ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<strncpy>:
  endbr64
  mov    %edx,%ecx
  cld
L12:
  dec    %ecx
  js     L0
  lods   %ds:(%rsi),%al
  stos   %al,%es:(%rdi)
  test   %al,%al
  jne    L12
  rep stos %al,%es:(%rdi)
L0:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, int count) {
  __asm__("cld\n"
          "1:\tdecl %2\n\t"
          "js 2f\n\t"
          "lodsb\n\t"
          "stosb\n\t"
          "testb %%al,%%al\n\t"
          "jne 1b\n\t"
          "rep\n\t"
          "stosb\n"
          "2:" ::"S"(src),
          "D"(dest), "c"(count));
  return dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  mov    %edx,%ecx
  cld
L4:
  dec    %ecx
  js     L2
  lods   %ds:(%rsi),%al
  stos   %al,%es:(%rdi)
  test   %al,%al
  jne    L4
  rep stos %al,%es:(%rdi)
L2:
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, int count) {
  __asm__("cld\n"
          "1:\tdecl %2\n\t"
          "js 2f\n\t"
          "lodsb\n\t"
          "stosb\n\t"
          "testb %%al,%%al\n\t"
          "jne 1b\n\t"
          "rep\n\t"
          "stosb\n"
          "2:" ::"S"(src),
          "D"(dest), "c"(count));
  return dest;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  mov    %edx,%ecx
  cld
L10:
  dec    %ecx
  js     L2
  lods   %ds:(%rsi),%al
  stos   %al,%es:(%rdi)
  test   %al,%al
  jne    L10
  rep stos %al,%es:(%rdi)
L2:
  ret
```

assistant: ```c
char *strncpy(char *dest, const char *src, int count) {
  __asm__("cld\n"
          "1:\tdecl %2\n\t"
          "js 2f\n\t"
          "lodsb\n\t"
          "stosb\n\t"
          "testb %%al,%%al\n\t"
          "jne 1b\n\t"
          "rep\n\t"
          "stosb\n"
          "2:" ::"S"(src),
          "D"(dest), "c"(count));
  return dest;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<asshuku>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L28
  mov    $0x7f,%eax
  jmp    L23
L28:
  cmpl   $0x4,-0x14(%rbp)
  jg     L8
  mov    -0x14(%rbp),%eax
  jmp    L23
L8:
  shll   -0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <isqrt@plt>
  add    $0x1,%eax
  and    $0xfffffffe,%eax
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  cltd
  idiv   %ebx
  sub    $0x2,%eax
  add    %eax,%ebx
  mov    $0x7e,%eax
  cmp    %eax,%ebx
  cmovle %ebx,%eax
L23:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int asshuku(hin)
int hin;
{
  register int c;
  if (hin == -1)
    return (127);
  if (hin <= 4)
    return (hin);
  c = (isqrt((hin <<= 1) + 1) + 1) & ~1;
  c += hin / c - 2;
  return (c < 126 ? c : 126);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<asshuku>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L1
  mov    %edi,%eax
  cmp    $0x4,%edi
  jg     L16
  ret
L16:
  push   %rbx
  lea    (%rdi,%rdi,1),%ebx
  lea    0x1(%rbx),%edi
  mov    $0x0,%eax
  call   <isqrt@plt>
  lea    0x1(%rax),%ecx
  and    $0xfffffffe,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  lea    -0x2(%rcx,%rax,1),%eax
  mov    $0x7e,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  pop    %rbx
  ret
L1:
  mov    $0x7f,%eax
  ret
```

assistant: ```c
int asshuku(hin)
int hin;
{
  register int c;
  if (hin == -1)
    return (127);
  if (hin <= 4)
    return (hin);
  c = (isqrt((hin <<= 1) + 1) + 1) & ~1;
  c += hin / c - 2;
  return (c < 126 ? c : 126);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<asshuku>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L14
  mov    %edi,%eax
  cmp    $0x4,%edi
  jg     L6
  ret
  nopl   0x0(%rax)
L6:
  push   %rbx
  lea    (%rdi,%rdi,1),%ebx
  xor    %eax,%eax
  lea    0x1(%rbx),%edi
  call   <isqrt@plt>
  lea    0x1(%rax),%ecx
  mov    %ebx,%eax
  pop    %rbx
  and    $0xfffffffe,%ecx
  cltd
  idiv   %ecx
  mov    $0x7e,%edx
  lea    -0x2(%rcx,%rax,1),%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  ret
  nopl   0x0(%rax)
L14:
  mov    $0x7f,%eax
  ret
```

assistant: ```c
int asshuku(hin)
int hin;
{
  register int c;
  if (hin == -1)
    return (127);
  if (hin <= 4)
    return (hin);
  c = (isqrt((hin <<= 1) + 1) + 1) & ~1;
  c += hin / c - 2;
  return (c < 126 ? c : 126);
}
```
user: What is the c source code of the assembly code below:

```asm
<asshuku>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L8
  mov    %edi,%eax
  cmp    $0x4,%edi
  jg     L5
  ret
  nopl   0x0(%rax)
L5:
  push   %rbx
  lea    (%rdi,%rdi,1),%ebx
  xor    %eax,%eax
  lea    0x1(%rbx),%edi
  call   <isqrt@plt>
  lea    0x1(%rax),%ecx
  mov    %ebx,%eax
  pop    %rbx
  and    $0xfffffffe,%ecx
  cltd
  idiv   %ecx
  mov    $0x7e,%edx
  lea    -0x2(%rcx,%rax,1),%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  ret
  nopl   0x0(%rax)
L8:
  mov    $0x7f,%eax
  ret
```

assistant: ```c
int asshuku(hin)
int hin;
{
  register int c;
  if (hin == -1)
    return (127);
  if (hin <= 4)
    return (hin);
  c = (isqrt((hin <<= 1) + 1) + 1) & ~1;
  c += hin / c - 2;
  return (c < 126 ? c : 126);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<quaddobl_monodromy_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x54(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L15:
  cmp    %rdx,%rsp
  je     L2
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L15
L2:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L39
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L39:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x30(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2a0,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x4c(%rbp)
  mov    -0x54(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x30(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2a1,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x4c(%rbp)
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x60(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2a3,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int quaddobl_monodromy_permutation(int d, int *done) {
  int *a, *b, fail, i;
  int permutation[d];
  int nf[2];
  double *c;
  fail = _ada_use_c2phc(672, a, permutation, c);
  if (0 > 0) {
    printf("the permutation :");
    for (i = 0; i < d; i++)
      printf(" %d", permutation[i]);
  }
  nf[0] = d;
  nf[1] = 0;
  fail = _ada_use_c2phc(673, nf, permutation, c);
  if (0 > 0)
    printf(" : %d -> %d\n", nf[0], nf[1]);
  fail = _ada_use_c2phc(675, done, b, c);
  return fail;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quaddobl_monodromy_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L20:
  cmp    %rdx,%rsp
  je     L1
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L20
L1:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L18
  orq    $0x0,-0x8(%rsp,%rax,1)
L18:
  mov    %rsp,%r14
  mov    $0x0,%r13d
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    $0x0,%esi
  mov    $0x2a0,%edi
  call   <_ada_use_c2phc@plt>
  mov    %ebx,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  lea    -0x30(%rbp),%rsi
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    $0x2a1,%edi
  call   <_ada_use_c2phc@plt>
  mov    %r13,%rcx
  mov    $0x0,%edx
  mov    %r12,%rsi
  mov    $0x2a3,%edi
  call   <_ada_use_c2phc@plt>
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int quaddobl_monodromy_permutation(int d, int *done) {
  int *a, *b, fail, i;
  int permutation[d];
  int nf[2];
  double *c;
  fail = _ada_use_c2phc(672, a, permutation, c);
  if (0 > 0) {
    printf("the permutation :");
    for (i = 0; i < d; i++)
      printf(" %d", permutation[i]);
  }
  nf[0] = d;
  nf[1] = 0;
  fail = _ada_use_c2phc(673, nf, permutation, c);
  if (0 > 0)
    printf(" : %d -> %d\n", nf[0], nf[1]);
  fail = _ada_use_c2phc(675, done, b, c);
  return fail;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<quaddobl_monodromy_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L41
L27:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L27
L41:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L36
L16:
  xor    %r13d,%r13d
  mov    %rsp,%r14
  xor    %esi,%esi
  mov    $0x2a0,%edi
  mov    %r13,%rcx
  mov    %r14,%rdx
  call   <_ada_use_c2phc@plt>
  lea    -0x30(%rbp),%rsi
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    $0x2a1,%edi
  mov    %ebx,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  call   <_ada_use_c2phc@plt>
  xor    %edx,%edx
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    $0x2a3,%edi
  call   <_ada_use_c2phc@plt>
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L34
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax)
L36:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L16
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int quaddobl_monodromy_permutation(int d, int *done) {
  int *a, *b, fail, i;
  int permutation[d];
  int nf[2];
  double *c;
  fail = _ada_use_c2phc(672, a, permutation, c);
  if (0 > 0) {
    printf("the permutation :");
    for (i = 0; i < d; i++)
      printf(" %d", permutation[i]);
  }
  nf[0] = d;
  nf[1] = 0;
  fail = _ada_use_c2phc(673, nf, permutation, c);
  if (0 > 0)
    printf(" : %d -> %d\n", nf[0], nf[1]);
  fail = _ada_use_c2phc(675, done, b, c);
  return fail;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<quaddobl_monodromy_permutation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L6
L1:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L1
L6:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L8
L18:
  xor    %r13d,%r13d
  mov    %rsp,%r14
  xor    %esi,%esi
  mov    $0x2a0,%edi
  mov    %r13,%rcx
  mov    %r14,%rdx
  call   <_ada_use_c2phc@plt>
  lea    -0x30(%rbp),%rsi
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    $0x2a1,%edi
  mov    %ebx,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  call   <_ada_use_c2phc@plt>
  xor    %edx,%edx
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    $0x2a3,%edi
  call   <_ada_use_c2phc@plt>
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L35
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax)
L8:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L18
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int quaddobl_monodromy_permutation(int d, int *done) {
  int *a, *b, fail, i;
  int permutation[d];
  int nf[2];
  double *c;
  fail = _ada_use_c2phc(672, a, permutation, c);
  if (0 > 0) {
    printf("the permutation :");
    for (i = 0; i < d; i++)
      printf(" %d", permutation[i]);
  }
  nf[0] = d;
  nf[1] = 0;
  fail = _ada_use_c2phc(673, nf, permutation, c);
  if (0 > 0)
    printf(" : %d -> %d\n", nf[0], nf[1]);
  fail = _ada_use_c2phc(675, done, b, c);
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<problem006>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L8
L2:
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L8:
  cmpl   $0x64,-0x4(%rbp)
  jle    L2
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type: string

tool: D3: %d


assistant: ```c
int problem006() {
  int sum = 0;
  int sum2 = 0;
  int i = 0;
  for (i = 1; i < 101; i++) {
    sum += i * i;
    sum2 += i;
  }
  printf("%d\n", (sum2 * sum2) - sum);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<problem006>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%eax
L3:
  sub    $0x1,%eax
  jne    L3
  mov    $0x17ff976,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D3
type: string

tool: D3: %d


assistant: ```c
int problem006() {
  int sum = 0;
  int sum2 = 0;
  int i = 0;
  for (i = 1; i < 101; i++) {
    sum += i * i;
    sum2 += i;
  }
  printf("%d\n", (sum2 * sum2) - sum);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<problem006>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x17ff976,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D7: %d


assistant: ```c
int problem006() {
  int sum = 0;
  int sum2 = 0;
  int i = 0;
  for (i = 1; i < 101; i++) {
    sum += i * i;
    sum2 += i;
  }
  printf("%d\n", (sum2 * sum2) - sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<problem006>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x17ff976,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string

tool: D1: %d


assistant: ```c
int problem006() {
  int sum = 0;
  int sum2 = 0;
  int i = 0;
  for (i = 1; i < 101; i++) {
    sum += i * i;
    sum2 += i;
  }
  printf("%d\n", (sum2 * sum2) - sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<send_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %esi,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x6c(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x30(%rbp)
  lea    -0x60(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  movl   $0x1,-0x50(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  movl   $0x18,-0x40(%rbp)
  lea    -0x6c(%rbp),%rax
  mov    %rax,-0x60(%rbp)
  movl   $0x4,-0x58(%rbp)
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,-0x68(%rbp)
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,0x8(%rax)
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,0x4(%rax)
  mov    -0x68(%rbp),%rax
  movl   $0x14,(%rax)
  mov    -0x68(%rbp),%rax
  mov    %rax,%rdi
  call   <CMSG_DATA@plt>
  mov    %eax,%ecx
  lea    -0x78(%rbp),%rax
  mov    $0x4,%edx
  mov    %rax,%rsi
  mov    %ecx,%edi
  call   <memcpy@plt>
  mov    -0x68(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x40(%rbp)
  lea    -0x50(%rbp),%rcx
  mov    -0x74(%rbp),%eax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sendmsg@plt>
  cmp    $0x4,%eax
  jne    L23
  mov    $0x0,%eax
  jmp    L14
L23:
  mov    $0xffffffff,%eax
L14:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int send_fd(int p, int fd) {
  struct msghdr hdr;
  struct iovec iov;
  struct cmsghdr *cmsg;
  char buf[((((sizeof(struct cmsghdr)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)) + (((sizeof(fd)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)))];
  int val = 0;
  hdr.msg_name = NULL;
  hdr.msg_namelen = 0;
  hdr.msg_iov = &iov;
  hdr.msg_iovlen = 1;
  hdr.msg_control = buf;
  hdr.msg_controllen = sizeof(buf);
  iov.iov_base = &val;
  iov.iov_len = sizeof(val);
  cmsg = CMSG_FIRSTHDR(&hdr);
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  cmsg->cmsg_len = ((((sizeof(struct cmsghdr)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)) + (sizeof(fd)));
  memcpy(CMSG_DATA(cmsg), &fd, sizeof(fd));
  hdr.msg_controllen = cmsg->cmsg_len;
  return sendmsg(p, &hdr, 0) == sizeof(val) ? 0 : -1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<send_fd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %edi,%ebp
  mov    %esi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x1c(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x0,0x50(%rsp)
  lea    0x20(%rsp),%rax
  mov    %rax,0x48(%rsp)
  movl   $0x1,0x30(%rsp)
  lea    0x60(%rsp),%rax
  mov    %rax,0x38(%rsp)
  movl   $0x18,0x40(%rsp)
  lea    0x1c(%rsp),%rax
  mov    %rax,0x20(%rsp)
  movl   $0x4,0x28(%rsp)
  lea    0x30(%rsp),%r12
  mov    %r12,%rdi
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%rbx
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%rbx)
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x4(%rbx)
  movl   $0x14,(%rbx)
  mov    %rbx,%rdi
  call   <CMSG_DATA@plt>
  mov    %eax,%edi
  lea    0xc(%rsp),%rsi
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    (%rbx),%eax
  mov    %eax,0x40(%rsp)
  mov    $0x0,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <sendmsg@plt>
  cmp    $0x4,%eax
  setne  %al
  movzbl %al,%eax
  neg    %eax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int send_fd(int p, int fd) {
  struct msghdr hdr;
  struct iovec iov;
  struct cmsghdr *cmsg;
  char buf[((((sizeof(struct cmsghdr)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)) + (((sizeof(fd)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)))];
  int val = 0;
  hdr.msg_name = NULL;
  hdr.msg_namelen = 0;
  hdr.msg_iov = &iov;
  hdr.msg_iovlen = 1;
  hdr.msg_control = buf;
  hdr.msg_controllen = sizeof(buf);
  iov.iov_base = &val;
  iov.iov_len = sizeof(val);
  cmsg = CMSG_FIRSTHDR(&hdr);
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  cmsg->cmsg_len = ((((sizeof(struct cmsghdr)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)) + (sizeof(fd)));
  memcpy(CMSG_DATA(cmsg), &fd, sizeof(fd));
  hdr.msg_controllen = cmsg->cmsg_len;
  return sendmsg(p, &hdr, 0) == sizeof(val) ? 0 : -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<send_fd>:
  endbr64
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %esi,0xc(%rsp)
  lea    0x30(%rsp),%r12
  mov    %r12,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rax
  movl   $0x0,0x1c(%rsp)
  mov    %rax,0x48(%rsp)
  lea    0x60(%rsp),%rax
  mov    %rax,0x38(%rsp)
  lea    0x1c(%rsp),%rax
  movq   $0x0,0x58(%rsp)
  movq   $0x0,0x50(%rsp)
  movl   $0x1,0x30(%rsp)
  movl   $0x18,0x40(%rsp)
  mov    %rax,0x20(%rsp)
  movl   $0x4,0x28(%rsp)
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%rbx
  mov    D12(%rip),%rax
  mov    %rbx,%rdi
  mov    (%rax),%eax
  mov    %eax,0x8(%rbx)
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  movl   $0x14,(%rbx)
  mov    %eax,0x4(%rbx)
  call   <CMSG_DATA@plt>
  lea    0xc(%rsp),%rsi
  mov    $0x4,%edx
  mov    %eax,%edi
  call   <memcpy@plt>
  mov    (%rbx),%eax
  xor    %edx,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %eax,0x40(%rsp)
  call   <sendmsg@plt>
  cmp    $0x4,%eax
  setne  %al
  movzbl %al,%eax
  neg    %eax
  mov    0x78(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int send_fd(int p, int fd) {
  struct msghdr hdr;
  struct iovec iov;
  struct cmsghdr *cmsg;
  char buf[((((sizeof(struct cmsghdr)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)) + (((sizeof(fd)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)))];
  int val = 0;
  hdr.msg_name = NULL;
  hdr.msg_namelen = 0;
  hdr.msg_iov = &iov;
  hdr.msg_iovlen = 1;
  hdr.msg_control = buf;
  hdr.msg_controllen = sizeof(buf);
  iov.iov_base = &val;
  iov.iov_len = sizeof(val);
  cmsg = CMSG_FIRSTHDR(&hdr);
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  cmsg->cmsg_len = ((((sizeof(struct cmsghdr)) + sizeof(intptr_t) - 1) & ~(sizeof(intptr_t) - 1)) + (sizeof(fd)));
  memcpy(CMSG_DATA(cmsg), &fd, sizeof(fd));
  hdr.msg_controllen = cmsg->cmsg_len;
  return sendmsg(p, &hdr, 0) == sizeof(val) ? 0 : -1;
}
```