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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x5,-0x24(%rbp)
  movl   $0x7,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  imul   -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  imul   %edx,%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x2,-0x2c(%rbp)
  jmp    L33
L36:
  mov    -0x18(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <egcd@plt>
  cmp    $0x1,%eax
  je     L23
  addl   $0x1,-0x2c(%rbp)
L33:
  mov    -0x2c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L36
  jmp    L60
L23:
  nop
L60:
  movl   $0x2,-0x28(%rbp)
  jmp    L66
L48:
  mov    -0x2c(%rbp),%eax
  imul   -0x28(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <fmod@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L22
  addl   $0x1,-0x28(%rbp)
L66:
  mov    -0x28(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L48
  jmp    L89
L22:
  nop
L89:
  mov    -0x30(%rbp),%eax
  mov    -0x2c(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <fmod@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x28(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <fmod@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D28
type_guessed: string
name: read_data
arguments: label: D10
type_guessed: string
name: read_data
arguments: label: D27
type_guessed: string
name: read_data
arguments: label: D33
type_guessed: string

tool: D28: Enter numeric text message : 

tool: D10: %d

tool: D27: Cipher = %d


tool: D33: Message = %d


assistant: ```c
int main() {
  int m, p, q, phi, e, d, temp, n, c, message;
  printf("Enter numeric text message : ");
  scanf("%d", &m);
  p = 5;
  q = 7;
  n = p * q;
  phi = (p - 1) * (q - 1);
  e = 2;
  while (e < phi) {
    if (egcd(e, phi) == 1)
      break;
    e++;
  }
  d = 2;
  while (d < phi) {
    temp = fmod(((e * d) - 1), phi);
    if (temp == 0)
      break;
    d++;
  }
  c = fmod(pow(m, e), n);
  printf("Cipher = %d\n", c);
  message = fmod(pow(c, d), n);
  printf("Message = %d\n", message);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  mov    $0x2,%ebx
L22:
  mov    $0x18,%esi
  mov    %ebx,%edi
  call   <egcd@plt>
  cmp    $0x1,%eax
  je     L4
  add    $0x1,%ebx
  cmp    $0x18,%ebx
  jne    L22
L4:
  mov    %ebx,%r13d
  lea    -0x1(%rbx,%rbx,1),%r12d
  mov    $0x2,%ebp
L6:
  mov    $0x18,%esi
  mov    %r12d,%edi
  call   <fmod@plt>
  test   %eax,%eax
  je     L49
  add    $0x1,%ebp
  add    %r13d,%r12d
  cmp    $0x18,%ebp
  jne    L6
L49:
  mov    %ebx,%esi
  mov    0x4(%rsp),%edi
  call   <pow@plt>
  mov    %eax,%edi
  mov    $0x23,%esi
  call   <fmod@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <pow@plt>
  mov    %eax,%edi
  mov    $0x23,%esi
  call   <fmod@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
data_type: string
name: retrieve_data
arguments: data_label: D33
data_type: string
name: retrieve_data
arguments: data_label: D18
data_type: string
name: retrieve_data
arguments: data_label: D12
data_type: string

tool: D8: Enter numeric text message : 

tool: D33: %d

tool: D18: Cipher = %d


tool: D12: Message = %d


assistant: ```c
int main() {
  int m, p, q, phi, e, d, temp, n, c, message;
  printf("Enter numeric text message : ");
  scanf("%d", &m);
  p = 5;
  q = 7;
  n = p * q;
  phi = (p - 1) * (q - 1);
  e = 2;
  while (e < phi) {
    if (egcd(e, phi) == 1)
      break;
    e++;
  }
  d = 2;
  while (d < phi) {
    temp = fmod(((e * d) - 1), phi);
    if (temp == 0)
      break;
    d++;
  }
  c = fmod(pow(m, e), n);
  printf("Cipher = %d\n", c);
  message = fmod(pow(c, d), n);
  printf("Message = %d\n", message);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  lea    D9(%rip),%rdi
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  jmp    L55
L52:
  add    $0x1,%ebp
  cmp    $0x18,%ebp
  je     L43
L55:
  mov    $0x18,%esi
  mov    %ebp,%edi
  call   <egcd@plt>
  cmp    $0x1,%eax
  jne    L52
L43:
  lea    -0x1(%rbp,%rbp,1),%ebx
  mov    $0x2,%r12d
  jmp    L0
  nopl   (%rax)
L23:
  add    $0x1,%r12d
  add    %ebp,%ebx
  cmp    $0x18,%r12d
  je     L48
L0:
  mov    $0x18,%esi
  mov    %ebx,%edi
  call   <fmod@plt>
  test   %eax,%eax
  jne    L23
L48:
  mov    0x4(%rsp),%edi
  mov    %ebp,%esi
  call   <pow@plt>
  mov    $0x23,%esi
  mov    %eax,%edi
  call   <fmod@plt>
  lea    D39(%rip),%rdi
  mov    %eax,%esi
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <pow@plt>
  mov    $0x23,%esi
  mov    %eax,%edi
  call   <fmod@plt>
  lea    D37(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_guessed: string
name: retrieve_data
arguments: data_tag: D10
type_guessed: string
name: retrieve_data
arguments: data_tag: D39
type_guessed: string
name: retrieve_data
arguments: data_tag: D37
type_guessed: string

tool: D9: Enter numeric text message : 

tool: D10: %d

tool: D39: Cipher = %d


tool: D37: Message = %d


assistant: ```c
int main() {
  int m, p, q, phi, e, d, temp, n, c, message;
  printf("Enter numeric text message : ");
  scanf("%d", &m);
  p = 5;
  q = 7;
  n = p * q;
  phi = (p - 1) * (q - 1);
  e = 2;
  while (e < phi) {
    if (egcd(e, phi) == 1)
      break;
    e++;
  }
  d = 2;
  while (d < phi) {
    temp = fmod(((e * d) - 1), phi);
    if (temp == 0)
      break;
    d++;
  }
  c = fmod(pow(m, e), n);
  printf("Cipher = %d\n", c);
  message = fmod(pow(c, d), n);
  printf("Message = %d\n", message);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D25(%rip),%rdi
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  jmp    L31
L35:
  add    $0x1,%ebp
  cmp    $0x18,%ebp
  je     L23
L31:
  mov    $0x18,%esi
  mov    %ebp,%edi
  call   <egcd@plt>
  cmp    $0x1,%eax
  jne    L35
L23:
  lea    -0x1(%rbp,%rbp,1),%ebx
  mov    $0x2,%r12d
  jmp    L27
  nopl   (%rax)
L46:
  add    $0x1,%r12d
  add    %ebp,%ebx
  cmp    $0x18,%r12d
  je     L22
L27:
  mov    $0x18,%esi
  mov    %ebx,%edi
  call   <fmod@plt>
  test   %eax,%eax
  jne    L46
L22:
  mov    0x4(%rsp),%edi
  mov    %ebp,%esi
  call   <pow@plt>
  mov    $0x23,%esi
  mov    %eax,%edi
  call   <fmod@plt>
  lea    D35(%rip),%rdi
  mov    %eax,%esi
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <pow@plt>
  mov    $0x23,%esi
  mov    %eax,%edi
  call   <fmod@plt>
  lea    D15(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D25
type_infered: string
name: retrieve_data
arguments: label: D21
type_infered: string
name: retrieve_data
arguments: label: D35
type_infered: string
name: retrieve_data
arguments: label: D15
type_infered: string

tool: D25: Enter numeric text message : 

tool: D21: %d

tool: D35: Cipher = %d


tool: D15: Message = %d


assistant: ```c
int main() {
  int m, p, q, phi, e, d, temp, n, c, message;
  printf("Enter numeric text message : ");
  scanf("%d", &m);
  p = 5;
  q = 7;
  n = p * q;
  phi = (p - 1) * (q - 1);
  e = 2;
  while (e < phi) {
    if (egcd(e, phi) == 1)
      break;
    e++;
  }
  d = 2;
  while (d < phi) {
    temp = fmod(((e * d) - 1), phi);
    if (temp == 0)
      break;
    d++;
  }
  c = fmod(pow(m, e), n);
  printf("Cipher = %d\n", c);
  message = fmod(pow(c, d), n);
  printf("Message = %d\n", message);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<run_tests_with_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <parse_file@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L39
L3:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <run_test_belief_propagation@plt>
  addl   $0x1,-0xc(%rbp)
L39:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jb     L3
  movl   $0x0,-0xc(%rbp)
  jmp    L29
L22:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <run_test_loopy_belief_propagation@plt>
  addl   $0x1,-0xc(%rbp)
L29:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jb     L22
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <delete_expression@plt>
  nop
  leave
  ret
```

assistant: ```c
void run_tests_with_file(const char *file_name, unsigned int num_iterations) {
  unsigned int i;
  struct expression *expr;
  expr = parse_file(file_name);
  for (i = 0; i < num_iterations; ++i) {
    run_test_belief_propagation(expr, file_name);
  }
  for (i = 0; i < num_iterations; ++i) {
    run_test_loopy_belief_propagation(expr, file_name);
  }
  delete_expression(expr);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<run_tests_with_file>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%r14d
  call   <parse_file@plt>
  mov    %rax,%rbp
  test   %r14d,%r14d
  je     L1
  mov    $0x0,%ebx
L28:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <run_test_belief_propagation@plt>
  mov    %ebx,%r13d
  add    $0x1,%ebx
  cmp    %ebx,%r14d
  jne    L28
  mov    $0x0,%ebx
L21:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <run_test_loopy_belief_propagation@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%r13d
  jne    L21
L1:
  mov    %rbp,%rdi
  call   <delete_expression@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void run_tests_with_file(const char *file_name, unsigned int num_iterations) {
  unsigned int i;
  struct expression *expr;
  expr = parse_file(file_name);
  for (i = 0; i < num_iterations; ++i) {
    run_test_belief_propagation(expr, file_name);
  }
  for (i = 0; i < num_iterations; ++i) {
    run_test_loopy_belief_propagation(expr, file_name);
  }
  delete_expression(expr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<run_tests_with_file>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <parse_file@plt>
  mov    %rax,%r12
  test   %r14d,%r14d
  je     L4
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L9:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %ebx,%ebp
  call   <run_test_belief_propagation@plt>
  lea    0x1(%rbx),%ebx
  cmp    %ebx,%r14d
  jne    L9
  xor    %ebx,%ebx
  nop
L15:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <run_test_loopy_belief_propagation@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L15
L4:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <delete_expression@plt>
```

assistant: ```c
void run_tests_with_file(const char *file_name, unsigned int num_iterations) {
  unsigned int i;
  struct expression *expr;
  expr = parse_file(file_name);
  for (i = 0; i < num_iterations; ++i) {
    run_test_belief_propagation(expr, file_name);
  }
  for (i = 0; i < num_iterations; ++i) {
    run_test_loopy_belief_propagation(expr, file_name);
  }
  delete_expression(expr);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<run_tests_with_file>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <parse_file@plt>
  mov    %rax,%r12
  test   %r14d,%r14d
  je     L22
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L2:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %ebx,%ebp
  call   <run_test_belief_propagation@plt>
  lea    0x1(%rbx),%ebx
  cmp    %ebx,%r14d
  jne    L2
  xor    %ebx,%ebx
  nop
L26:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <run_test_loopy_belief_propagation@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L26
L22:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <delete_expression@plt>
```

assistant: ```c
void run_tests_with_file(const char *file_name, unsigned int num_iterations) {
  unsigned int i;
  struct expression *expr;
  expr = parse_file(file_name);
  for (i = 0; i < num_iterations; ++i) {
    run_test_belief_propagation(expr, file_name);
  }
  for (i = 0; i < num_iterations; ++i) {
    run_test_loopy_belief_propagation(expr, file_name);
  }
  delete_expression(expr);
}
```
user: What is the c source code of the assembly code below?

```asm
<ltrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L12
L3:
  addq   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L3
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ltrim(char *s) {
  while (isspace(*s))
    s++;
  return s;
}
```
user: What is the c source code of the assembly code below?

```asm
<ltrim>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L11
L4:
  add    $0x1,%rbx
L11:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L4
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *ltrim(char *s) {
  while (isspace(*s))
    s++;
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ltrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  jmp    L12
  nopl   0x0(%rax,%rax,1)
L9:
  add    $0x1,%r12
L12:
  movsbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L9
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *ltrim(char *s) {
  while (isspace(*s))
    s++;
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ltrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L13:
  add    $0x1,%r12
L3:
  movsbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L13
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *ltrim(char *s) {
  while (isspace(*s))
    s++;
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L57
  cmpq   $0x0,-0x30(%rbp)
  jne    L60
L57:
  mov    $0x0,%eax
  jmp    L59
L58:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jl     L87
  movl   $0x0,-0x14(%rbp)
  jmp    L12
L87:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jge    L12
  movl   $0x1,-0x14(%rbp)
L12:
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x30(%rbp)
L60:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbx
  jle    L1
  mov    $0x1,%eax
  jmp    L59
L1:
  mov    -0x14(%rbp),%eax
L59:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  if (!str1 || !str2)
    return 0;
  int check;
  while (*str1) {
    if (*str1 >= *str2)
      check = 0;
    else if (*str1 < *str2)
      check = 1;
    *str1++;
    *str2++;
  }
  if (strlen(str1) > strlen(str2)) {
    return 1;
  }
  return check;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L10
L6:
  add    $0x1,%rdi
  mov    %rbx,%rcx
  add    $0x1,%rbx
  mov    %eax,%edx
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L6
  cmp    %dl,(%rcx)
  setg   %r12b
  movzbl %r12b,%r12d
L3:
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbp
  mov    $0x1,%eax
  cmovg  %eax,%r12d
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  mov    $0x0,%r12d
  jmp    L3
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  if (!str1 || !str2)
    return 0;
  int check;
  while (*str1) {
    if (*str1 >= *str2)
      check = 0;
    else if (*str1 < *str2)
      check = 1;
    *str1++;
    *str2++;
  }
  if (strlen(str1) > strlen(str2)) {
    return 1;
  }
  return check;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strcmp>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L7
  nopw   0x0(%rax,%rax,1)
L4:
  add    $0x1,%rdi
  mov    %eax,%edx
  mov    %rbp,%rcx
  add    $0x1,%rbp
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L4
  xor    %r12d,%r12d
  cmp    %dl,(%rcx)
  setg   %r12b
L8:
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  cmp    %rax,%rbx
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  cmovg  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  xor    %r12d,%r12d
  jmp    L8
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  if (!str1 || !str2)
    return 0;
  int check;
  while (*str1) {
    if (*str1 >= *str2)
      check = 0;
    else if (*str1 < *str2)
      check = 1;
    *str1++;
    *str2++;
  }
  if (strlen(str1) > strlen(str2)) {
    return 1;
  }
  return check;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcmp>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movzbl (%rdi),%eax
  test   %al,%al
  je     L28
  nopw   0x0(%rax,%rax,1)
L8:
  add    $0x1,%rdi
  mov    %eax,%edx
  mov    %rbp,%rcx
  add    $0x1,%rbp
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L8
  xor    %r12d,%r12d
  cmp    %dl,(%rcx)
  setg   %r12b
L9:
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  cmp    %rax,%rbx
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  cmovg  %eax,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L28:
  xor    %r12d,%r12d
  jmp    L9
```

assistant: ```c
int strcmp(char *str1, char *str2) {
  if (!str1 || !str2)
    return 0;
  int check;
  while (*str1) {
    if (*str1 >= *str2)
      check = 0;
    else if (*str1 < *str2)
      check = 1;
    *str1++;
    *str2++;
  }
  if (strlen(str1) > strlen(str2)) {
    return 1;
  }
  return check;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<affirmative>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fflush@plt>
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fflush@plt>
  nop
L56:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0x20,-0x8(%rbp)
  je     L56
  cmpl   $0x79,-0x8(%rbp)
  je     L17
  cmpl   $0x59,-0x8(%rbp)
  jne    L2
L17:
  mov    $0x1,%eax
  jmp    L58
L2:
  mov    $0x0,%eax
L58:
  mov    %eax,-0x4(%rbp)
  jmp    L47
L64:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
L47:
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L25
  cmpl   $0xa,-0x8(%rbp)
  jne    L64
L25:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int affirmative(void) {
  int c;
  int ok;
  fflush(stdout);
  fflush(stderr);
  while ((c = getchar()) == ' ') {
  }
  ok = (c == 'y' || c == 'Y');
  while (c != EOF && c != '\n')
    c = getchar();
  return ok;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<affirmative>:
  endbr64
  push   %rbx
  mov    D18(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  mov    D28(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
L20:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0x20,%eax
  je     L20
  mov    %eax,%edx
  and    $0xffffffdf,%edx
  cmp    $0x59,%edx
  sete   %bl
  movzbl %bl,%ebx
  mov    D29(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L13
  cmp    $0xa,%eax
  je     L13
L19:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D29(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L13
  cmp    $0xa,%eax
  jne    L19
L13:
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int affirmative(void) {
  int c;
  int ok;
  fflush(stdout);
  fflush(stderr);
  while ((c = getchar()) == ' ') {
  }
  ok = (c == 'y' || c == 'Y');
  while (c != EOF && c != '\n')
    c = getchar();
  return ok;
}
```
user: Translate the below assembly code into C programming language:

```asm
<affirmative>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  mov    D15(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x20,%eax
  je     L8
  mov    %eax,%edx
  mov    D5(%rip),%rbx
  xor    %r12d,%r12d
  and    $0xffffffdf,%edx
  cmp    $0x59,%edx
  sete   %r12b
  cmp    %eax,(%rbx)
  jne    L30
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  je     L24
L30:
  cmp    $0xa,%eax
  jne    L5
L24:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int affirmative(void) {
  int c;
  int ok;
  fflush(stdout);
  fflush(stderr);
  while ((c = getchar()) == ' ') {
  }
  ok = (c == 'y' || c == 'Y');
  while (c != EOF && c != '\n')
    c = getchar();
  return ok;
}
```