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

```asm
<merge>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  mov    %edx,%esi
  mov    %ecx,%r9d
  lea    0x1(%rdx),%eax
  mov    %ecx,%ebp
  sub    %ebx,%ebp
  cmp    %edx,%ebx
  jg     L29
  cmp    %ecx,%eax
  jg     L29
  mov    %ebx,%ecx
  mov    $0x1,%edx
  jmp    L0
L62:
  mov    %r10d,-0x4(%r8,%rdx,4)
  add    $0x1,%eax
L19:
  mov    %edx,%r10d
  add    $0x1,%rdx
  cmp    %esi,%ecx
  jg     L76
  cmp    %r9d,%eax
  jg     L76
L0:
  movslq %ecx,%r10
  mov    (%rdi,%r10,4),%r11d
  movslq %eax,%r10
  mov    (%rdi,%r10,4),%r10d
  cmp    %r10d,%r11d
  jg     L62
  mov    %r11d,-0x4(%r8,%rdx,4)
  add    $0x1,%ecx
  jmp    L19
L29:
  mov    %ebx,%ecx
  mov    $0x0,%r10d
L76:
  cmp    %ecx,%esi
  jl     L104
  movslq %ecx,%rdx
  movslq %r10d,%r11
  sub    %rdx,%r11
  lea    (%r8,%r11,4),%r12
L38:
  mov    (%rdi,%rdx,4),%r11d
  mov    %r11d,(%r12,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%esi
  jge    L38
  lea    0x1(%r10,%rsi,1),%r10d
  sub    %ecx,%r10d
L104:
  cmp    %eax,%r9d
  jl     L8
  cltq
  movslq %r10d,%r10
  sub    %rax,%r10
  lea    (%r8,%r10,4),%rcx
L79:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r9d
  jge    L79
L8:
  test   %ebp,%ebp
  js     L97
  lea    0x1(%rbp),%ecx
  shl    $0x2,%rcx
  movslq %ebx,%rbx
  lea    (%rdi,%rbx,4),%rsi
  mov    $0x0,%eax
L118:
  mov    (%r8,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L118
L97:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void merge(int *array, int begining, int half, int end, int *aux_array) {
  int idx_first_array = begining, idx_second_array = half + 1, idx_new_array = 0, length_new_array = end - begining + 1, i;
  while (idx_first_array <= half && idx_second_array <= end) {
    if (array[idx_first_array] <= array[idx_second_array]) {
      aux_array[idx_new_array] = array[idx_first_array];
      idx_first_array++;
    } else {
      aux_array[idx_new_array] = array[idx_second_array];
      idx_second_array++;
    }
    idx_new_array++;
  }
  while (idx_first_array <= half) {
    aux_array[idx_new_array] = array[idx_first_array];
    idx_first_array++;
    idx_new_array++;
  }
  while (idx_second_array <= end) {
    aux_array[idx_new_array] = array[idx_second_array];
    idx_second_array++;
    idx_new_array++;
  }
  for (i = 0; i < length_new_array; i++) {
    array[begining + i] = aux_array[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<merge>:
  endbr64
  push   %r13
  mov    %edx,%r9d
  mov    %ecx,%r10d
  push   %r12
  lea    0x1(%rdx),%r12d
  push   %rbp
  mov    %ecx,%ebp
  mov    %r12d,%eax
  push   %rbx
  movslq %esi,%rbx
  sub    %ebx,%ebp
  mov    %ebx,%esi
  cmp    %edx,%ebx
  jg     L103
  cmp    %ecx,%r12d
  jg     L103
  mov    $0x1,%edx
  jmp    L19
L93:
  add    $0x1,%esi
L88:
  mov    %ecx,-0x4(%r8,%rdx,4)
  mov    %edx,%r11d
  add    $0x1,%rdx
  cmp    %r9d,%esi
  jg     L100
  cmp    %r10d,%eax
  jg     L100
L19:
  movslq %esi,%rcx
  movslq %eax,%r11
  mov    (%rdi,%rcx,4),%ecx
  mov    (%rdi,%r11,4),%r11d
  cmp    %r11d,%ecx
  jle    L93
  add    $0x1,%eax
  mov    %r11d,%ecx
  jmp    L88
  nopl   0x0(%rax,%rax,1)
L103:
  xor    %r11d,%r11d
  nopl   0x0(%rax,%rax,1)
L100:
  cmp    %esi,%r9d
  jl     L122
  movslq %esi,%rdx
  movslq %r11d,%rcx
  sub    %rdx,%rcx
  lea    (%r8,%rcx,4),%r13
  nopw   0x0(%rax,%rax,1)
L81:
  mov    (%rdi,%rdx,4),%ecx
  mov    %ecx,0x0(%r13,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,%r9d
  jge    L81
  add    %r11d,%r12d
  mov    %r12d,%r11d
  sub    %esi,%r11d
L122:
  cmp    %eax,%r10d
  jl     L80
  cltq
  movslq %r11d,%r11
  sub    %rax,%r11
  lea    (%r8,%r11,4),%rcx
  nopl   0x0(%rax,%rax,1)
L20:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r10d
  jge    L20
L80:
  test   %ebp,%ebp
  js     L78
  movslq %ebp,%rbp
  lea    (%rdi,%rbx,4),%rcx
  xor    %eax,%eax
  lea    0x4(,%rbp,4),%rsi
  nopl   0x0(%rax)
L32:
  mov    (%r8,%rax,1),%edx
  mov    %edx,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L32
L78:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void merge(int *array, int begining, int half, int end, int *aux_array) {
  int idx_first_array = begining, idx_second_array = half + 1, idx_new_array = 0, length_new_array = end - begining + 1, i;
  while (idx_first_array <= half && idx_second_array <= end) {
    if (array[idx_first_array] <= array[idx_second_array]) {
      aux_array[idx_new_array] = array[idx_first_array];
      idx_first_array++;
    } else {
      aux_array[idx_new_array] = array[idx_second_array];
      idx_second_array++;
    }
    idx_new_array++;
  }
  while (idx_first_array <= half) {
    aux_array[idx_new_array] = array[idx_first_array];
    idx_first_array++;
    idx_new_array++;
  }
  while (idx_second_array <= end) {
    aux_array[idx_new_array] = array[idx_second_array];
    idx_second_array++;
    idx_new_array++;
  }
  for (i = 0; i < length_new_array; i++) {
    array[begining + i] = aux_array[i];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rsi
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  imul   %edx,%eax
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  imul   %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type_guessed: string
name: read_data
arguments: tag: D4
type_guessed: string

tool: D0: %d%d%d%d

tool: D4: DIFERENCA = %d


assistant: ```c
int main() {
  int A, B, C, D, diferenca;
  scanf("%d%d%d%d", &A, &B, &C, &D);
  diferenca = A * B - C * D;
  printf("DIFERENCA = %d\n", diferenca);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x14(%rsp),%r8
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  imul   0xc(%rsp),%esi
  mov    0x10(%rsp),%eax
  imul   0x14(%rsp),%eax
  sub    %eax,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D16
datatype: string
name: retrieve_data
arguments: label: D11
datatype: string

tool: D16: %d%d%d%d

tool: D11: DIFERENCA = %d


assistant: ```c
int main() {
  int A, B, C, D, diferenca;
  scanf("%d%d%d%d", &A, &B, &C, &D);
  diferenca = A * B - C * D;
  printf("DIFERENCA = %d\n", diferenca);
  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    $0x28,%rsp
  lea    D0(%rip),%rdi
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  mov    0x10(%rsp),%eax
  lea    D3(%rip),%rdi
  imul   0xc(%rsp),%esi
  imul   0x14(%rsp),%eax
  sub    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
datatype: string
name: extract_symbolic_data
arguments: label: D3
datatype: string

tool: D0: %d%d%d%d

tool: D3: DIFERENCA = %d


assistant: ```c
int main() {
  int A, B, C, D, diferenca;
  scanf("%d%d%d%d", &A, &B, &C, &D);
  diferenca = A * B - C * D;
  printf("DIFERENCA = %d\n", diferenca);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D11(%rip),%rdi
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  mov    0x10(%rsp),%eax
  lea    D5(%rip),%rdi
  imul   0xc(%rsp),%esi
  imul   0x14(%rsp),%eax
  sub    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D11
type_infered: string
name: parse_data
arguments: data_tag: D5
type_infered: string

tool: D11: %d%d%d%d

tool: D5: DIFERENCA = %d


assistant: ```c
int main() {
  int A, B, C, D, diferenca;
  scanf("%d%d%d%d", &A, &B, &C, &D);
  diferenca = A * B - C * D;
  printf("DIFERENCA = %d\n", diferenca);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<LOG_SYSTEM>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x50(%rbp),%rax
  lea    D21(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x50(%rbp),%rax
  lea    D15(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  lea    -0x50(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <Streams_pushBordered@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D21
type: string
name: read_data
arguments: tag: D15
type: string
name: read_data
arguments: tag: D8
type: string

tool: D21: Compiled as C

tool: D15:  for 

tool: D8: Linux.

assistant: ```c
int LOG_SYSTEM(void) {
  char szBuffered[66];
  strcpy(szBuffered, "Compiled as C");
  strcat(szBuffered, " for ");
  strcat(szBuffered, "Linux.");
  return Streams_pushBordered(szBuffered);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<LOG_SYSTEM>:
  endbr64
  push   %rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D29(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  lea    D6(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  lea    D2(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  mov    %rbx,%rdi
  call   <Streams_pushBordered@plt>
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x50,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D29
data_type: string
name: extract_symbolic_data
arguments: tag: D6
data_type: string
name: extract_symbolic_data
arguments: tag: D2
data_type: string

tool: D29: Compiled as C

tool: D6:  for 

tool: D2: Linux.

assistant: ```c
int LOG_SYSTEM(void) {
  char szBuffered[66];
  strcpy(szBuffered, "Compiled as C");
  strcat(szBuffered, " for ");
  strcat(szBuffered, "Linux.");
  return Streams_pushBordered(szBuffered);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<LOG_SYSTEM>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rsi
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  lea    D23(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %rbp,%rdi
  lea    D24(%rip),%rsi
  call   <strcat@plt>
  mov    %rbp,%rdi
  call   <Streams_pushBordered@plt>
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x50,%rsp
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D9
type: string
name: read_data
arguments: tag: D23
type: string
name: read_data
arguments: tag: D24
type: string

tool: D9: Compiled as C

tool: D23:  for 

tool: D24: Linux.

assistant: ```c
int LOG_SYSTEM(void) {
  char szBuffered[66];
  strcpy(szBuffered, "Compiled as C");
  strcat(szBuffered, " for ");
  strcat(szBuffered, "Linux.");
  return Streams_pushBordered(szBuffered);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<LOG_SYSTEM>:
  endbr64
  push   %rbp
  lea    D29(%rip),%rsi
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  lea    D10(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %rbp,%rdi
  lea    D28(%rip),%rsi
  call   <strcat@plt>
  mov    %rbp,%rdi
  call   <Streams_pushBordered@plt>
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x50,%rsp
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D29
data_type: string
name: disassemble_data
arguments: tag: D10
data_type: string
name: disassemble_data
arguments: tag: D28
data_type: string

tool: D29: Compiled as C

tool: D10:  for 

tool: D28: Linux.

assistant: ```c
int LOG_SYSTEM(void) {
  char szBuffered[66];
  strcpy(szBuffered, "Compiled as C");
  strcat(szBuffered, " for ");
  strcat(szBuffered, "Linux.");
  return Streams_pushBordered(szBuffered);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dSign>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jb     L11
  movsd  D12(%rip),%xmm0
  jmp    L2
L11:
  movsd  D4(%rip),%xmm0
L2:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
datatype: f32[2]
name: extract_symbolic_data
arguments: data_label: D4
datatype: f32[2]

tool: {"D12": [0.0,1.875]}

tool: {"D4": [0.0,-1.875]}

assistant: ```c
double dSign(const double number) {
  if (number >= 0.0) {
    return 1;
  } else {
    return -1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dSign>:
  endbr64
  comisd D12(%rip),%xmm0
  jb     L6
  movsd  D7(%rip),%xmm0
  ret
L6:
  movsd  D11(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D12
type_infered: f64
name: disassemble_data
arguments: data_label: D7
type_infered: byte[8]
name: disassemble_data
arguments: data_label: D11
type_infered: byte[8]

tool: D12: 0

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

assistant: ```c
double dSign(const double number) {
  if (number >= 0.0) {
    return 1;
  } else {
    return -1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dSign>:
  endbr64
  comisd D21(%rip),%xmm0
  jb     L6
  movsd  D5(%rip),%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  movsd  D14(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D21
data_type: f64
name: extract_symbolic_data
arguments: data_label: D5
data_type: byte[8]
name: extract_symbolic_data
arguments: data_label: D14
data_type: byte[8]

tool: D21: 0

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

assistant: ```c
double dSign(const double number) {
  if (number >= 0.0) {
    return 1;
  } else {
    return -1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dSign>:
  endbr64
  comisd D0(%rip),%xmm0
  jb     L1
  movsd  D16(%rip),%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  movsd  D21(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_guessed: f64
name: retrieve_data
arguments: data_label: D16
type_guessed: byte[8]
name: retrieve_data
arguments: data_label: D21
type_guessed: byte[8]

tool: D0: 0

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D21": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

assistant: ```c
double dSign(const double number) {
  if (number >= 0.0) {
    return 1;
  } else {
    return -1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dfs_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  jmp    L131
L116:
  cmpl   $0x8,-0x20(%rbp)
  jne    L14
  movl   $0x0,-0x20(%rbp)
  addl   $0x1,-0x1c(%rbp)
  jmp    L131
L14:
  addl   $0x1,-0x20(%rbp)
L131:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L17
  cmpl   $0x7,-0x1c(%rbp)
  jle    L116
  cmpl   $0x7,-0x20(%rbp)
  jle    L116
L17:
  cmpl   $0x8,-0x1c(%rbp)
  jne    L83
  cmpl   $0x8,-0x20(%rbp)
  jne    L83
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L83
  mov    $0x1,%eax
  jmp    L117
L83:
  movl   $0x1,-0x4(%rbp)
  jmp    L53
L82:
  mov    -0x4(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <judge@plt>
  test   %rax,%rax
  je     L81
  mov    -0x4(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <dfs_1>
  test   %eax,%eax
  je     L10
  mov    $0x1,%eax
  jmp    L117
L10:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rdx,%rax
  movb   $0x2e,(%rax)
L81:
  addl   $0x1,-0x4(%rbp)
L53:
  cmpl   $0x9,-0x4(%rbp)
  jle    L82
  mov    $0x0,%eax
L117:
  leave
  ret
```

assistant: ```c
int dfs_1(char **board, int x, int y) {
  int i;
  while (board[x][y] != '.' && (x < 8 || y < 8)) {
    if (y == 8)
      y = 0, x++;
    else
      y++;
  }
  if (x == 8 && y == 8 && board[x][y] != '.')
    return 1;
  for (i = 1; i <= 9; i++) {
    if (judge(board, x, y, i + '0')) {
      board[x][y] = i + '0';
      if (dfs_1(board, x, y))
        return 1;
      board[x][y] = '.';
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dfs_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %esi,%ebp
  mov    %edx,%ebx
  movslq %esi,%rax
  lea    (%rdi,%rax,8),%r13
  movslq %edx,%r15
  mov    0x0(%r13),%rax
  movzbl (%rax,%r15,1),%eax
  cmp    $0x2e,%al
  je     L33
  cmp    $0x7,%esi
  jle    L54
  cmp    $0x7,%edx
  jle    L54
  jmp    L31
L128:
  add    $0x1,%ebp
  movslq %ebp,%rax
  lea    (%r14,%rax,8),%r13
  mov    0x0(%r13),%rax
  cmpb   $0x2e,(%rax)
  je     L61
  mov    $0x0,%ebx
L54:
  cmp    $0x8,%ebx
  je     L128
  add    $0x1,%ebx
  movslq %ebp,%rax
  lea    (%r14,%rax,8),%r13
  movslq %ebx,%r15
  mov    0x0(%r13),%rax
  movzbl (%rax,%r15,1),%eax
  cmp    $0x2e,%al
  je     L33
  cmp    $0x7,%ebp
  jle    L54
  cmp    $0x7,%ebx
  jle    L54
L31:
  cmp    $0x8,%ebp
  sete   %cl
  cmp    $0x8,%ebx
  sete   %dl
  test   %dl,%cl
  je     L33
  mov    $0x1,%edx
  cmp    $0x2e,%al
  jne    L166
  jmp    L33
L61:
  mov    $0x0,%ebx
  mov    $0x0,%r15d
L33:
  mov    $0x31,%r12d
  jmp    L137
L100:
  mov    0x0(%r13),%rax
  movb   $0x2e,(%rax,%r15,1)
L139:
  add    $0x1,%r12d
  cmp    $0x3a,%r12d
  je     L111
L137:
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <judge@plt>
  test   %rax,%rax
  je     L139
  mov    0x0(%r13),%rax
  mov    %r12b,(%rax,%r15,1)
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <dfs_1>
  test   %eax,%eax
  je     L100
  mov    $0x1,%edx
  jmp    L166
L111:
  mov    $0x0,%edx
L166:
  mov    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int dfs_1(char **board, int x, int y) {
  int i;
  while (board[x][y] != '.' && (x < 8 || y < 8)) {
    if (y == 8)
      y = 0, x++;
    else
      y++;
  }
  if (x == 8 && y == 8 && board[x][y] != '.')
    return 1;
  for (i = 1; i <= 9; i++) {
    if (judge(board, x, y, i + '0')) {
      board[x][y] = i + '0';
      if (dfs_1(board, x, y))
        return 1;
      board[x][y] = '.';
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dfs_1>:
  endbr64
  push   %r15
  movslq %edx,%r15
  movslq %esi,%rdx
  push   %r14
  lea    (%rdi,%rdx,8),%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %r15,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%r14),%rax
  cmpb   $0x2e,(%rax,%r15,1)
  je     L137
  cmp    $0x7,%esi
  jle    L44
  cmp    $0x7,%r15d
  jg     L64
L44:
  cmp    $0x8,%r12d
  jne    L66
  add    $0x1,%ebp
  movslq %ebp,%rdx
  lea    0x0(%r13,%rdx,8),%r14
  mov    (%r14),%rax
  cmpb   $0x2e,(%rax)
  je     L131
  cmpb   $0x2e,0x1(%rax)
  je     L23
  cmpb   $0x2e,0x2(%rax)
  mov    $0x2,%r12d
  je     L159
L66:
  lea    0x0(%r13,%rdx,8),%r14
  add    $0x1,%r12d
  mov    (%r14),%rax
  movslq %r12d,%r15
  cmpb   $0x2e,(%rax,%r15,1)
  je     L137
  cmp    $0x7,%ebp
  jle    L44
  cmp    $0x7,%r12d
  jle    L44
L64:
  cmp    $0x8,%ebp
  jne    L137
  cmp    $0x8,%r12d
  jne    L137
L106:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L159:
  mov    $0x2,%r15d
L137:
  mov    $0x31,%ebx
  jmp    L140
  nopl   (%rax)
L129:
  add    $0x1,%ebx
  cmp    $0x3a,%ebx
  je     L110
L140:
  mov    %ebx,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <judge@plt>
  test   %rax,%rax
  je     L129
  mov    (%r14),%rax
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %bl,(%rax,%r15,1)
  call   <dfs_1>
  test   %eax,%eax
  jne    L106
  mov    (%r14),%rax
  add    $0x1,%ebx
  movb   $0x2e,(%rax,%r15,1)
  cmp    $0x3a,%ebx
  jne    L140
L110:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L131:
  xor    %r12d,%r12d
  xor    %r15d,%r15d
  jmp    L137
L23:
  mov    $0x1,%r12d
  mov    $0x1,%r15d
  jmp    L137
```

assistant: ```c
int dfs_1(char **board, int x, int y) {
  int i;
  while (board[x][y] != '.' && (x < 8 || y < 8)) {
    if (y == 8)
      y = 0, x++;
    else
      y++;
  }
  if (x == 8 && y == 8 && board[x][y] != '.')
    return 1;
  for (i = 1; i <= 9; i++) {
    if (judge(board, x, y, i + '0')) {
      board[x][y] = i + '0';
      if (dfs_1(board, x, y))
        return 1;
      board[x][y] = '.';
    }
  }
  return 0;
}
```