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

```asm
<setvtxregister>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <writeregisteraddress@plt>
  mov    $0x1,%edi
  call   <writebit@plt>
  nopl   (%rax)
L3:
  mov    %ebx,%ecx
  mov    %ebp,%edi
  add    $0x1,%ebx
  sar    %cl,%edi
  and    $0x1,%edi
  call   <writebit@plt>
  cmp    $0x14,%ebx
  jne    L3
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void setvtxregister(uint8_t which, uint32_t value) {
  int i;
  writeregisteraddress(which);
  writebit(true);
  for (i = 0; i < 20; i++) {
    writebit((value >> i) & 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<thr_fn2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <sleep@plt>
  mov    $0x7d0,%edi
  call   <pthread_exit@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
data_type: string

tool: D6: thread 2 exiting


assistant: ```c
void *thr_fn2(void *arg) {
  printf("thread 2 exiting\n");
  sleep(3);
  pthread_exit((void *)2000);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<thr_fn2>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <sleep@plt>
  mov    $0x7d0,%edi
  call   <pthread_exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
datatype: string

tool: D5: thread 2 exiting


assistant: ```c
void *thr_fn2(void *arg) {
  printf("thread 2 exiting\n");
  sleep(3);
  pthread_exit((void *)2000);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<thr_fn2>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <sleep@plt>
  mov    $0x7d0,%edi
  call   <pthread_exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
datatype: string

tool: D8: thread 2 exiting


assistant: ```c
void *thr_fn2(void *arg) {
  printf("thread 2 exiting\n");
  sleep(3);
  pthread_exit((void *)2000);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<thr_fn2>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <sleep@plt>
  mov    $0x7d0,%edi
  call   <pthread_exit@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D3: thread 2 exiting


assistant: ```c
void *thr_fn2(void *arg) {
  printf("thread 2 exiting\n");
  sleep(3);
  pthread_exit((void *)2000);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L20
L5:
  movsd  -0x18(%rbp),%xmm1
  movsd  D31(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,-0x28(%rbp)
  movsd  -0x10(%rbp),%xmm1
  movsd  D9(%rip),%xmm0
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <sin@plt>
  movsd  -0x28(%rbp),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L20:
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  je     L5
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D31
data_type: f64
name: retrieve_data
arguments: data_label: D9
data_type: byte[8]
name: retrieve_data
arguments: data_label: D39
data_type: string
name: retrieve_data
arguments: data_label: D15
data_type: string

tool: D31: 0.5

tool: {"D9": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

tool: D39: %.3lf


tool: D15: %lf %lf

assistant: ```c
int main() {
  double r, n;
  while (scanf("%lf %lf", &r, &n) == 2) {
    printf("%.3lf\n", 0.5 * r * sin(2 * 3.1415926535897932384626433832795 / n) * r * n);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D9(%rip),%rbx
  lea    D38(%rip),%rbp
  jmp    L11
L3:
  movsd  0x18(%rsp),%xmm1
  mulsd  D24(%rip),%xmm1
  movsd  %xmm1,0x8(%rsp)
  movsd  D3(%rip),%xmm0
  divsd  0x20(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <sin@plt>
  mulsd  0x8(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  mulsd  0x20(%rsp),%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
L11:
  lea    0x20(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  je     L3
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
data_type: string
name: retrieve_data
arguments: data_tag: D38
data_type: string
name: retrieve_data
arguments: data_tag: D24
data_type: f64
name: retrieve_data
arguments: data_tag: D3
data_type: byte[8]

tool: D9: %lf %lf

tool: D38: %.3lf


tool: D24: 0.5

tool: {"D3": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

assistant: ```c
int main() {
  double r, n;
  while (scanf("%lf %lf", &r, &n) == 2) {
    printf("%.3lf\n", 0.5 * r * sin(2 * 3.1415926535897932384626433832795 / n) * r * n);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D11(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D22(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r12
  lea    0x18(%rsp),%rbp
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L22:
  movsd  D1(%rip),%xmm1
  mulsd  0x18(%rsp),%xmm1
  movsd  D35(%rip),%xmm0
  divsd  0x20(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  movsd  %xmm1,0x8(%rsp)
  call   <sin@plt>
  mulsd  0x8(%rsp),%xmm0
  mov    %r13,%rdi
  mulsd  0x18(%rsp),%xmm0
  mulsd  0x20(%rsp),%xmm0
  call   <printf@plt>
L15:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  je     L22
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D11
datatype: string
name: unpack_data
arguments: data_tag: D22
datatype: string
name: unpack_data
arguments: data_tag: D1
datatype: f64
name: unpack_data
arguments: data_tag: D35
datatype: byte[8]

tool: D11: %.3lf


tool: D22: %lf %lf

tool: D1: 0.5

tool: {"D35": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

assistant: ```c
int main() {
  double r, n;
  while (scanf("%lf %lf", &r, &n) == 2) {
    printf("%.3lf\n", 0.5 * r * sin(2 * 3.1415926535897932384626433832795 / n) * r * n);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D17(%rip),%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r12
  lea    0x18(%rsp),%rbp
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L14:
  movsd  D22(%rip),%xmm1
  mulsd  0x18(%rsp),%xmm1
  movsd  D1(%rip),%xmm0
  divsd  0x20(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  movsd  %xmm1,0x8(%rsp)
  call   <sin@plt>
  mulsd  0x8(%rsp),%xmm0
  mov    %r13,%rdi
  mulsd  0x18(%rsp),%xmm0
  mulsd  0x20(%rsp),%xmm0
  call   <printf@plt>
L28:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  je     L14
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D29
type: string
name: retrieve_data
arguments: data_tag: D17
type: string
name: retrieve_data
arguments: data_tag: D22
type: f64
name: retrieve_data
arguments: data_tag: D1
type: byte[8]

tool: D29: %.3lf


tool: D17: %lf %lf

tool: D22: 0.5

tool: {"D1": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

assistant: ```c
int main() {
  double r, n;
  while (scanf("%lf %lf", &r, &n) == 2) {
    printf("%.3lf\n", 0.5 * r * sin(2 * 3.1415926535897932384626433832795 / n) * r * n);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xxx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void xxx(int argc) {
  int x;
#pragma omp atomic read
  argc = x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<xxx>:
  endbr64
  ret
```

assistant: ```c
void xxx(int argc) {
  int x;
#pragma omp atomic read
  argc = x;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<xxx>:
  endbr64
  ret
```

assistant: ```c
void xxx(int argc) {
  int x;
#pragma omp atomic read
  argc = x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L18
L3:
  mov    -0x18(%rbp),%rax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  sar    $0x14,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  sar    $0xc,%eax
  xor    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x7,%eax
  xor    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  sar    $0x4,%eax
  xor    %edx,%eax
L18:
  pop    %rbp
  ret
```

assistant: ```c
uint64_t hash(void *data) {
  if (data == NULL) {
    return 0;
  }
  uint64_t h = (uintptr_t)data;
  uint64_t res = (h >> 20) ^ (h >> 12);
  return res ^ (res >> 7) ^ (res >> 4);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hash>:
  endbr64
  test   %rdi,%rdi
  je     L8
  mov    %edi,%edx
  sar    $0x14,%edx
  sar    $0xc,%edi
  xor    %edi,%edx
  mov    %edx,%eax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x4,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  ret
L8:
  mov    $0x0,%eax
  ret
```

assistant: ```c
uint64_t hash(void *data) {
  if (data == NULL) {
    return 0;
  }
  uint64_t h = (uintptr_t)data;
  uint64_t res = (h >> 20) ^ (h >> 12);
  return res ^ (res >> 7) ^ (res >> 4);
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash>:
  endbr64
  test   %rdi,%rdi
  je     L2
  mov    %edi,%edx
  sar    $0xc,%edi
  sar    $0x14,%edx
  xor    %edi,%edx
  mov    %edx,%eax
  mov    %edx,%ecx
  sar    $0x7,%eax
  sar    $0x4,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
uint64_t hash(void *data) {
  if (data == NULL) {
    return 0;
  }
  uint64_t h = (uintptr_t)data;
  uint64_t res = (h >> 20) ^ (h >> 12);
  return res ^ (res >> 7) ^ (res >> 4);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hash>:
  endbr64
  test   %rdi,%rdi
  je     L2
  mov    %edi,%edx
  sar    $0xc,%edi
  sar    $0x14,%edx
  xor    %edi,%edx
  mov    %edx,%eax
  mov    %edx,%ecx
  sar    $0x7,%eax
  sar    $0x4,%ecx
  xor    %ecx,%eax
  xor    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
uint64_t hash(void *data) {
  if (data == NULL) {
    return 0;
  }
  uint64_t h = (uintptr_t)data;
  uint64_t res = (h >> 20) ^ (h >> 12);
  return res ^ (res >> 7) ^ (res >> 4);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x8,-0x14(%rbp)
  movl   $0x0,-0x24(%rbp)
  lea    -0x24(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <solveNQueens@plt>
  cltq
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L36
L26:
  movl   $0x0,-0x1c(%rbp)
  jmp    L59
L34:
  movl   $0x0,-0x18(%rbp)
  jmp    L38
L41:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L38:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L41
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x1c(%rbp)
L59:
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L34
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x20(%rbp)
L36:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jl     L26
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L68
  call   <__stack_chk_fail@plt>
L68:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D27
type_guessed: string
name: unpack_data
arguments: label: D4
type_guessed: string
name: unpack_data
arguments: label: D29
type_guessed: string

tool: D27: size = %d


tool: D4: %c 

tool: D29: pause

assistant: ```c
int main() {
  int n = 8;
  int returnSize = 0;
  char ***res = solveNQueens(n, &returnSize);
  printf("size = %d\n", returnSize);
  for (int i = 0; i < returnSize; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        printf("%c ", res[i][j][k]);
      }
      printf("\n");
      free(res[i][j]);
    }
    printf("\n");
    free(res[i]);
  }
  free(res);
  system("pause");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  mov    $0x8,%edi
  call   <solveNQueens@plt>
  mov    %eax,%r14d
  cltq
  mov    %rax,0x8(%rsp)
  mov    0x14(%rsp),%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L32
  movslq %r14d,%r14
  mov    $0x0,%r15d
  lea    D25(%rip),%r13
  jmp    L7
L25:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    (%r12),%rdi
  call   <free@plt>
  add    $0x1,%r15d
  add    $0x8,%r14
  cmp    %r15d,0x14(%rsp)
  jle    L32
L7:
  mov    %r14,%r12
  mov    $0x0,%ebp
L41:
  mov    $0x0,%ebx
L48:
  mov    (%r12),%rax
  mov    (%rax,%rbp,1),%rax
  movsbl (%rax,%rbx,1),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x8,%rbx
  jne    L48
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    (%r12),%rax
  mov    (%rax,%rbp,1),%rdi
  call   <free@plt>
  add    $0x8,%rbp
  cmp    $0x40,%rbp
  jne    L41
  jmp    L25
L32:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <system@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type_guessed: string
name: retrieve_data
arguments: data_label: D25
type_guessed: string
name: retrieve_data
arguments: data_label: D20
type_guessed: string

tool: D9: size = %d


tool: D25: %c 

tool: D20: pause

assistant: ```c
int main() {
  int n = 8;
  int returnSize = 0;
  char ***res = solveNQueens(n, &returnSize);
  printf("size = %d\n", returnSize);
  for (int i = 0; i < returnSize; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        printf("%c ", res[i][j][k]);
      }
      printf("\n");
      free(res[i][j]);
    }
    printf("\n");
    free(res[i]);
  }
  free(res);
  system("pause");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x8,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  movl   $0x0,0x4(%rsp)
  call   <solveNQueens@plt>
  mov    0x4(%rsp),%esi
  lea    D18(%rip),%rdi
  movslq %eax,%r15
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L18
  mov    %r15,%r12
  xor    %r14d,%r14d
  lea    D3(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L34:
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L29:
  mov    (%r12),%rax
  mov    %r13,%rdi
  mov    (%rax,%rbp,1),%rax
  movsbl (%rax,%rbx,1),%esi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x8,%rbx
  jne    L29
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    (%r12),%rax
  mov    (%rax,%rbp,1),%rdi
  add    $0x8,%rbp
  call   <free@plt>
  cmp    $0x40,%rbp
  jne    L4
  mov    $0xa,%edi
  add    $0x1,%r14d
  add    $0x8,%r12
  call   <putchar@plt>
  mov    -0x8(%r12),%rdi
  call   <free@plt>
  cmp    %r14d,0x4(%rsp)
  jg     L34
L18:
  mov    %r15,%rdi
  call   <free@plt>
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D18
datatype: string
name: read_data
arguments: label: D3
datatype: string
name: read_data
arguments: label: D15
datatype: string

tool: D18: size = %d


tool: D3: %c 

tool: D15: pause

assistant: ```c
int main() {
  int n = 8;
  int returnSize = 0;
  char ***res = solveNQueens(n, &returnSize);
  printf("size = %d\n", returnSize);
  for (int i = 0; i < returnSize; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        printf("%c ", res[i][j][k]);
      }
      printf("\n");
      free(res[i][j]);
    }
    printf("\n");
    free(res[i]);
  }
  free(res);
  system("pause");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x8,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  movl   $0x0,0x4(%rsp)
  call   <solveNQueens@plt>
  mov    0x4(%rsp),%esi
  lea    D27(%rip),%rdi
  movslq %eax,%r15
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L28
  mov    %r15,%r12
  xor    %r14d,%r14d
  lea    D26(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L34:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L3:
  mov    (%r12),%rax
  mov    %r13,%rdi
  mov    (%rax,%rbp,1),%rax
  movsbl (%rax,%rbx,1),%esi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x8,%rbx
  jne    L3
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    (%r12),%rax
  mov    (%rax,%rbp,1),%rdi
  add    $0x8,%rbp
  call   <free@plt>
  cmp    $0x40,%rbp
  jne    L34
  mov    $0xa,%edi
  add    $0x1,%r14d
  add    $0x8,%r12
  call   <putchar@plt>
  mov    -0x8(%r12),%rdi
  call   <free@plt>
  cmp    %r14d,0x4(%rsp)
  jg     L8
L28:
  mov    %r15,%rdi
  call   <free@plt>
  xor    %eax,%eax
  lea    D25(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D27
type: string
name: unpack_data
arguments: data_tag: D26
type: string
name: unpack_data
arguments: data_tag: D25
type: string

tool: D27: size = %d


tool: D26: %c 

tool: D25: pause

assistant: ```c
int main() {
  int n = 8;
  int returnSize = 0;
  char ***res = solveNQueens(n, &returnSize);
  printf("size = %d\n", returnSize);
  for (int i = 0; i < returnSize; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        printf("%c ", res[i][j][k]);
      }
      printf("\n");
      free(res[i][j]);
    }
    printf("\n");
    free(res[i]);
  }
  free(res);
  system("pause");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gf8mul_01101100>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  pxor   %xmm7,%xmm6
  pxor   %xmm7,%xmm3
  pxor   %xmm7,%xmm2
  pxor   %xmm7,%xmm0
  pxor   %xmm6,%xmm7
  pxor   %xmm6,%xmm4
  pxor   %xmm6,%xmm0
  pxor   %xmm5,%xmm7
  pxor   %xmm5,%xmm6
  pxor   %xmm5,%xmm2
  pxor   %xmm5,%xmm0
  pxor   %xmm6,%xmm5
  pxor   %xmm4,%xmm7
  pxor   %xmm4,%xmm6
  pxor   %xmm4,%xmm3
  pxor   %xmm4,%xmm2
  pxor   %xmm4,%xmm1
  pxor   %xmm4,%xmm0
  pxor   %xmm3,%xmm7
  pxor   %xmm3,%xmm6
  pxor   %xmm3,%xmm5
  pxor   %xmm3,%xmm2
  pxor   %xmm2,%xmm6
  pxor   %xmm2,%xmm5
  pxor   %xmm2,%xmm4
  pxor   %xmm2,%xmm1
  pxor   %xmm1,%xmm7
  pxor   %xmm1,%xmm6
  pxor   %xmm1,%xmm4
  pxor   %xmm1,%xmm3
  pxor   %xmm3,%xmm1
  pxor   %xmm0,%xmm6
  pxor   %xmm0,%xmm5
  pxor   %xmm0,%xmm3
  pxor   %xmm0,%xmm2
  pxor   %xmm2,%xmm0
  nop
  pop    %rbp
  ret
```

assistant: ```c
void gf8mul_01101100(void) {
  __asm__ __volatile__("\tpxor    %xmm7, %xmm6\n"
                       "\tpxor    %xmm7, %xmm3\n"
                       "\tpxor    %xmm7, %xmm2\n"
                       "\tpxor    %xmm7, %xmm0\n"
                       "\tpxor    %xmm6, %xmm7\n"
                       "\tpxor    %xmm6, %xmm4\n"
                       "\tpxor    %xmm6, %xmm0\n"
                       "\tpxor    %xmm5, %xmm7\n"
                       "\tpxor    %xmm5, %xmm6\n"
                       "\tpxor    %xmm5, %xmm2\n"
                       "\tpxor    %xmm5, %xmm0\n"
                       "\tpxor    %xmm6, %xmm5\n"
                       "\tpxor    %xmm4, %xmm7\n"
                       "\tpxor    %xmm4, %xmm6\n"
                       "\tpxor    %xmm4, %xmm3\n"
                       "\tpxor    %xmm4, %xmm2\n"
                       "\tpxor    %xmm4, %xmm1\n"
                       "\tpxor    %xmm4, %xmm0\n"
                       "\tpxor    %xmm3, %xmm7\n"
                       "\tpxor    %xmm3, %xmm6\n"
                       "\tpxor    %xmm3, %xmm5\n"
                       "\tpxor    %xmm3, %xmm2\n"
                       "\tpxor    %xmm2, %xmm6\n"
                       "\tpxor    %xmm2, %xmm5\n"
                       "\tpxor    %xmm2, %xmm4\n"
                       "\tpxor    %xmm2, %xmm1\n"
                       "\tpxor    %xmm1, %xmm7\n"
                       "\tpxor    %xmm1, %xmm6\n"
                       "\tpxor    %xmm1, %xmm4\n"
                       "\tpxor    %xmm1, %xmm3\n"
                       "\tpxor    %xmm3, %xmm1\n"
                       "\tpxor    %xmm0, %xmm6\n"
                       "\tpxor    %xmm0, %xmm5\n"
                       "\tpxor    %xmm0, %xmm3\n"
                       "\tpxor    %xmm0, %xmm2\n"
                       "\tpxor    %xmm2, %xmm0\n");
}
```