ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<buf2hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x30(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L3
L29:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  sar    $0x4,%al
  movsbl %al,%edx
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,-0x14(%rbp)
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <nibble_to_hex@plt>
  mov    %al,(%rbx)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,-0x14(%rbp)
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <nibble_to_hex@plt>
  mov    %al,(%rbx)
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  addl   $0x1,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L26
  mov    -0x30(%rbp),%eax
  sub    $0x4,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L29
L26:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void buf2hex(char *buf, int buf_len, char *hexbuf, int hexbuf_sz) {
  int i, j;
  for (i = 0, j = 0; i < buf_len && j < hexbuf_sz - 4; i++) {
    hexbuf[j++] = nibble_to_hex(buf[i] >> 4);
    hexbuf[j++] = nibble_to_hex(buf[i]);
    hexbuf[j++] = ' ';
  }
  hexbuf[j] = 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<buf2hex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%r15
  test   %esi,%esi
  jle    L15
  lea    -0x4(%rcx),%r13d
  test   %r13d,%r13d
  jle    L14
  mov    %rdi,%rbp
  mov    %rdx,%rbx
  lea    (%rsi,%rsi,2),%r14d
  mov    $0x0,%r12d
L41:
  movzbl 0x0(%rbp),%edi
  sar    $0x4,%dil
  movsbl %dil,%edi
  call   <nibble_to_hex@plt>
  mov    %al,(%rbx)
  movsbl 0x0(%rbp),%edi
  call   <nibble_to_hex@plt>
  mov    %al,0x1(%rbx)
  add    $0x3,%r12d
  movb   $0x20,0x2(%rbx)
  cmp    %r14d,%r12d
  je     L49
  add    $0x1,%rbp
  add    $0x3,%rbx
  cmp    %r13d,%r12d
  jl     L41
  mov    %r12d,%r14d
L49:
  movslq %r14d,%r14
  movb   $0x0,(%r15,%r14,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  mov    $0x0,%r14d
  jmp    L49
L14:
  mov    $0x0,%r14d
  jmp    L49
```

assistant: ```c
void buf2hex(char *buf, int buf_len, char *hexbuf, int hexbuf_sz) {
  int i, j;
  for (i = 0, j = 0; i < buf_len && j < hexbuf_sz - 4; i++) {
    hexbuf[j++] = nibble_to_hex(buf[i] >> 4);
    hexbuf[j++] = nibble_to_hex(buf[i]);
    hexbuf[j++] = ' ';
  }
  hexbuf[j] = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<buf2hex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L0
  lea    -0x4(%rcx),%ebp
  test   %ebp,%ebp
  jle    L0
  mov    %rdi,%r14
  mov    %rdx,%rbx
  lea    (%rsi,%rsi,2),%r12d
  xor    %r15d,%r15d
  jmp    L24
  nop
L33:
  add    $0x1,%r14
  add    $0x3,%rbx
  cmp    %ebp,%r15d
  jge    L36
L24:
  movzbl (%r14),%edi
  add    $0x3,%r15d
  sar    $0x4,%dil
  movsbl %dil,%edi
  call   <nibble_to_hex@plt>
  mov    %al,(%rbx)
  movsbl (%r14),%edi
  call   <nibble_to_hex@plt>
  movb   $0x20,0x2(%rbx)
  mov    %al,0x1(%rbx)
  cmp    %r15d,%r12d
  jne    L33
  movslq %r12d,%r12
  add    %r12,%r13
L0:
  movb   $0x0,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L36:
  movslq %r15d,%r15
  add    %r15,%r13
  jmp    L0
```

assistant: ```c
void buf2hex(char *buf, int buf_len, char *hexbuf, int hexbuf_sz) {
  int i, j;
  for (i = 0, j = 0; i < buf_len && j < hexbuf_sz - 4; i++) {
    hexbuf[j++] = nibble_to_hex(buf[i] >> 4);
    hexbuf[j++] = nibble_to_hex(buf[i]);
    hexbuf[j++] = ' ';
  }
  hexbuf[j] = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<buf2hex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L35
  lea    -0x4(%rcx),%ebp
  test   %ebp,%ebp
  jle    L35
  mov    %rdi,%r14
  mov    %rdx,%rbx
  lea    (%rsi,%rsi,2),%r12d
  xor    %r15d,%r15d
  jmp    L38
  nop
L46:
  add    $0x1,%r14
  add    $0x3,%rbx
  cmp    %ebp,%r15d
  jge    L25
L38:
  movzbl (%r14),%edi
  add    $0x3,%r15d
  sar    $0x4,%dil
  movsbl %dil,%edi
  call   <nibble_to_hex@plt>
  mov    %al,(%rbx)
  movsbl (%r14),%edi
  call   <nibble_to_hex@plt>
  movb   $0x20,0x2(%rbx)
  mov    %al,0x1(%rbx)
  cmp    %r15d,%r12d
  jne    L46
  movslq %r12d,%r12
  add    %r12,%r13
L35:
  movb   $0x0,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  movslq %r15d,%r15
  add    %r15,%r13
  jmp    L35
```

assistant: ```c
void buf2hex(char *buf, int buf_len, char *hexbuf, int hexbuf_sz) {
  int i, j;
  for (i = 0, j = 0; i < buf_len && j < hexbuf_sz - 4; i++) {
    hexbuf[j++] = nibble_to_hex(buf[i] >> 4);
    hexbuf[j++] = nibble_to_hex(buf[i]);
    hexbuf[j++] = ' ';
  }
  hexbuf[j] = 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
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L90
L121:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L57
  addl   $0x1,-0x8(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jg     L148
  addl   $0x1,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cmp    $0x2,%eax
  jle    L22
  addl   $0x1,-0x8(%rbp)
  jmp    L90
L22:
  addl   $0x1,-0x4(%rbp)
  jmp    L90
L148:
  subl   $0x1,-0x4(%rbp)
  jmp    L90
L57:
  cmpl   $0x0,-0x8(%rbp)
  jle    L101
  cmpl   $0x0,-0x4(%rbp)
  jns    L101
  subl   $0x1,-0x8(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jge    L30
  addl   $0x1,-0x4(%rbp)
  jmp    L90
L30:
  addl   $0x1,-0x8(%rbp)
  jmp    L90
L101:
  addl   $0x1,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  cmp    %eax,-0x8(%rbp)
  jge    L68
  subl   $0x1,-0x4(%rbp)
  jmp    L90
L68:
  addl   $0x1,-0x4(%rbp)
L90:
  cmpl   $0x0,-0x8(%rbp)
  jg     L121
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int i;
  int j;
  i = __VERIFIER_nondet_int();
  j = i;
  while (i > 0) {
    if (i >= j) {
      i = i + 1;
      if (j < 5) {
        j = j + 1;
        if (i - j > 2) {
          i = i + 1;
        } else {
          j = j + 1;
        }
      } else {
        j = j - 1;
      }
    } else {
      if (i > 0 && j < 0) {
        i = i - 1;
        if (j < -1) {
          j = j + 1;
        } else {
          i = i + 1;
        }
      } else {
        i = i + 1;
        if (j * 2 > i) {
          j = j - 1;
        } else {
          j = j + 1;
        }
      }
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L85
  mov    %eax,%edx
  jmp    L82
L117:
  add    $0x2,%edx
  mov    %ecx,%eax
  jmp    L5
L110:
  sub    $0x1,%edx
  mov    %ecx,%eax
L5:
  test   %eax,%eax
  jle    L85
L82:
  cmp    %edx,%eax
  jl     L29
  lea    0x1(%rax),%ecx
  cmp    $0x4,%edx
  jg     L110
  mov    %eax,%esi
  sub    %edx,%esi
  cmp    $0x2,%esi
  jle    L117
  add    $0x1,%edx
  add    $0x2,%eax
  jmp    L5
L29:
  test   %edx,%edx
  jns    L33
  cmp    $0xffffffff,%edx
  jge    L5
  sub    $0x1,%eax
  add    $0x1,%edx
  jmp    L5
L33:
  add    $0x1,%eax
  lea    (%rdx,%rdx,1),%esi
  lea    -0x1(%rdx),%ecx
  add    $0x1,%edx
  cmp    %eax,%esi
  cmovg  %ecx,%edx
  jmp    L5
L85:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int i;
  int j;
  i = __VERIFIER_nondet_int();
  j = i;
  while (i > 0) {
    if (i >= j) {
      i = i + 1;
      if (j < 5) {
        j = j + 1;
        if (i - j > 2) {
          i = i + 1;
        } else {
          j = j + 1;
        }
      } else {
        j = j - 1;
      }
    } else {
      if (i > 0 && j < 0) {
        i = i - 1;
        if (j < -1) {
          j = j + 1;
        } else {
          i = i + 1;
        }
      } else {
        i = i + 1;
        if (j * 2 > i) {
          j = j - 1;
        } else {
          j = j + 1;
        }
      }
    }
  }
  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
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L9
L11:
  jmp    L11
  nopl   (%rax)
L9:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int i;
  int j;
  i = __VERIFIER_nondet_int();
  j = i;
  while (i > 0) {
    if (i >= j) {
      i = i + 1;
      if (j < 5) {
        j = j + 1;
        if (i - j > 2) {
          i = i + 1;
        } else {
          j = j + 1;
        }
      } else {
        j = j - 1;
      }
    } else {
      if (i > 0 && j < 0) {
        i = i - 1;
        if (j < -1) {
          j = j + 1;
        } else {
          i = i + 1;
        }
      } else {
        i = i + 1;
        if (j * 2 > i) {
          j = j - 1;
        } else {
          j = j + 1;
        }
      }
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L8
L4:
  jmp    L4
  nopl   (%rax)
L8:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int i;
  int j;
  i = __VERIFIER_nondet_int();
  j = i;
  while (i > 0) {
    if (i >= j) {
      i = i + 1;
      if (j < 5) {
        j = j + 1;
        if (i - j > 2) {
          i = i + 1;
        } else {
          j = j + 1;
        }
      } else {
        j = j - 1;
      }
    } else {
      if (i > 0 && j < 0) {
        i = i - 1;
        if (j < -1) {
          j = j + 1;
        } else {
          i = i + 1;
        }
      } else {
        i = i + 1;
        if (j * 2 > i) {
          j = j - 1;
        } else {
          j = j + 1;
        }
      }
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<merge_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L9
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <merge_sort>
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <merge_sort>
  mov    -0x8(%rbp),%rdi
  mov    -0x20(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <merge@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void merge_sort(int *arr, int start, int end) {
  int mid = (start + end) / 2;
  int *result;
  result = (int *)malloc(sizeof(int) * (end - start + 1));
  if (start < end) {
    merge_sort(arr, start, mid);
    merge_sort(arr, mid + 1, end);
    merge(arr, start, mid, end, result);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<merge_sort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %edx,%ebp
  mov    %edx,%eax
  sub    %esi,%eax
  lea    0x4(,%rax,4),%edi
  call   <malloc@plt>
  cmp    %ebp,%ebx
  jl     L10
L19:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  mov    %rax,%r13
  lea    (%rbx,%rbp,1),%eax
  mov    %eax,%r14d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  mov    %r14d,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <merge_sort>
  lea    0x1(%r14),%esi
  mov    %ebp,%edx
  mov    %r12,%rdi
  call   <merge_sort>
  mov    %r13,%r8
  mov    %ebp,%ecx
  mov    %r14d,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <merge@plt>
  jmp    L19
```

assistant: ```c
void merge_sort(int *arr, int start, int end) {
  int mid = (start + end) / 2;
  int *result;
  result = (int *)malloc(sizeof(int) * (end - start + 1));
  if (start < end) {
    merge_sort(arr, start, mid);
    merge_sort(arr, mid + 1, end);
    merge(arr, start, mid, end, result);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<merge_sort>:
  endbr64
  push   %r14
  mov    %edx,%eax
  push   %r13
  sub    %esi,%eax
  mov    %rdi,%r13
  push   %r12
  lea    0x4(,%rax,4),%edi
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <malloc@plt>
  cmp    %r12d,%ebp
  jl     L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %rax,%rbx
  lea    0x0(%rbp,%r12,1),%eax
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %eax,%r14d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  mov    %r14d,%edx
  call   <merge_sort>
  lea    0x1(%r14),%esi
  mov    %r12d,%edx
  mov    %r13,%rdi
  call   <merge_sort>
  mov    %rbx,%r8
  mov    %r12d,%ecx
  pop    %rbx
  mov    %r14d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <merge@plt>
```

assistant: ```c
void merge_sort(int *arr, int start, int end) {
  int mid = (start + end) / 2;
  int *result;
  result = (int *)malloc(sizeof(int) * (end - start + 1));
  if (start < end) {
    merge_sort(arr, start, mid);
    merge_sort(arr, mid + 1, end);
    merge(arr, start, mid, end, result);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<status_count_file_lines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L16
  mov    $0xffffffff,%eax
  jmp    L24
L38:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jne    L16
  addl   $0x1,-0x8(%rbp)
L16:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L38
  cmpl   $0x0,-0x8(%rbp)
  jne    L8
  mov    $0xfffffffe,%eax
  jmp    L24
L8:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
L24:
  leave
  ret
```

assistant: ```c
int status_count_file_lines(FILE *fp) {
  int count = 0;
  int ch = 0;
  if (!fp)
    return -1;
  while (!feof(fp)) {
    ch = fgetc(fp);
    if (ch == '\n')
      count++;
  }
  if (count == 0)
    return -2;
  else
    return count - 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<status_count_file_lines>:
  endbr64
  test   %rdi,%rdi
  je     L44
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    $0x0,%ebp
  jmp    L9
L26:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ebp
L9:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L26
  test   %ebp,%ebp
  je     L16
  lea    -0x1(%rbp),%eax
L32:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L44:
  mov    $0xffffffff,%eax
  ret
L16:
  mov    $0xfffffffe,%eax
  jmp    L32
```

assistant: ```c
int status_count_file_lines(FILE *fp) {
  int count = 0;
  int ch = 0;
  if (!fp)
    return -1;
  while (!feof(fp)) {
    ch = fgetc(fp);
    if (ch == '\n')
      count++;
  }
  if (count == 0)
    return -2;
  else
    return count - 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<status_count_file_lines>:
  endbr64
  test   %rdi,%rdi
  je     L41
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ebp
L6:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L19
  test   %ebp,%ebp
  je     L18
  lea    -0x1(%rbp),%eax
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    $0xfffffffe,%eax
  jmp    L22
L41:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int status_count_file_lines(FILE *fp) {
  int count = 0;
  int ch = 0;
  if (!fp)
    return -1;
  while (!feof(fp)) {
    ch = fgetc(fp);
    if (ch == '\n')
      count++;
  }
  if (count == 0)
    return -2;
  else
    return count - 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<status_count_file_lines>:
  endbr64
  test   %rdi,%rdi
  je     L16
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L43
  cs nopw 0x0(%rax,%rax,1)
L28:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%ebp
L43:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L28
  test   %ebp,%ebp
  je     L10
  lea    -0x1(%rbp),%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    $0xfffffffe,%eax
  jmp    L3
L16:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int status_count_file_lines(FILE *fp) {
  int count = 0;
  int ch = 0;
  if (!fp)
    return -1;
  while (!feof(fp)) {
    ch = fgetc(fp);
    if (ch == '\n')
      count++;
  }
  if (count == 0)
    return -2;
  else
    return count - 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<mult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x2,%esi
  mov    $0xa,%edi
  call   <__CrestCall@plt>
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x9,%edi
  call   <__CrestStore@plt>
  mov    $0x6,%edx
  mov    $0x0,%esi
  mov    $0xd,%edi
  call   <__CrestLoad@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xc,%edi
  call   <__CrestLoad@plt>
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  cltq
  mov    %rax,%rdx
  mov    $0x2,%esi
  mov    $0xb,%edi
  call   <__CrestApply2@plt>
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xe,%edi
  call   <__CrestStore@plt>
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xf,%edi
  call   <__CrestLoad@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <inc@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  mov    %rax,%rsi
  mov    $0x11,%edi
  call   <__CrestHandleReturn@plt>
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x10,%edi
  call   <__CrestStore@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x4(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x12,%edi
  call   <__CrestLoad@plt>
  mov    $0x13,%edi
  call   <__CrestReturn@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int mult(int y) {
  {
    __CrestCall(10, 2);
    __CrestStore(9, (unsigned long)(&y));
    __CrestLoad(13, (unsigned long)0, (long long)6);
    __CrestLoad(12, (unsigned long)(&y), (long long)y);
    __CrestApply2(11, 2, (long long)(6 * y));
    __CrestStore(14, (unsigned long)(&y));
    y = 6 * y;
    __CrestLoad(15, (unsigned long)(&y), (long long)y);
    y = inc(y);
    __CrestHandleReturn(17, (long long)y);
    __CrestStore(16, (unsigned long)(&y));
    {
      __CrestLoad(18, (unsigned long)(&y), (long long)y);
      __CrestReturn(19);
      return (y);
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mult>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,0xc(%rsp)
  mov    $0x2,%esi
  mov    $0xa,%edi
  call   <__CrestCall@plt>
  lea    0xc(%rsp),%rbx
  mov    %rbx,%rsi
  mov    $0x9,%edi
  call   <__CrestStore@plt>
  mov    $0x6,%edx
  mov    $0x0,%esi
  mov    $0xd,%edi
  call   <__CrestLoad@plt>
  movslq 0xc(%rsp),%rdx
  mov    %rbx,%rsi
  mov    $0xc,%edi
  call   <__CrestLoad@plt>
  mov    0xc(%rsp),%eax
  lea    (%rax,%rax,2),%edx
  add    %edx,%edx
  movslq %edx,%rdx
  mov    $0x2,%esi
  mov    $0xb,%edi
  call   <__CrestApply2@plt>
  mov    %rbx,%rsi
  mov    $0xe,%edi
  call   <__CrestStore@plt>
  mov    0xc(%rsp),%eax
  lea    (%rax,%rax,2),%edx
  add    %edx,%edx
  mov    %edx,0xc(%rsp)
  movslq %edx,%rdx
  mov    %rbx,%rsi
  mov    $0xf,%edi
  call   <__CrestLoad@plt>
  mov    0xc(%rsp),%edi
  call   <inc@plt>
  mov    %eax,0xc(%rsp)
  movslq %eax,%rsi
  mov    $0x11,%edi
  call   <__CrestHandleReturn@plt>
  mov    %rbx,%rsi
  mov    $0x10,%edi
  call   <__CrestStore@plt>
  movslq 0xc(%rsp),%rdx
  mov    %rbx,%rsi
  mov    $0x12,%edi
  call   <__CrestLoad@plt>
  mov    $0x13,%edi
  call   <__CrestReturn@plt>
  mov    0xc(%rsp),%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
```

assistant: ```c
int mult(int y) {
  {
    __CrestCall(10, 2);
    __CrestStore(9, (unsigned long)(&y));
    __CrestLoad(13, (unsigned long)0, (long long)6);
    __CrestLoad(12, (unsigned long)(&y), (long long)y);
    __CrestApply2(11, 2, (long long)(6 * y));
    __CrestStore(14, (unsigned long)(&y));
    y = 6 * y;
    __CrestLoad(15, (unsigned long)(&y), (long long)y);
    y = inc(y);
    __CrestHandleReturn(17, (long long)y);
    __CrestStore(16, (unsigned long)(&y));
    {
      __CrestLoad(18, (unsigned long)(&y), (long long)y);
      __CrestReturn(19);
      return (y);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mult>:
  endbr64
  push   %rbp
  mov    $0x2,%esi
  sub    $0x10,%rsp
  lea    0xc(%rsp),%rbp
  mov    %edi,0xc(%rsp)
  mov    $0xa,%edi
  call   <__CrestCall@plt>
  mov    %rbp,%rsi
  mov    $0x9,%edi
  call   <__CrestStore@plt>
  mov    $0x6,%edx
  xor    %esi,%esi
  mov    $0xd,%edi
  call   <__CrestLoad@plt>
  movslq 0xc(%rsp),%rdx
  mov    %rbp,%rsi
  mov    $0xc,%edi
  call   <__CrestLoad@plt>
  mov    0xc(%rsp),%eax
  mov    $0x2,%esi
  mov    $0xb,%edi
  lea    (%rax,%rax,2),%edx
  add    %edx,%edx
  movslq %edx,%rdx
  call   <__CrestApply2@plt>
  mov    %rbp,%rsi
  mov    $0xe,%edi
  call   <__CrestStore@plt>
  mov    0xc(%rsp),%eax
  mov    %rbp,%rsi
  mov    $0xf,%edi
  lea    (%rax,%rax,2),%edx
  add    %edx,%edx
  mov    %edx,0xc(%rsp)
  movslq %edx,%rdx
  call   <__CrestLoad@plt>
  mov    0xc(%rsp),%edi
  call   <inc@plt>
  mov    $0x11,%edi
  movslq %eax,%rsi
  mov    %eax,0xc(%rsp)
  call   <__CrestHandleReturn@plt>
  mov    %rbp,%rsi
  mov    $0x10,%edi
  call   <__CrestStore@plt>
  movslq 0xc(%rsp),%rdx
  mov    %rbp,%rsi
  mov    $0x12,%edi
  call   <__CrestLoad@plt>
  mov    $0x13,%edi
  call   <__CrestReturn@plt>
  mov    0xc(%rsp),%eax
  add    $0x10,%rsp
  pop    %rbp
  ret
```

assistant: ```c
int mult(int y) {
  {
    __CrestCall(10, 2);
    __CrestStore(9, (unsigned long)(&y));
    __CrestLoad(13, (unsigned long)0, (long long)6);
    __CrestLoad(12, (unsigned long)(&y), (long long)y);
    __CrestApply2(11, 2, (long long)(6 * y));
    __CrestStore(14, (unsigned long)(&y));
    y = 6 * y;
    __CrestLoad(15, (unsigned long)(&y), (long long)y);
    y = inc(y);
    __CrestHandleReturn(17, (long long)y);
    __CrestStore(16, (unsigned long)(&y));
    {
      __CrestLoad(18, (unsigned long)(&y), (long long)y);
      __CrestReturn(19);
      return (y);
    }
  }
}
```