ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x80000000,%edi
  call   <sub1@plt>
  test   %rax,%rax
  jne    L26
L6:
  mov    $0x80000000,%edi
  call   <sub2@plt>
  test   %rax,%rax
  jne    L49
L44:
  mov    $0x80000000,%edi
  call   <sub3@plt>
  test   %rax,%rax
  jne    L29
L48:
  mov    $0x80000000,%edi
  call   <sub4@plt>
  test   %rax,%rax
  jne    L16
L56:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L26:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L6
L16:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L56
L29:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L48
L49:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L44
```

assistant: ```c
main() {
  if (sub1(0x80000000ULL))
    abort();
  if (sub2(0x80000000ULL))
    abort();
  if (sub3(0x80000000ULL))
    abort();
  if (sub4(0x80000000ULL))
    abort();
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x80000000,%edi
  call   <sub1@plt>
  test   %rax,%rax
  jne    L62
L10:
  mov    $0x80000000,%edi
  call   <sub2@plt>
  test   %rax,%rax
  jne    L66
L51:
  mov    $0x80000000,%edi
  call   <sub3@plt>
  test   %rax,%rax
  jne    L2
L34:
  mov    $0x80000000,%edi
  call   <sub4@plt>
  test   %rax,%rax
  jne    L35
L74:
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L62:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L10
L35:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L74
L2:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L34
L66:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L51
```

assistant: ```c
main() {
  if (sub1(0x80000000ULL))
    abort();
  if (sub2(0x80000000ULL))
    abort();
  if (sub3(0x80000000ULL))
    abort();
  if (sub4(0x80000000ULL))
    abort();
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ulaw2linear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  notl   -0x14(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0x3,%eax
  and    $0x78,%eax
  add    $0x84,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x4,%eax
  and    $0x7,%eax
  mov    %eax,%ecx
  shll   %cl,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x80,%eax
  test   %eax,%eax
  je     L16
  mov    $0x84,%eax
  sub    -0x4(%rbp),%eax
  jmp    L3
L16:
  mov    -0x4(%rbp),%eax
  sub    $0x84,%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
int ulaw2linear(int u_val) {
  int t;
  u_val = ~u_val;
  t = ((u_val & (0xf)) << 3) + (0x84);
  t <<= (u_val & (0x70)) >> (4);
  return ((u_val & (0x80)) ? ((0x84) - t) : (t - (0x84)));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ulaw2linear>:
  endbr64
  not    %edi
  lea    0x0(,%rdi,8),%eax
  and    $0x78,%eax
  add    $0x84,%eax
  mov    %edi,%ecx
  sar    $0x4,%ecx
  and    $0x7,%ecx
  shl    %cl,%eax
  mov    $0x84,%edx
  sub    %eax,%edx
  sub    $0x84,%eax
  test   $0x80,%dil
  cmovne %edx,%eax
  ret
```

assistant: ```c
int ulaw2linear(int u_val) {
  int t;
  u_val = ~u_val;
  t = ((u_val & (0xf)) << 3) + (0x84);
  t <<= (u_val & (0x70)) >> (4);
  return ((u_val & (0x80)) ? ((0x84) - t) : (t - (0x84)));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ulaw2linear>:
  endbr64
  not    %edi
  mov    $0x84,%edx
  lea    0x0(,%rdi,8),%eax
  mov    %edi,%ecx
  and    $0x78,%eax
  sar    $0x4,%ecx
  and    $0x7,%ecx
  add    $0x84,%eax
  shl    %cl,%eax
  sub    %eax,%edx
  sub    $0x84,%eax
  and    $0x80,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int ulaw2linear(int u_val) {
  int t;
  u_val = ~u_val;
  t = ((u_val & (0xf)) << 3) + (0x84);
  t <<= (u_val & (0x70)) >> (4);
  return ((u_val & (0x80)) ? ((0x84) - t) : (t - (0x84)));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ulaw2linear>:
  endbr64
  not    %edi
  mov    $0x84,%edx
  lea    0x0(,%rdi,8),%eax
  mov    %edi,%ecx
  and    $0x78,%eax
  sar    $0x4,%ecx
  and    $0x7,%ecx
  add    $0x84,%eax
  shl    %cl,%eax
  sub    %eax,%edx
  sub    $0x84,%eax
  and    $0x80,%edi
  cmovne %edx,%eax
  ret
```

assistant: ```c
int ulaw2linear(int u_val) {
  int t;
  u_val = ~u_val;
  t = ((u_val & (0xf)) << 3) + (0x84);
  t <<= (u_val & (0x70)) >> (4);
  return ((u_val & (0x80)) ? ((0x84) - t) : (t - (0x84)));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hasVowel>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L18
L5:
  addq   $0x1,-0x8(%rbp)
L18:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <isConsonant@plt>
  test   %rax,%rax
  jne    L5
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  setne  %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int hasVowel(const char *z) {
  while (isConsonant(z)) {
    z++;
  }
  return *z != 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hasVowel>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L13
L19:
  add    $0x1,%rbx
L13:
  mov    %rbx,%rdi
  call   <isConsonant@plt>
  test   %rax,%rax
  jne    L19
  cmpb   $0x0,(%rbx)
  setne  %al
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int hasVowel(const char *z) {
  while (isConsonant(z)) {
    z++;
  }
  return *z != 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<hasVowel>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L16:
  add    $0x1,%rbx
L17:
  mov    %rbx,%rdi
  call   <isConsonant@plt>
  test   %rax,%rax
  jne    L16
  xor    %eax,%eax
  cmpb   $0x0,(%rbx)
  pop    %rbx
  setne  %al
  ret
```

assistant: ```c
int hasVowel(const char *z) {
  while (isConsonant(z)) {
    z++;
  }
  return *z != 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<hasVowel>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L8:
  add    $0x1,%rbx
L14:
  mov    %rbx,%rdi
  call   <isConsonant@plt>
  test   %rax,%rax
  jne    L8
  xor    %eax,%eax
  cmpb   $0x0,(%rbx)
  pop    %rbx
  setne  %al
  ret
```

assistant: ```c
int hasVowel(const char *z) {
  while (isConsonant(z)) {
    z++;
  }
  return *z != 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<path_008_expected_8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  pop    %rbp
  ret
```

assistant: ```c
void path_008_expected_8(int a, int b, int c) {
  if (a) {
  }
  if (b) {
  }
  if (c) {
    return;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<path_008_expected_8>:
  endbr64
  ret
```

assistant: ```c
void path_008_expected_8(int a, int b, int c) {
  if (a) {
  }
  if (b) {
  }
  if (c) {
    return;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<path_008_expected_8>:
  endbr64
  ret
```

assistant: ```c
void path_008_expected_8(int a, int b, int c) {
  if (a) {
  }
  if (b) {
  }
  if (c) {
    return;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<path_008_expected_8>:
  endbr64
  ret
```

assistant: ```c
void path_008_expected_8(int a, int b, int c) {
  if (a) {
  }
  if (b) {
  }
  if (c) {
    return;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x14(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D40(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x2,0xc(%rsp)
  je     L46
L127:
  lea    0x14(%rsp),%rdi
  mov    0x10(%rsp),%esi
  call   <str_to_val@plt>
  mov    %rax,%rbp
  mov    $0x2,%ebx
  lea    0x20(%rsp),%r12
  cmp    $0xffffffffffffffff,%rax
  jne    L114
  lea    D49(%rip),%rdi
  call   <puts@plt>
  jmp    L92
L46:
  lea    0x14(%rsp),%r12
  lea    0x2c(%rsp),%rbp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  lea    0x20(%rsp),%rbx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L127
L126:
  add    $0x1,%ebx
  cmp    $0x3e8,%ebx
  je     L88
L114:
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <str_to_val@plt>
  cmp    %rbp,%rax
  jne    L126
  cmp    $0xffffffffffffffff,%rax
  je     L126
  mov    %ebx,%esi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L92:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L88:
  lea    0x20(%rsp),%rdi
  call   <strlen@plt>
  cmp    $0x1,%eax
  je     L48
  mov    %eax,%edi
  call   <max_radix@plt>
  mov    %eax,%esi
  lea    0x20(%rsp),%rdi
  mov    %rbp,%rdx
  call   <bs_radix@plt>
  mov    %rax,%rsi
L24:
  cmp    $0xffffffffffffffff,%rsi
  je     L34
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L92
L48:
  lea    0x20(%rsp),%rdi
  mov    %rbp,%rdx
  mov    $0x50,%esi
  call   <bs_radix@plt>
  mov    %rax,%rsi
  jmp    L24
L34:
  lea    D49(%rip),%rdi
  call   <puts@plt>
  jmp    L92
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D40
datatype: string
name: disassemble_data
arguments: data_tag: D49
datatype: string
name: disassemble_data
arguments: data_tag: D8
datatype: string
name: disassemble_data
arguments: data_tag: D38
datatype: string

tool: D40: %s%s%d%d

tool: D49: Impossible

tool: D8: %d


tool: D38: %lld


assistant: ```c
int main() {
  char s1[12], s2[12], temp[12];
  int tag, radix;
  scanf("%s%s%d%d", s1, s2, &tag, &radix);
  if (tag == 2) {
    strcpy(temp, s1);
    strcpy(s1, s2);
    strcpy(s2, temp);
  }
  long long val = str_to_val(s1, radix);
  if (val == -1) {
    printf("Impossible\n");
    return 0;
  }
  int is_find = 0;
  long long a[1000];
  for (int i = 2; i < 1000; ++i) {
    a[i] = str_to_val(s2, i);
    if (a[i] == -1)
      continue;
    if (a[i] == val) {
      is_find = 1;
      printf("%d\n", i);
      break;
    }
  }
  if (is_find)
    return 0;
  int len = strlen(s2);
  long long ans;
  if (len == 1)
    ans = bs_radix(s2, 80, val);
  else
    ans = bs_radix(s2, max_radix(len), val);
  if (ans == -1) {
    printf("Impossible\n");
  } else {
    printf("%lld\n", ans);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D17(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x40,%rsp
  lea    0x20(%rsp),%r12
  lea    0x14(%rsp),%rbp
  lea    0xc(%rsp),%rcx
  lea    0x10(%rsp),%r8
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <scanf@plt>
  cmpl   $0x2,0xc(%rsp)
  je     L106
L85:
  mov    0x10(%rsp),%esi
  mov    %rbp,%rdi
  mov    $0x2,%ebp
  call   <str_to_val@plt>
  mov    %rax,%r13
  cmp    $0xffffffffffffffff,%rax
  jne    L24
  jmp    L109
  nopl   0x0(%rax)
L104:
  add    $0x1,%ebp
  cmp    $0x3e8,%ebp
  je     L66
L24:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <str_to_val@plt>
  cmp    $0xffffffffffffffff,%rax
  je     L104
  cmp    %r13,%rax
  jne    L104
  mov    %ebp,%esi
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L48:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L106:
  lea    0x2c(%rsp),%r13
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L85
L109:
  lea    D7(%rip),%rdi
  call   <puts@plt>
  jmp    L48
L66:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  je     L116
  call   <max_radix@plt>
  mov    %r13,%rdx
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <bs_radix@plt>
  mov    %rax,%rsi
L108:
  cmp    $0xffffffffffffffff,%rsi
  je     L109
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L48
L116:
  mov    $0x50,%esi
  mov    %r13,%rdx
  mov    %r12,%rdi
  call   <bs_radix@plt>
  mov    %rax,%rsi
  jmp    L108
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D34
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string

tool: D17: %s%s%d%d

tool: D34: %d


tool: D7: Impossible

tool: D8: %lld


assistant: ```c
int main() {
  char s1[12], s2[12], temp[12];
  int tag, radix;
  scanf("%s%s%d%d", s1, s2, &tag, &radix);
  if (tag == 2) {
    strcpy(temp, s1);
    strcpy(s1, s2);
    strcpy(s2, temp);
  }
  long long val = str_to_val(s1, radix);
  if (val == -1) {
    printf("Impossible\n");
    return 0;
  }
  int is_find = 0;
  long long a[1000];
  for (int i = 2; i < 1000; ++i) {
    a[i] = str_to_val(s2, i);
    if (a[i] == -1)
      continue;
    if (a[i] == val) {
      is_find = 1;
      printf("%d\n", i);
      break;
    }
  }
  if (is_find)
    return 0;
  int len = strlen(s2);
  long long ans;
  if (len == 1)
    ans = bs_radix(s2, 80, val);
  else
    ans = bs_radix(s2, max_radix(len), val);
  if (ans == -1) {
    printf("Impossible\n");
  } else {
    printf("%lld\n", ans);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  lea    D27(%rip),%rdi
  push   %r12
  push   %rbp
  sub    $0x40,%rsp
  lea    0x20(%rsp),%r12
  lea    0x14(%rsp),%rbp
  lea    0xc(%rsp),%rcx
  lea    0x10(%rsp),%r8
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <scanf@plt>
  cmpl   $0x2,0xc(%rsp)
  je     L65
L54:
  mov    0x10(%rsp),%esi
  mov    %rbp,%rdi
  mov    $0x2,%ebp
  call   <str_to_val@plt>
  mov    %rax,%r13
  cmp    $0xffffffffffffffff,%rax
  jne    L115
  jmp    L93
  nopl   0x0(%rax)
L116:
  add    $0x1,%ebp
  cmp    $0x3e8,%ebp
  je     L100
L115:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <str_to_val@plt>
  cmp    $0xffffffffffffffff,%rax
  je     L116
  cmp    %r13,%rax
  jne    L116
  mov    %ebp,%esi
  lea    D33(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L97:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L65:
  lea    0x2c(%rsp),%r13
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L54
L93:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  jmp    L97
L100:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  je     L74
  call   <max_radix@plt>
  mov    %r13,%rdx
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <bs_radix@plt>
  mov    %rax,%rsi
L56:
  cmp    $0xffffffffffffffff,%rsi
  je     L93
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L97
L74:
  mov    $0x50,%esi
  mov    %r13,%rdx
  mov    %r12,%rdi
  call   <bs_radix@plt>
  mov    %rax,%rsi
  jmp    L56
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D27
data_type: string
name: extract_symbolic_data
arguments: label: D33
data_type: string
name: extract_symbolic_data
arguments: label: D17
data_type: string
name: extract_symbolic_data
arguments: label: D6
data_type: string

tool: D27: %s%s%d%d

tool: D33: %d


tool: D17: Impossible

tool: D6: %lld


assistant: ```c
int main() {
  char s1[12], s2[12], temp[12];
  int tag, radix;
  scanf("%s%s%d%d", s1, s2, &tag, &radix);
  if (tag == 2) {
    strcpy(temp, s1);
    strcpy(s1, s2);
    strcpy(s2, temp);
  }
  long long val = str_to_val(s1, radix);
  if (val == -1) {
    printf("Impossible\n");
    return 0;
  }
  int is_find = 0;
  long long a[1000];
  for (int i = 2; i < 1000; ++i) {
    a[i] = str_to_val(s2, i);
    if (a[i] == -1)
      continue;
    if (a[i] == val) {
      is_find = 1;
      printf("%d\n", i);
      break;
    }
  }
  if (is_find)
    return 0;
  int len = strlen(s2);
  long long ans;
  if (len == 1)
    ans = bs_radix(s2, 80, val);
  else
    ans = bs_radix(s2, max_radix(len), val);
  if (ans == -1) {
    printf("Impossible\n");
  } else {
    printf("%lld\n", ans);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<consultarTxt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x128(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x118(%rbp)
  jmp    L28
L6:
  lea    -0x110(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L28:
  mov    -0x118(%rbp),%rdx
  lea    -0x110(%rbp),%rax
  mov    $0xff,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L6
  mov    -0x118(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
data_type: string
name: disassemble_data
arguments: label: D9
data_type: string

tool: D4: r

tool: D9: %s


assistant: ```c
void consultarTxt(char *filename) {
  FILE *fp = fopen(filename, "r");
  char buffer[255];
  while (fgets(buffer, sizeof(buffer), fp)) {
    printf("%s\n", buffer);
  }
  fclose(fp);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<consultarTxt>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbx
  lea    D18(%rip),%rbp
  jmp    L26
L29:
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
L26:
  mov    %rsp,%rdi
  mov    %rbx,%rdx
  mov    $0xff,%esi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L29
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
type_guessed: string
name: parse_data
arguments: tag: D18
type_guessed: string

tool: D10: r

tool: D18: %s


assistant: ```c
void consultarTxt(char *filename) {
  FILE *fp = fopen(filename, "r");
  char buffer[255];
  while (fgets(buffer, sizeof(buffer), fp)) {
    printf("%s\n", buffer);
  }
  fclose(fp);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<consultarTxt>:
  endbr64
  push   %r12
  lea    D3(%rip),%rsi
  lea    D14(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <fopen@plt>
  mov    %rax,%rbp
  jmp    L15
  nopl   (%rax)
L9:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
L15:
  mov    %rbp,%rdx
  mov    $0xff,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L9
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
data_type: string
name: extract_symbolic_data
arguments: data_tag: D14
data_type: string

tool: D3: r

tool: D14: %s


assistant: ```c
void consultarTxt(char *filename) {
  FILE *fp = fopen(filename, "r");
  char buffer[255];
  while (fgets(buffer, sizeof(buffer), fp)) {
    printf("%s\n", buffer);
  }
  fclose(fp);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<consultarTxt>:
  endbr64
  push   %r12
  lea    D16(%rip),%rsi
  lea    D13(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <fopen@plt>
  mov    %rax,%rbp
  jmp    L9
  nopl   (%rax)
L20:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
L9:
  mov    %rbp,%rdx
  mov    $0xff,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L20
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D16
type_infered: string
name: disassemble_data
arguments: tag: D13
type_infered: string

tool: D16: r

tool: D13: %s


assistant: ```c
void consultarTxt(char *filename) {
  FILE *fp = fopen(filename, "r");
  char buffer[255];
  while (fgets(buffer, sizeof(buffer), fp)) {
    printf("%s\n", buffer);
  }
  fclose(fp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<extract_busid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x140,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %rsi,-0x140(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x138(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x128(%rbp)
  mov    $0x100,%eax
  sub    -0x128(%rbp),%eax
  cmp    $0x6,%eax
  ja     L50
  mov    $0x0,%eax
  jmp    L46
L50:
  mov    -0x128(%rbp),%edx
  mov    -0x138(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x120(%rbp)
  mov    -0x120(%rbp),%rax
  lea    D1(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x110(%rbp),%rcx
  mov    -0x138(%rbp),%rax
  mov    $0xff,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <readlink@plt>
  mov    %eax,-0x124(%rbp)
  cmpl   $0x0,-0x124(%rbp)
  jns    L30
  mov    $0xffffffff,%eax
  jmp    L46
L30:
  mov    -0x124(%rbp),%eax
  cltq
  movb   $0x0,-0x110(%rbp,%rax,1)
  lea    -0x110(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x118(%rbp)
  cmpq   $0x0,-0x118(%rbp)
  jne    L8
  mov    $0xffffffff,%eax
  jmp    L46
L8:
  addq   $0x1,-0x118(%rbp)
  mov    -0x118(%rbp),%rdx
  mov    -0x140(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    $0x0,%eax
L46:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D1
type_guessed: string

tool: D1: device

assistant: ```c
int extract_busid(char *name, char *busid) {
  int count;
  unsigned int namelen;
  char linkbuffer[256];
  char *start, *end;
  namelen = strlen(name);
  if ((256 - namelen) < sizeof("device"))
    return 0;
  end = name + namelen;
  strcpy(end, "device");
  count = readlink(name, linkbuffer, 256 - 1);
  if (count < 0)
    return -1;
  linkbuffer[count] = 0;
  start = strrchr(linkbuffer, '/');
  if (!start)
    return -1;
  start++;
  strcpy(busid, start);
  return 0;
}
```