__stack_chk_fail@plt>
L57:
  leave
  ret
```

assistant: ```c
void ft_putchar(int c) {
  int result;
  if (c < 0x80)
    write(1, &c, 1);
  else if (c < 0x800) {
    result = ((c >> 6) | 0xC0) + (((c & 0x3F) | 0x80) << 8);
    write(1, &result, 2);
  } else if (c < 0x10000) {
    result = ((c >> 12) | 0xE0) + ((((c >> 6) & 0x3F) | 0x80) << 8);
    result += ((c & 0x3F) | 0x80) << 16;
    write(1, &result, 3);
  } else if (c < 0x110000) {
    result = ((c >> 18) | 0xF0) + ((((c >> 12) & 0x3F) | 0x80) << 8);
    result += (((c >> 6) & 0x3F) | 0x80) << 16;
    result += ((c & 0x3F) | 0x80) << 24;
    write(1, &result, 4);
  }
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_putchar>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,0xc(%rsp)
  mov    %fs:0x28,%rdx
  mov    %rdx,0x18(%rsp)
  xor    %edx,%edx
  cmp    $0x7f,%edi
  jle    L71
  mov    %edi,%eax
  cmp    $0x7ff,%edi
  jle    L62
  cmp    $0xffff,%edi
  jle    L0
  cmp    $0x10ffff,%edi
  jg     L70
  mov    %edi,%edx
  shr    $0x4,%edx
  and    $0x3f00,%edx
  or     $0x80,%dh
  mov    %edi,%ecx
  sar    $0x12,%ecx
  or     $0xf0,%cl
  add    %ecx,%edx
  mov    %edi,%ecx
  shl    $0xa,%ecx
  and    $0x3f0000,%ecx
  or     $0x800000,%ecx
  add    %ecx,%edx
  shl    $0x18,%eax
  and    $0x3f000000,%eax
  or     $0x80000000,%eax
  add    %edx,%eax
  mov    %eax,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x4,%edx
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L70
L71:
  lea    0xc(%rsp),%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
L70:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L74
  add    $0x28,%rsp
  ret
L62:
  mov    %edi,%edx
  shl    $0x8,%edx
  and    $0x3f00,%edx
  or     $0x80,%dh
  sar    $0x6,%eax
  or     $0xc0,%al
  add    %edx,%eax
  mov    %eax,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x2,%edx
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L70
L0:
  lea    0x0(,%rdi,4),%edx
  and    $0x3f00,%edx
  or     $0x80,%dh
  mov    %edi,%ecx
  sar    $0xc,%ecx
  or     $0xe0,%cl
  add    %ecx,%edx
  shl    $0x10,%eax
  and    $0x3f0000,%eax
  or     $0x800000,%eax
  add    %edx,%eax
  mov    %eax,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x3,%edx
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L70
L74:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ft_putchar(int c) {
  int result;
  if (c < 0x80)
    write(1, &c, 1);
  else if (c < 0x800) {
    result = ((c >> 6) | 0xC0) + (((c & 0x3F) | 0x80) << 8);
    write(1, &result, 2);
  } else if (c < 0x10000) {
    result = ((c >> 12) | 0xE0) + ((((c >> 6) & 0x3F) | 0x80) << 8);
    result += ((c & 0x3F) | 0x80) << 16;
    write(1, &result, 3);
  } else if (c < 0x110000) {
    result = ((c >> 18) | 0xF0) + ((((c >> 12) & 0x3F) | 0x80) << 8);
    result += (((c >> 6) & 0x3F) | 0x80) << 16;
    result += ((c & 0x3F) | 0x80) << 24;
    write(1, &result, 4);
  }
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_putchar>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,0xc(%rsp)
  mov    %fs:0x28,%rdx
  mov    %rdx,0x18(%rsp)
  xor    %edx,%edx
  cmp    $0x7f,%edi
  jle    L30
  mov    %edi,%eax
  cmp    $0x7ff,%edi
  jle    L73
  cmp    $0xffff,%edi
  jle    L48
  cmp    $0x10ffff,%edi
  jle    L57
L39:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x28,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L73:
  mov    %edi,%edx
  sar    $0x6,%eax
  lea    0x14(%rsp),%rsi
  mov    $0x1,%edi
  shl    $0x8,%edx
  or     $0xc0,%al
  and    $0x3f00,%edx
  or     $0x80,%dh
  add    %edx,%eax
  mov    $0x2,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L39
  nop
L30:
  lea    0xc(%rsp),%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L39
  cs nopw 0x0(%rax,%rax,1)
L48:
  lea    0x0(,%rdi,4),%edx
  mov    %edi,%ecx
  shl    $0x10,%eax
  lea    0x14(%rsp),%rsi
  and    $0x3f00,%edx
  sar    $0xc,%ecx
  and    $0x3f0000,%eax
  mov    $0x1,%edi
  or     $0x80,%dh
  or     $0xe0,%cl
  or     $0x800000,%eax
  add    %ecx,%edx
  add    %edx,%eax
  mov    $0x3,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L39
  cs nopw 0x0(%rax,%rax,1)
L57:
  mov    %edi,%edx
  mov    %edi,%ecx
  shl    $0x18,%eax
  lea    0x14(%rsp),%rsi
  shr    $0x4,%edx
  sar    $0x12,%ecx
  and    $0x3f000000,%eax
  and    $0x3f00,%edx
  or     $0xf0,%cl
  or     $0x80000000,%eax
  or     $0x80,%dh
  add    %ecx,%edx
  mov    %edi,%ecx
  mov    $0x1,%edi
  shl    $0xa,%ecx
  and    $0x3f0000,%ecx
  or     $0x800000,%ecx
  add    %ecx,%edx
  add    %edx,%eax
  mov    $0x4,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L39
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ft_putchar(int c) {
  int result;
  if (c < 0x80)
    write(1, &c, 1);
  else if (c < 0x800) {
    result = ((c >> 6) | 0xC0) + (((c & 0x3F) | 0x80) << 8);
    write(1, &result, 2);
  } else if (c < 0x10000) {
    result = ((c >> 12) | 0xE0) + ((((c >> 6) & 0x3F) | 0x80) << 8);
    result += ((c & 0x3F) | 0x80) << 16;
    write(1, &result, 3);
  } else if (c < 0x110000) {
    result = ((c >> 18) | 0xF0) + ((((c >> 12) & 0x3F) | 0x80) << 8);
    result += (((c >> 6) & 0x3F) | 0x80) << 16;
    result += ((c & 0x3F) | 0x80) << 24;
    write(1, &result, 4);
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ft_putchar>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,0xc(%rsp)
  mov    %fs:0x28,%rdx
  mov    %rdx,0x18(%rsp)
  xor    %edx,%edx
  cmp    $0x7f,%edi
  jle    L45
  mov    %edi,%eax
  cmp    $0x7ff,%edi
  jle    L42
  cmp    $0xffff,%edi
  jle    L81
  cmp    $0x10ffff,%edi
  jle    L40
L47:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  add    $0x28,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L42:
  mov    %edi,%edx
  sar    $0x6,%eax
  lea    0x14(%rsp),%rsi
  mov    $0x1,%edi
  shl    $0x8,%edx
  or     $0xc0,%al
  and    $0x3f00,%edx
  or     $0x80,%dh
  add    %edx,%eax
  mov    $0x2,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L47
  nop
L45:
  lea    0xc(%rsp),%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L47
  cs nopw 0x0(%rax,%rax,1)
L81:
  lea    0x0(,%rdi,4),%edx
  mov    %edi,%ecx
  shl    $0x10,%eax
  lea    0x14(%rsp),%rsi
  and    $0x3f00,%edx
  sar    $0xc,%ecx
  and    $0x3f0000,%eax
  mov    $0x1,%edi
  or     $0x80,%dh
  or     $0xe0,%cl
  or     $0x800000,%eax
  add    %ecx,%edx
  add    %edx,%eax
  mov    $0x3,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L47
  cs nopw 0x0(%rax,%rax,1)
L40:
  mov    %edi,%edx
  mov    %edi,%ecx
  shl    $0x18,%eax
  lea    0x14(%rsp),%rsi
  shr    $0x4,%edx
  sar    $0x12,%ecx
  and    $0x3f000000,%eax
  and    $0x3f00,%edx
  or     $0xf0,%cl
  or     $0x80000000,%eax
  or     $0x80,%dh
  add    %ecx,%edx
  mov    %edi,%ecx
  mov    $0x1,%edi
  shl    $0xa,%ecx
  and    $0x3f0000,%ecx
  or     $0x800000,%ecx
  add    %ecx,%edx
  add    %edx,%eax
  mov    $0x4,%edx
  mov    %eax,0x14(%rsp)
  call   <write@plt>
  jmp    L47
L70:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ft_putchar(int c) {
  int result;
  if (c < 0x80)
    write(1, &c, 1);
  else if (c < 0x800) {
    result = ((c >> 6) | 0xC0) + (((c & 0x3F) | 0x80) << 8);
    write(1, &result, 2);
  } else if (c < 0x10000) {
    result = ((c >> 12) | 0xE0) + ((((c >> 6) & 0x3F) | 0x80) << 8);
    result += ((c & 0x3F) | 0x80) << 16;
    write(1, &result, 3);
  } else if (c < 0x110000) {
    result = ((c >> 18) | 0xF0) + ((((c >> 12) & 0x3F) | 0x80) << 8);
    result += (((c >> 6) & 0x3F) | 0x80) << 16;
    result += ((c & 0x3F) | 0x80) << 24;
    write(1, &result, 4);
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<putbits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L38
L22:
  subl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  and    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x7,%eax
  jne    L24
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%ecx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L21
L24:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%esi
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%ecx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  lea    (%rsi,%rcx,1),%edx
  mov    %dl,(%rax)
L21:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L5
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x30(%rbp),%rax
  movl   $0x7,(%rax)
  jmp    L38
L5:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
L38:
  cmpl   $0x0,-0x8(%rbp)
  jg     L22
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void putbits(int data, unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits;
  while (nr > 0) {
    nr--;
    bit = (data & (1 << nr)) >> nr;
    if (*bit_ptr == 7)
      buffer[*byte_ptr] = (unsigned char)(bit << *bit_ptr);
    else
      buffer[*byte_ptr] = (unsigned char)(buffer[*byte_ptr] + (bit << *bit_ptr));
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<putbits>:
  endbr64
  test   %edx,%edx
  jle    L35
  push   %rbx
  mov    %rsi,%r9
  mov    %rcx,%rsi
  mov    $0x1,%r10d
  mov    $0x7,%r11d
  jmp    L10
L27:
  movslq (%rsi),%rcx
  shl    $0x7,%eax
  mov    %al,(%r9,%rcx,1)
  jmp    L9
L24:
  mov    %ecx,(%r8)
  test   %edx,%edx
  je     L40
L10:
  sub    $0x1,%edx
  mov    %r10d,%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  and    %edi,%eax
  sar    %cl,%eax
  mov    (%r8),%ecx
  cmp    $0x7,%ecx
  je     L27
  movslq (%rsi),%rbx
  shl    %cl,%eax
  add    %al,(%r9,%rbx,1)
L9:
  mov    (%r8),%eax
  lea    -0x1(%rax),%ecx
  test   %eax,%eax
  jne    L24
  addl   $0x1,(%rsi)
  mov    %r11d,%ecx
  jmp    L24
L40:
  pop    %rbx
  ret
L35:
  ret
```

assistant: ```c
void putbits(int data, unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits;
  while (nr > 0) {
    nr--;
    bit = (data & (1 << nr)) >> nr;
    if (*bit_ptr == 7)
      buffer[*byte_ptr] = (unsigned char)(bit << *bit_ptr);
    else
      buffer[*byte_ptr] = (unsigned char)(buffer[*byte_ptr] + (bit << *bit_ptr));
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<putbits>:
  endbr64
  mov    %rsi,%r9
  mov    %rcx,%r10
  mov    $0x1,%r11d
  test   %edx,%edx
  jg     L37
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
L16:
  shl    %cl,%eax
  add    (%rsi),%al
L24:
  mov    %al,(%rsi)
  mov    (%r8),%eax
  lea    -0x1(%rax),%ecx
  test   %eax,%eax
  jne    L50
  addl   $0x1,(%r10)
  mov    $0x7,%ecx
L50:
  mov    %ecx,(%r8)
  test   %edx,%edx
  je     L4
L37:
  sub    $0x1,%edx
  mov    %r11d,%eax
  movslq (%r10),%rsi
  mov    %edx,%ecx
  shl    %cl,%eax
  add    %r9,%rsi
  and    %edi,%eax
  sar    %cl,%eax
  mov    (%r8),%ecx
  cmp    $0x7,%ecx
  jne    L16
  shl    $0x7,%eax
  jmp    L24
  nopl   0x0(%rax)
L4:
  ret
```

assistant: ```c
void putbits(int data, unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits;
  while (nr > 0) {
    nr--;
    bit = (data & (1 << nr)) >> nr;
    if (*bit_ptr == 7)
      buffer[*byte_ptr] = (unsigned char)(bit << *bit_ptr);
    else
      buffer[*byte_ptr] = (unsigned char)(buffer[*byte_ptr] + (bit << *bit_ptr));
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
}
```