ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<PrintCalendarMonth>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <MonthName@plt>
  mov    %r13d,%edx
  lea    D12(%rip),%rdi
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <MonthDays@plt>
  mov    %r12d,%edi
  mov    %r13d,%esi
  mov    %eax,%ebp
  call   <FirstDayOfMonth@plt>
  mov    %eax,%edi
  mov    %eax,%ebx
  call   <IndentFirstLine@plt>
  test   %ebp,%ebp
  jle    L44
  mov    $0x1,%r13d
  lea    D26(%rip),%r12
  jmp    L27
  nopl   (%rax)
L56:
  lea    0x1(%rbx),%eax
  add    $0x1,%r13d
  movslq %eax,%rbx
  cltd
  imul   $0xffffffff92492493,%rbx,%rbx
  shr    $0x20,%rbx
  add    %eax,%ebx
  sar    $0x2,%ebx
  sub    %edx,%ebx
  lea    0x0(,%rbx,8),%edx
  sub    %ebx,%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  cmp    %r13d,%ebp
  jl     L44
L27:
  xor    %eax,%eax
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0x6,%ebx
  jne    L56
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <putchar@plt>
  cmp    %r13d,%ebp
  jl     L69
  mov    %r13d,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  jmp    L56
  nopl   (%rax)
L44:
  test   %ebx,%ebx
  jne    L65
L69:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L65:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D12
type_guessed: string
name: unpack_data
arguments: data_tag: D3
type_guessed: string
name: unpack_data
arguments: data_tag: D26
type_guessed: string

tool: D12:      %s %d


tool: D3:  Su Mo Tu We Th Fr Sa

tool: D26:  %2d

assistant: ```c
void PrintCalendarMonth(int month, int year) {
  int weekday, nDays, day;
  printf("     %s %d\n", MonthName(month), year);
  printf(" Su Mo Tu We Th Fr Sa\n");
  nDays = MonthDays(month, year);
  weekday = FirstDayOfMonth(month, year);
  IndentFirstLine(weekday);
  for (day = 1; day <= nDays; day++) {
    printf(" %2d", day);
    if (weekday == 6)
      printf("\n");
    weekday = (weekday + 1) % 7;
  }
  if (weekday != 0)
    printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<permuteUnique>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  mov    %r8d,-0x3c(%rbp)
  mov    -0x34(%rbp),%eax
  sub    -0x38(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x3c(%rbp)
  jne    L32
  movl   $0x0,-0x14(%rbp)
  jmp    L69
L41:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  je     L108
  mov    -0x14(%rbp),%eax
  lea    0x31(%rax),%ecx
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L108:
  addl   $0x1,-0x14(%rbp)
L69:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L41
  jmp    L43
L32:
  movl   $0x1,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L55
L48:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x10(%rbp)
L55:
  mov    -0xc(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L48
  mov    -0x3c(%rbp),%eax
  cltd
  idivl  -0x10(%rbp)
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L57
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  je     L45
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  test   %eax,%eax
  jne    L45
  mov    -0x4(%rbp),%eax
  lea    0x31(%rax),%ecx
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
L45:
  addl   $0x1,-0x4(%rbp)
L57:
  mov    -0x4(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L18
  mov    -0x3c(%rbp),%eax
  cltd
  idivl  -0x10(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    -0x38(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x3c(%rbp),%edi
  mov    -0x34(%rbp),%edx
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <permuteUnique>
  nop
L43:
  leave
  ret
```

assistant: ```c
void permuteUnique(char *res, int *used, int size, int n, int k) {
  int resIdx = size - n;
  if (k == 0) {
    for (int i = 0; i < size; i++) {
      if (used[i] != 1) {
        res[resIdx++] = '0' + i + 1;
      }
    }
    return;
  }
  int temp = 1;
  int i = 1;
  while (i < n) {
    temp *= i++;
  }
  int idx = k / temp;
  for (int i = 0; i < size; i++) {
    if (used[i] != 1 && idx-- == 0) {
      res[resIdx] = '0' + i + 1;
      used[i] = 1;
    }
  }
  k %= temp;
  permuteUnique(res, used, size, n - 1, k);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<permuteUnique>:
  endbr64
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rdi
  mov    %edx,%r10d
  mov    %edx,%ebx
  sub    %ecx,%ebx
  test   %r8d,%r8d
  je     L37
  mov    $0x1,%edx
  mov    $0x1,%esi
  cmp    $0x1,%ecx
  jle    L36
L131:
  mov    %edx,%r9d
  add    $0x1,%edx
  imul   %r9d,%esi
  cmp    %edx,%ecx
  jne    L131
L36:
  mov    %r8d,%eax
  cltd
  idiv   %esi
  test   %r10d,%r10d
  jle    L77
  mov    %r10d,%r9d
  mov    $0x0,%edx
  movslq %ebx,%rbx
  jmp    L40
L37:
  test   %edx,%edx
  jle    L68
  mov    %edx,%r10d
  mov    $0x0,%eax
  jmp    L124
L111:
  add    $0x1,%rax
  cmp    %r10,%rax
  je     L68
L124:
  cmpl   $0x1,(%rdi,%rax,4)
  je     L111
  movslq %ebx,%rdx
  lea    0x31(%rax),%ecx
  mov    %cl,(%r11,%rdx,1)
  lea    0x1(%rbx),%ebx
  jmp    L111
L6:
  lea    0x31(%rdx),%eax
  mov    %al,(%r11,%rbx,1)
  movl   $0x1,(%rdi,%rdx,4)
  mov    $0xffffffff,%eax
L80:
  add    $0x1,%rdx
  cmp    %r9,%rdx
  je     L77
L40:
  cmpl   $0x1,(%rdi,%rdx,4)
  je     L80
  test   %eax,%eax
  je     L6
  sub    $0x1,%eax
  jmp    L80
L77:
  mov    %r8d,%eax
  cltd
  idiv   %esi
  mov    %edx,%r8d
  sub    $0x1,%ecx
  mov    %r10d,%edx
  mov    %rdi,%rsi
  mov    %r11,%rdi
  call   <permuteUnique>
L68:
  pop    %rbx
  ret
```

assistant: ```c
void permuteUnique(char *res, int *used, int size, int n, int k) {
  int resIdx = size - n;
  if (k == 0) {
    for (int i = 0; i < size; i++) {
      if (used[i] != 1) {
        res[resIdx++] = '0' + i + 1;
      }
    }
    return;
  }
  int temp = 1;
  int i = 1;
  while (i < n) {
    temp *= i++;
  }
  int idx = k / temp;
  for (int i = 0; i < size; i++) {
    if (used[i] != 1 && idx-- == 0) {
      res[resIdx] = '0' + i + 1;
      used[i] = 1;
    }
  }
  k %= temp;
  permuteUnique(res, used, size, n - 1, k);
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<permuteUnique>:
  endbr64
  push   %rbx
  movslq %edx,%rbx
  mov    %rdi,%r11
  mov    %ecx,%edi
  mov    %ebx,%ecx
  mov    %rsi,%r9
  sub    %edi,%ecx
  test   %r8d,%r8d
  je     L20
  movslq %ecx,%rcx
  mov    %r8d,%edx
  movslq %ebx,%r10
  lea    (%r11,%rcx,1),%r8
  nopw   0x0(%rax,%rax,1)
L37:
  mov    $0x1,%eax
  mov    $0x1,%ecx
  cmp    $0x1,%edi
  jle    L129
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %eax,%esi
  add    $0x1,%eax
  imul   %esi,%ecx
  cmp    %edi,%eax
  jne    L19
  mov    %edx,%eax
  cltd
  idiv   %ecx
  test   %ebx,%ebx
  jle    L38
L146:
  xor    %ecx,%ecx
  nop
L8:
  cmpl   $0x1,(%r9,%rcx,4)
  je     L140
  test   %eax,%eax
  jne    L135
  lea    0x31(%rcx),%eax
  mov    %al,(%r8)
  movl   $0x1,(%r9,%rcx,4)
  add    $0x1,%rcx
  cmp    %r10,%rcx
  je     L38
  cmpl   $0x1,(%r9,%rcx,4)
  mov    $0xffffffff,%eax
  je     L140
L135:
  sub    $0x1,%eax
L140:
  add    $0x1,%rcx
  cmp    %r10,%rcx
  jne    L8
L38:
  sub    $0x1,%edi
  mov    %ebx,%ecx
  add    $0x1,%r8
  sub    %edi,%ecx
  test   %edx,%edx
  jne    L37
L20:
  test   %ebx,%ebx
  jle    L106
  xor    %eax,%eax
  nop
L122:
  cmpl   $0x1,(%r9,%rax,4)
  je     L1
  movslq %ecx,%rdx
  lea    0x31(%rax),%esi
  add    $0x1,%ecx
  mov    %sil,(%r11,%rdx,1)
L1:
  add    $0x1,%rax
  cmp    %rax,%rbx
  jne    L122
L106:
  pop    %rbx
  ret
  nop
L129:
  test   %ebx,%ebx
  jle    L106
  mov    %edx,%eax
  xor    %edx,%edx
  jmp    L146
```

assistant: ```c
void permuteUnique(char *res, int *used, int size, int n, int k) {
  int resIdx = size - n;
  if (k == 0) {
    for (int i = 0; i < size; i++) {
      if (used[i] != 1) {
        res[resIdx++] = '0' + i + 1;
      }
    }
    return;
  }
  int temp = 1;
  int i = 1;
  while (i < n) {
    temp *= i++;
  }
  int idx = k / temp;
  for (int i = 0; i < size; i++) {
    if (used[i] != 1 && idx-- == 0) {
      res[resIdx] = '0' + i + 1;
      used[i] = 1;
    }
  }
  k %= temp;
  permuteUnique(res, used, size, n - 1, k);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<permuteUnique>:
  endbr64
  push   %rbx
  movslq %edx,%rbx
  mov    %rdi,%r11
  mov    %ecx,%edi
  mov    %ebx,%ecx
  mov    %rsi,%r9
  sub    %edi,%ecx
  test   %r8d,%r8d
  je     L53
  movslq %ecx,%rcx
  mov    %r8d,%edx
  movslq %ebx,%r10
  lea    (%r11,%rcx,1),%r8
  nopw   0x0(%rax,%rax,1)
L123:
  mov    $0x1,%eax
  mov    $0x1,%ecx
  cmp    $0x1,%edi
  jle    L23
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %eax,%esi
  add    $0x1,%eax
  imul   %esi,%ecx
  cmp    %edi,%eax
  jne    L11
  mov    %edx,%eax
  cltd
  idiv   %ecx
  test   %ebx,%ebx
  jle    L118
L135:
  xor    %ecx,%ecx
  nop
L57:
  cmpl   $0x1,(%r9,%rcx,4)
  je     L58
  test   %eax,%eax
  jne    L32
  lea    0x31(%rcx),%eax
  mov    %al,(%r8)
  movl   $0x1,(%r9,%rcx,4)
  add    $0x1,%rcx
  cmp    %r10,%rcx
  je     L118
  cmpl   $0x1,(%r9,%rcx,4)
  mov    $0xffffffff,%eax
  je     L58
L32:
  sub    $0x1,%eax
L58:
  add    $0x1,%rcx
  cmp    %rcx,%r10
  jne    L57
L118:
  sub    $0x1,%edi
  mov    %ebx,%ecx
  add    $0x1,%r8
  sub    %edi,%ecx
  test   %edx,%edx
  jne    L123
L53:
  test   %ebx,%ebx
  jle    L19
  xor    %eax,%eax
  nop
L84:
  cmpl   $0x1,(%r9,%rax,4)
  je     L106
  movslq %ecx,%rdx
  lea    0x31(%rax),%esi
  add    $0x1,%ecx
  mov    %sil,(%r11,%rdx,1)
L106:
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L84
L19:
  pop    %rbx
  ret
  nop
L23:
  test   %ebx,%ebx
  jle    L19
  mov    %edx,%eax
  xor    %edx,%edx
  jmp    L135
```

assistant: ```c
void permuteUnique(char *res, int *used, int size, int n, int k) {
  int resIdx = size - n;
  if (k == 0) {
    for (int i = 0; i < size; i++) {
      if (used[i] != 1) {
        res[resIdx++] = '0' + i + 1;
      }
    }
    return;
  }
  int temp = 1;
  int i = 1;
  while (i < n) {
    temp *= i++;
  }
  int idx = k / temp;
  for (int i = 0; i < size; i++) {
    if (used[i] != 1 && idx-- == 0) {
      res[resIdx] = '0' + i + 1;
      used[i] = 1;
    }
  }
  k %= temp;
  permuteUnique(res, used, size, n - 1, k);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<EndSUMOProof>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D22(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  mov    -0x8(%rbp),%rax
  lea    D27(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  mov    -0x8(%rbp),%rax
  lea    D34(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  mov    -0x8(%rbp),%rax
  lea    D2(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D22
data_type: string
name: parse_data
arguments: tag: D27
data_type: string
name: parse_data
arguments: tag: D34
data_type: string
name: parse_data
arguments: tag: D2
data_type: string

tool: D22:     </proof>


tool: D27:   </answer>


tool: D34:   <summary proofs='1'/>


tool: D2: </queryResponse>


assistant: ```c
void EndSUMOProof(FILE *Stream) {
  fprintf(Stream, "    </proof>\n");
  fprintf(Stream, "  </answer>\n");
  fprintf(Stream, "  <summary proofs='1'/>\n");
  fprintf(Stream, "</queryResponse>\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<EndSUMOProof>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D26(%rip),%rsi
  call   <fprintf@plt>
  lea    D4(%rip),%rsi
  mov    %rbx,%rdi
  call   <fprintf@plt>
  lea    D29(%rip),%rsi
  mov    %rbx,%rdi
  call   <fprintf@plt>
  lea    D36(%rip),%rsi
  mov    %rbx,%rdi
  call   <fprintf@plt>
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D26
type_infered: string
name: extract_symbolic_data
arguments: tag: D4
type_infered: string
name: extract_symbolic_data
arguments: tag: D29
type_infered: string
name: extract_symbolic_data
arguments: tag: D36
type_infered: string

tool: D26:     </proof>


tool: D4:   </answer>


tool: D29:   <summary proofs='1'/>


tool: D36: </queryResponse>


assistant: ```c
void EndSUMOProof(FILE *Stream) {
  fprintf(Stream, "    </proof>\n");
  fprintf(Stream, "  </answer>\n");
  fprintf(Stream, "  <summary proofs='1'/>\n");
  fprintf(Stream, "</queryResponse>\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<EndSUMOProof>:
  endbr64
  push   %rbp
  lea    D22(%rip),%rsi
  mov    %rdi,%rbp
  call   <fprintf@plt>
  mov    %rbp,%rdi
  lea    D34(%rip),%rsi
  call   <fprintf@plt>
  mov    %rbp,%rdi
  lea    D2(%rip),%rsi
  call   <fprintf@plt>
  mov    %rbp,%rdi
  lea    D39(%rip),%rsi
  pop    %rbp
  jmp    1050 <fprintf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D22
type: string
name: unpack_data
arguments: tag: D34
type: string
name: unpack_data
arguments: tag: D2
type: string
name: unpack_data
arguments: tag: D39
type: string

tool: D22:     </proof>


tool: D34:   </answer>


tool: D2:   <summary proofs='1'/>


tool: D39: </queryResponse>


assistant: ```c
void EndSUMOProof(FILE *Stream) {
  fprintf(Stream, "    </proof>\n");
  fprintf(Stream, "  </answer>\n");
  fprintf(Stream, "  <summary proofs='1'/>\n");
  fprintf(Stream, "</queryResponse>\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<EndSUMOProof>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rsi
  mov    %rdi,%rbp
  call   <fprintf@plt>
  mov    %rbp,%rdi
  lea    D27(%rip),%rsi
  call   <fprintf@plt>
  mov    %rbp,%rdi
  lea    D33(%rip),%rsi
  call   <fprintf@plt>
  mov    %rbp,%rdi
  lea    D18(%rip),%rsi
  pop    %rbp
  jmp    1050 <fprintf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D3
type_infered: string
name: unpack_data
arguments: tag: D27
type_infered: string
name: unpack_data
arguments: tag: D33
type_infered: string
name: unpack_data
arguments: tag: D18
type_infered: string

tool: D3:     </proof>


tool: D27:   </answer>


tool: D33:   <summary proofs='1'/>


tool: D18: </queryResponse>


assistant: ```c
void EndSUMOProof(FILE *Stream) {
  fprintf(Stream, "    </proof>\n");
  fprintf(Stream, "  </answer>\n");
  fprintf(Stream, "  <summary proofs='1'/>\n");
  fprintf(Stream, "</queryResponse>\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<calcDoubleDiv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L32
L2:
  movl   $0x1,-0x20(%rbp)
  jmp    L10
L34:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x20(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x20(%rbp)
L10:
  cmpl   $0x2710,-0x20(%rbp)
  jle    L34
  addl   $0x1,-0x24(%rbp)
L32:
  cmpl   $0x270f,-0x24(%rbp)
  jle    L2
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x18(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L6
  mov    D5(%rip),%rax
  mov    (%rax),%ecx
  mov    $0x3e8,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L6
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x18(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x18(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
long calcDoubleDiv(void) {
  long stTim, endTim;
  int i, j;
  double a;
  stTim = clock();
  a = 0.0;
  for (i = 0; i < 10000; i++) {
    for (j = 1; j < 10001; j++) {
      a /= (double)j;
    }
  }
  endTim = clock();
  endTim -= stTim;
  if (CLOCKS_PER_SEC > 0) {
    int time_dd = 1000 / CLOCKS_PER_SEC;
    if (time_dd > 0)
      endTim *= time_dd;
  }
  return endTim;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<calcDoubleDiv>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,%rbx
  mov    $0x2710,%edx
L0:
  mov    $0x2710,%eax
L14:
  sub    $0x1,%eax
  jne    L14
  sub    $0x1,%edx
  jne    L0
  mov    $0x0,%eax
  call   <clock@plt>
  sub    %rbx,%rax
  mov    %rax,%rcx
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  test   %esi,%esi
  jle    L33
  mov    $0x3e8,%eax
  cltd
  idiv   %esi
  test   %eax,%eax
  jle    L33
  cltq
  imul   %rax,%rcx
L33:
  mov    %rcx,%rax
  pop    %rbx
  ret
```

assistant: ```c
long calcDoubleDiv(void) {
  long stTim, endTim;
  int i, j;
  double a;
  stTim = clock();
  a = 0.0;
  for (i = 0; i < 10000; i++) {
    for (j = 1; j < 10001; j++) {
      a /= (double)j;
    }
  }
  endTim = clock();
  endTim -= stTim;
  if (CLOCKS_PER_SEC > 0) {
    int time_dd = 1000 / CLOCKS_PER_SEC;
    if (time_dd > 0)
      endTim *= time_dd;
  }
  return endTim;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<calcDoubleDiv>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <clock@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <clock@plt>
  sub    %rbx,%rax
  mov    %rax,%r8
  mov    D2(%rip),%rax
  mov    (%rax),%ecx
  test   %ecx,%ecx
  jle    L12
  mov    $0x3e8,%eax
  cltd
  idiv   %ecx
  test   %eax,%eax
  je     L12
  cltq
  imul   %rax,%r8
L12:
  mov    %r8,%rax
  pop    %rbx
  ret
```

assistant: ```c
long calcDoubleDiv(void) {
  long stTim, endTim;
  int i, j;
  double a;
  stTim = clock();
  a = 0.0;
  for (i = 0; i < 10000; i++) {
    for (j = 1; j < 10001; j++) {
      a /= (double)j;
    }
  }
  endTim = clock();
  endTim -= stTim;
  if (CLOCKS_PER_SEC > 0) {
    int time_dd = 1000 / CLOCKS_PER_SEC;
    if (time_dd > 0)
      endTim *= time_dd;
  }
  return endTim;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calcDoubleDiv>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <clock@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <clock@plt>
  sub    %rbx,%rax
  mov    %rax,%r8
  mov    D1(%rip),%rax
  mov    (%rax),%ecx
  test   %ecx,%ecx
  jle    L4
  mov    $0x3e8,%eax
  cltd
  idiv   %ecx
  test   %eax,%eax
  je     L4
  cltq
  imul   %rax,%r8
L4:
  mov    %r8,%rax
  pop    %rbx
  ret
```

assistant: ```c
long calcDoubleDiv(void) {
  long stTim, endTim;
  int i, j;
  double a;
  stTim = clock();
  a = 0.0;
  for (i = 0; i < 10000; i++) {
    for (j = 1; j < 10001; j++) {
      a /= (double)j;
    }
  }
  endTim = clock();
  endTim -= stTim;
  if (CLOCKS_PER_SEC > 0) {
    int time_dd = 1000 / CLOCKS_PER_SEC;
    if (time_dd > 0)
      endTim *= time_dd;
  }
  return endTim;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fill_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncat@plt>
  nop
  leave
  ret
```

assistant: ```c
void fill_str(char buff[], char *start, char *end) {
  buff[0] = '\0';
  strncat(buff, start, end - start);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fill_str>:
  endbr64
  sub    $0x8,%rsp
  movb   $0x0,(%rdi)
  sub    %esi,%edx
  call   <strncat@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void fill_str(char buff[], char *start, char *end) {
  buff[0] = '\0';
  strncat(buff, start, end - start);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fill_str>:
  endbr64
  movb   $0x0,(%rdi)
  sub    %esi,%edx
  jmp    1050 <strncat@plt>
```

assistant: ```c
void fill_str(char buff[], char *start, char *end) {
  buff[0] = '\0';
  strncat(buff, start, end - start);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fill_str>:
  endbr64
  movb   $0x0,(%rdi)
  sub    %esi,%edx
  jmp    1050 <strncat@plt>
```

assistant: ```c
void fill_str(char buff[], char *start, char *end) {
  buff[0] = '\0';
  strncat(buff, start, end - start);
}
```
user: What is the c source code of the assembly code below:

```asm
<part1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    $0x9,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L29
L12:
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <hash@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <interesting@plt>
  test   %rax,%rax
  je     L31
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  movzbl 0x5(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L45
L31:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L12
L45:
  addl   $0x1,-0x14(%rbp)
L29:
  cmpl   $0x7,-0x14(%rbp)
  jbe    L12
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *part1(char *doorid) {
  unsigned int n = 0;
  char *password = malloc(8 + 1);
  for (unsigned int i = 0; i < 8; i++) {
    for (;;) {
      char *s = hash(doorid, n++);
      if (interesting(s)) {
        password[i] = s[5];
        free(s);
        break;
      }
      free(s);
    }
  }
  password[8] = '\0';
  return password;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<part1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    $0x9,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  mov    %rax,%r13
  lea    0x8(%rax),%r14
  mov    $0x0,%ebp
  jmp    L28
L23:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %ebp,%esi
L7:
  lea    0x1(%rsi),%ebp
  mov    %r12,%rdi
  call   <hash@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <interesting@plt>
  test   %rax,%rax
  je     L23
  movzbl 0x5(%rbx),%eax
  mov    %al,0x0(%r13)
  mov    %rbx,%rdi
  call   <free@plt>
  add    $0x1,%r13
  cmp    %r14,%r13
  je     L20
L28:
  mov    %ebp,%esi
  jmp    L7
L20:
  movb   $0x0,0x8(%r15)
  mov    %r15,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *part1(char *doorid) {
  unsigned int n = 0;
  char *password = malloc(8 + 1);
  for (unsigned int i = 0; i < 8; i++) {
    for (;;) {
      char *s = hash(doorid, n++);
      if (interesting(s)) {
        password[i] = s[5];
        free(s);
        break;
      }
      free(s);
    }
  }
  password[8] = '\0';
  return password;
}
```