ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<bitToDecimal>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D18(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L3
  lea    D19(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  test   %edx,%edx
  je     L3
  lea    D21(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  cmp    $0x1,%eax
  mov    $0x2,%eax
  sbb    $0xffffffff,%eax
L3:
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D19
type: string
name: parse_data
arguments: data_tag: D21
type: string

tool: D18: 00

tool: D19: 01

tool: D21: 10

assistant: ```c
int bitToDecimal(char *bit) {
  int test = strcmp(bit, "00");
  if (!strcmp(bit, "00")) {
    return 0;
  } else if (!strcmp(bit, "01")) {
    return 1;
  } else if (!strcmp(bit, "10")) {
    return 2;
  } else {
    return 3;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bitToDecimal>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x30,%eax
  je     L44
L21:
  cmp    $0x31,%eax
  je     L20
L37:
  mov    $0x3,%r8d
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L44:
  cmpb   $0x30,0x1(%rdi)
  je     L49
L67:
  cmp    $0x30,%eax
  jne    L21
  cmpb   $0x31,0x1(%rdi)
  jne    L21
  cmpb   $0x0,0x2(%rdi)
  mov    $0x1,%r8d
  jne    L21
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L49:
  movzbl 0x2(%rdi),%r8d
  test   %r8d,%r8d
  jne    L67
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L20:
  cmpb   $0x30,0x1(%rdi)
  jne    L37
  cmpb   $0x0,0x2(%rdi)
  mov    $0x2,%r8d
  jne    L37
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitToDecimal(char *bit) {
  int test = strcmp(bit, "00");
  if (!strcmp(bit, "00")) {
    return 0;
  } else if (!strcmp(bit, "01")) {
    return 1;
  } else if (!strcmp(bit, "10")) {
    return 2;
  } else {
    return 3;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bitToDecimal>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x30,%eax
  je     L23
L6:
  cmp    $0x31,%eax
  je     L56
L71:
  mov    $0x3,%r8d
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  cmpb   $0x30,0x1(%rdi)
  je     L19
L32:
  cmp    $0x30,%eax
  jne    L6
  cmpb   $0x31,0x1(%rdi)
  jne    L6
  cmpb   $0x0,0x2(%rdi)
  mov    $0x1,%r8d
  jne    L6
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L19:
  movzbl 0x2(%rdi),%r8d
  test   %r8d,%r8d
  jne    L32
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L56:
  cmpb   $0x30,0x1(%rdi)
  jne    L71
  cmpb   $0x0,0x2(%rdi)
  mov    $0x2,%r8d
  jne    L71
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bitToDecimal(char *bit) {
  int test = strcmp(bit, "00");
  if (!strcmp(bit, "00")) {
    return 0;
  } else if (!strcmp(bit, "01")) {
    return 1;
  } else if (!strcmp(bit, "10")) {
    return 2;
  } else {
    return 3;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<syscon_initialize_number_of_multprec_Laurentials>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  lea    -0x24(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x23d,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
int syscon_initialize_number_of_multprec_Laurentials(int length) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(573, &length, b, c, 0);
  return fail;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<syscon_initialize_number_of_multprec_Laurentials>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x23d,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int syscon_initialize_number_of_multprec_Laurentials(int length) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(573, &length, b, c, 0);
  return fail;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<syscon_initialize_number_of_multprec_Laurentials>:
  endbr64
  sub    $0x18,%rsp
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x23d,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int syscon_initialize_number_of_multprec_Laurentials(int length) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(573, &length, b, c, 0);
  return fail;
}
```
user: Translate the below assembly code into C programming language:

```asm
<syscon_initialize_number_of_multprec_Laurentials>:
  endbr64
  sub    $0x18,%rsp
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %edi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    $0x23d,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
int syscon_initialize_number_of_multprec_Laurentials(int length) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(573, &length, b, c, 0);
  return fail;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<skip_white>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L34
L7:
  addl   $0x1,-0x8(%rbp)
L34:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x20,-0x4(%rbp)
  je     L7
  cmpl   $0x9,-0x4(%rbp)
  je     L7
  cmpl   $0xa,-0x4(%rbp)
  je     L7
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int skip_white(char *pin) {
  int c, len = 0;
  while ((c = *pin++) == ' ' || c == '\t' || c == '\n')
    len++;
  return (len);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<skip_white>:
  endbr64
  mov    $0x0,%edx
  jmp    L1
L14:
  add    $0x1,%rdx
L1:
  movzbl (%rdi,%rdx,1),%eax
  cmp    $0x20,%al
  je     L14
  movsbl %al,%eax
  sub    $0x9,%eax
  cmp    $0x1,%eax
  jbe    L14
  mov    %edx,%eax
  ret
```

assistant: ```c
int skip_white(char *pin) {
  int c, len = 0;
  while ((c = *pin++) == ' ' || c == '\t' || c == '\n')
    len++;
  return (len);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<skip_white>:
  endbr64
  xor    %edx,%edx
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rdx
L24:
  movsbl (%rdi,%rdx,1),%eax
  cmp    $0x20,%al
  je     L21
  sub    $0x9,%eax
  cmp    $0x1,%eax
  jbe    L21
  mov    %edx,%eax
  ret
```

assistant: ```c
int skip_white(char *pin) {
  int c, len = 0;
  while ((c = *pin++) == ' ' || c == '\t' || c == '\n')
    len++;
  return (len);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<skip_white>:
  endbr64
  xor    %edx,%edx
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rdx
L6:
  movsbl (%rdi,%rdx,1),%eax
  cmp    $0x20,%al
  je     L21
  sub    $0x9,%eax
  cmp    $0x1,%eax
  jbe    L21
  mov    %edx,%eax
  ret
```

assistant: ```c
int skip_white(char *pin) {
  int c, len = 0;
  while ((c = *pin++) == ' ' || c == '\t' || c == '\n')
    len++;
  return (len);
}
```
user: What is the c source code of the assembly code below?

```asm
<stage_wrapper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9d,-0x30(%rbp)
  cmpl   $0x0,-0x30(%rbp)
  je     L52
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <epilogue@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L52
  mov    $0x1,%eax
  jmp    L41
L52:
  mov    -0x30(%rbp),%edx
  mov    -0x38(%rbp),%rcx
  mov    -0x2c(%rbp),%eax
  mov    -0x20(%rbp),%r8
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   *%r8
L1:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <prologue@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L2
  cmpl   $0x2,-0x4(%rbp)
  jne    L52
  mov    -0x38(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
  jmp    L1
L2:
  nop
  mov    $0x0,%eax
L41:
  leave
  ret
```

assistant: ```c
int stage_wrapper(char *str, void (*stage)(int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode), void (*stage_dump)(struct eeprom_table *new_e2prom_tab_p), int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode) {
  unsigned int ret;
  if (dump_mode) {
    ret = epilogue(str);
    if (ret == 1)
      return 1;
  }
  while (1) {
    stage(device_fd, new_e2prom_tab_p, dump_mode);
  summary:
    ret = prologue(str);
    if (!ret)
      break;
    else if (ret == 2) {
      stage_dump(new_e2prom_tab_p);
      goto summary;
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<stage_wrapper>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    %rdx,%r12
  mov    %ecx,%r15d
  mov    %r8,%rbp
  mov    %r9d,%r13d
  test   %r9d,%r9d
  je     L4
  call   <epilogue@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  cmp    $0x1,%edx
  jne    L4
  jmp    L58
L13:
  mov    %rbp,%rdi
  call   *%r12
L25:
  mov    %rbx,%rdi
  call   <prologue@plt>
  test   %eax,%eax
  je     L56
  cmp    $0x2,%eax
  je     L13
L4:
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r15d,%edi
  call   *%r14
  jmp    L25
L56:
  mov    $0x0,%eax
L58:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int stage_wrapper(char *str, void (*stage)(int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode), void (*stage_dump)(struct eeprom_table *new_e2prom_tab_p), int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode) {
  unsigned int ret;
  if (dump_mode) {
    ret = epilogue(str);
    if (ret == 1)
      return 1;
  }
  while (1) {
    stage(device_fd, new_e2prom_tab_p, dump_mode);
  summary:
    ret = prologue(str);
    if (!ret)
      break;
    else if (ret == 2) {
      stage_dump(new_e2prom_tab_p);
      goto summary;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<stage_wrapper>:
  endbr64
  push   %r15
  mov    %ecx,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %r9d,%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %r9d,%r9d
  jne    L2
L10:
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r15d,%edi
  call   *%r14
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L11:
  cmp    $0x2,%eax
  jne    L10
  mov    %rbp,%rdi
  call   *%r12
L23:
  mov    %rbx,%rdi
  call   <prologue@plt>
  test   %eax,%eax
  jne    L11
  xor    %eax,%eax
L52:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  call   <epilogue@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  cmp    $0x1,%r8d
  jne    L10
  jmp    L52
```

assistant: ```c
int stage_wrapper(char *str, void (*stage)(int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode), void (*stage_dump)(struct eeprom_table *new_e2prom_tab_p), int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode) {
  unsigned int ret;
  if (dump_mode) {
    ret = epilogue(str);
    if (ret == 1)
      return 1;
  }
  while (1) {
    stage(device_fd, new_e2prom_tab_p, dump_mode);
  summary:
    ret = prologue(str);
    if (!ret)
      break;
    else if (ret == 2) {
      stage_dump(new_e2prom_tab_p);
      goto summary;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<stage_wrapper>:
  endbr64
  push   %r15
  mov    %ecx,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %r9d,%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  test   %r9d,%r9d
  jne    L4
L34:
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r15d,%edi
  call   *%r14
  jmp    L53
  nopw   0x0(%rax,%rax,1)
L30:
  cmp    $0x2,%eax
  jne    L34
  mov    %rbp,%rdi
  call   *%r12
L53:
  mov    %rbx,%rdi
  call   <prologue@plt>
  test   %eax,%eax
  jne    L30
  xor    %eax,%eax
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  call   <epilogue@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  cmp    $0x1,%r8d
  jne    L34
  jmp    L12
```

assistant: ```c
int stage_wrapper(char *str, void (*stage)(int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode), void (*stage_dump)(struct eeprom_table *new_e2prom_tab_p), int device_fd, struct eeprom_table *new_e2prom_tab_p, int dump_mode) {
  unsigned int ret;
  if (dump_mode) {
    ret = epilogue(str);
    if (ret == 1)
      return 1;
  }
  while (1) {
    stage(device_fd, new_e2prom_tab_p, dump_mode);
  summary:
    ret = prologue(str);
    if (!ret)
      break;
    else if (ret == 2) {
      stage_dump(new_e2prom_tab_p);
      goto summary;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    $0xfffffffe,%r14d
  mov    $0x0,%r13d
  lea    D46(%rip),%r15
  jmp    L78
L97:
  cmp    $0x3,%ebp
  jle    L68
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L69
L68:
  add    %r14d,%ebx
  add    $0x1,%ebp
L66:
  lea    -0x1(%rbp),%r12d
  mov    %ebx,%esi
  shr    $0x1f,%esi
  add    %ebx,%esi
  sar    %esi
  mov    %r12d,%edx
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x2,%r12d
  jg     L97
  mov    $0x2c,%edi
  call   <putchar@plt>
  jmp    L68
L109:
  mov    $0x6,%eax
  sub    %r13d,%eax
  mov    %eax,0xc(%rsp)
  mov    %eax,%ebx
  mov    $0x2,%ebp
  jmp    L117
L83:
  cmp    $0x3,%ebp
  jle    L11
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0xe,%r13d
  jg     L63
L69:
  lea    D43(%rip),%rdi
  call   <puts@plt>
L63:
  add    $0x1,%r13d
  add    $0x1,%r14d
  cmp    $0x8,%r13d
  je     L3
L78:
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x4,%r13d
  jg     L109
  mov    %r14d,%ebx
  mov    $0x2,%ebp
  jmp    L66
L11:
  add    0xc(%rsp),%ebx
  add    $0x1,%ebp
L117:
  lea    -0x1(%rbp),%r12d
  mov    %ebx,%edx
  shr    $0x1f,%edx
  add    %ebx,%edx
  sar    %edx
  mov    %r12d,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x2,%r12d
  jg     L83
  mov    $0x2c,%edi
  call   <putchar@plt>
  jmp    L11
L3:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
datatype: string
name: read_data
arguments: data_tag: D46
datatype: string
name: read_data
arguments: data_tag: D12
datatype: string
name: read_data
arguments: data_tag: D43
datatype: string
name: read_data
arguments: data_tag: D45
datatype: string
name: read_data
arguments: data_tag: D6
datatype: string

tool: D2: int direction_offsets_table[8][3] = {

tool: D46: %2i*OD_FILT_BSTRIDE + %2i

tool: D12:  }

tool: D43: ,

tool: D45:   {

tool: D6: 
};

assistant: ```c
int main(int argc, char **argv) {
  printf("int direction_offsets_table[8][3] = {\n");
  int offset[3];
  for (int dir = 0; dir < 8; dir++) {
    int k;
    int f;
    printf("  {");
    if (dir <= 4) {
      f = dir - 2;
      for (int k = 1; k <= 3; k++) {
        printf("%2i*OD_FILT_BSTRIDE + %2i", f * k / 2, k);
        if (k < 3) {
          printf(",");
        }
      }
    } else {
      f = 6 - dir;
      for (int k = 1; k <= 3; k++) {
        printf("%2i*OD_FILT_BSTRIDE + %2i", k, f * k / 2);
        if (k < 3) {
          printf(",");
        }
      }
    }
    printf(" }");
    if (dir < 15) {
      printf(",\n");
    }
  }
  printf("\n};\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  lea    D21(%rip),%rdi
  lea    D14(%rip),%r15
  push   %r14
  lea    D49(%rip),%r14
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  mov    $0x12,%r12d
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x18,%rsp
  call   <puts@plt>
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L14:
  lea    -0x3(%rbx),%r8d
  mov    $0x1,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %r8d,%esi
  mov    %r8d,0xc(%rsp)
  shr    $0x1f,%esi
  add    %r8d,%esi
  sar    %esi
  call   <printf@plt>
  mov    $0x2c,%edi
  call   <putchar@plt>
  mov    0xc(%rsp),%esi
  mov    $0x2,%edx
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x2c,%edi
  call   <putchar@plt>
  mov    $0xc,%eax
  mov    $0x3,%edx
  mov    %rbp,%rdi
  sub    %r12d,%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  xor    %eax,%eax
  sar    %esi
  call   <printf@plt>
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <puts@plt>
L10:
  add    $0x1,%ebx
  sub    $0x3,%r12d
L29:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x1(%rbx),%eax
  cmp    $0x4,%eax
  jbe    L14
  xor    %edx,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2c,%edi
  call   <putchar@plt>
  mov    $0x7,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  sub    %ebx,%edx
  mov    $0x2,%esi
  call   <printf@plt>
  mov    $0x2c,%edi
  call   <putchar@plt>
  mov    %r12d,%edx
  mov    $0x3,%esi
  mov    %rbp,%rdi
  shr    $0x1f,%edx
  xor    %eax,%eax
  add    %r12d,%edx
  sar    %edx
  call   <printf@plt>
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <puts@plt>
  cmp    $0x8,%ebx
  jne    L10
  lea    D10(%rip),%rdi
  call   <puts@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D21
datatype: string
name: parse_data
arguments: tag: D14
datatype: string
name: parse_data
arguments: tag: D49
datatype: string
name: parse_data
arguments: tag: D0
datatype: string
name: parse_data
arguments: tag: D6
datatype: string
name: parse_data
arguments: tag: D10
datatype: string

tool: D21: int direction_offsets_table[8][3] = {

tool: D14:   {

tool: D49:  }

tool: D0: ,

tool: D6: %2i*OD_FILT_BSTRIDE + %2i

tool: D10: 
};

assistant: ```c
int main(int argc, char **argv) {
  printf("int direction_offsets_table[8][3] = {\n");
  int offset[3];
  for (int dir = 0; dir < 8; dir++) {
    int k;
    int f;
    printf("  {");
    if (dir <= 4) {
      f = dir - 2;
      for (int k = 1; k <= 3; k++) {
        printf("%2i*OD_FILT_BSTRIDE + %2i", f * k / 2, k);
        if (k < 3) {
          printf(",");
        }
      }
    } else {
      f = 6 - dir;
      for (int k = 1; k <= 3; k++) {
        printf("%2i*OD_FILT_BSTRIDE + %2i", k, f * k / 2);
        if (k < 3) {
          printf(",");
        }
      }
    }
    printf(" }");
    if (dir < 15) {
      printf(",\n");
    }
  }
  printf("\n};\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parseLevels>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L15
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jne    L65
L15:
  cmpl   $0x0,-0x1c(%rbp)
  je     L37
  mov    $0x0,%eax
  jmp    L44
L37:
  mov    $0xffffffff,%eax
  jmp    L44
L57:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  je     L23
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x36,%al
  jle    L86
L23:
  mov    $0xffffffff,%eax
  jmp    L44
L86:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L65:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L57
  mov    -0xc(%rbp),%eax
L44:
  leave
  ret
```

assistant: ```c
int parseLevels(char *str, int emptyOk) {
  char *chptr = str;
  int rc = 0;
  if (!str || !strlen(str))
    return emptyOk ? 0 : -1;
  while (*chptr) {
    if (!isdigit(*chptr) || *chptr > '6')
      return -1;
    rc |= 1 << (*chptr - '0');
    chptr++;
  }
  return rc;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<parseLevels>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  test   %rdi,%rdi
  je     L73
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L73
  movzbl (%rbx),%edi
  mov    $0x0,%ebp
  mov    $0x1,%r12d
  test   %dil,%dil
  je     L5
L88:
  movsbl %dil,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  je     L60
  movzbl (%rbx),%ecx
  cmp    $0x36,%cl
  jg     L29
  sub    $0x30,%ecx
  mov    %r12d,%eax
  shl    %cl,%eax
  or     %eax,%ebp
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L88
  jmp    L5
L73:
  cmp    $0x1,%ebp
  sbb    %ebp,%ebp
L5:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L60:
  mov    $0xffffffff,%ebp
  jmp    L5
L29:
  mov    $0xffffffff,%ebp
  jmp    L5
```

assistant: ```c
int parseLevels(char *str, int emptyOk) {
  char *chptr = str;
  int rc = 0;
  if (!str || !strlen(str))
    return emptyOk ? 0 : -1;
  while (*chptr) {
    if (!isdigit(*chptr) || *chptr > '6')
      return -1;
    rc |= 1 << (*chptr - '0');
    chptr++;
  }
  return rc;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parseLevels>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  test   %rdi,%rdi
  je     L21
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L21
  movsbl (%rbx),%edi
  xor    %r12d,%r12d
  mov    $0x1,%ebp
  test   %dil,%dil
  jne    L36
  jmp    L3
  nopl   (%rax)
L66:
  movzbl (%rbx),%ecx
  cmp    $0x36,%cl
  jg     L6
  sub    $0x30,%ecx
  mov    %ebp,%eax
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  shl    %cl,%eax
  or     %eax,%r12d
  test   %dil,%dil
  je     L3
L36:
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L66
L6:
  mov    $0xffffffff,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L21:
  cmp    $0x1,%ebp
  sbb    %r12d,%r12d
L3:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parseLevels(char *str, int emptyOk) {
  char *chptr = str;
  int rc = 0;
  if (!str || !strlen(str))
    return emptyOk ? 0 : -1;
  while (*chptr) {
    if (!isdigit(*chptr) || *chptr > '6')
      return -1;
    rc |= 1 << (*chptr - '0');
    chptr++;
  }
  return rc;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parseLevels>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  test   %rdi,%rdi
  je     L26
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L26
  movsbl (%rbx),%edi
  xor    %r12d,%r12d
  mov    $0x1,%ebp
  test   %dil,%dil
  jne    L38
  jmp    L35
  nopl   (%rax)
L46:
  movzbl (%rbx),%ecx
  cmp    $0x36,%cl
  jg     L27
  sub    $0x30,%ecx
  mov    %ebp,%eax
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  shl    %cl,%eax
  or     %eax,%r12d
  test   %dil,%dil
  je     L35
L38:
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L46
L27:
  mov    $0xffffffff,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L26:
  cmp    $0x1,%ebp
  sbb    %r12d,%r12d
L35:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int parseLevels(char *str, int emptyOk) {
  char *chptr = str;
  int rc = 0;
  if (!str || !strlen(str))
    return emptyOk ? 0 : -1;
  while (*chptr) {
    if (!isdigit(*chptr) || *chptr > '6')
      return -1;
    rc |= 1 << (*chptr - '0');
    chptr++;
  }
  return rc;
}
```