ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<reverse_my>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L18
  mov    $0x1,%r12d
  lea    D11(%rip),%r13
  sub    %edi,%r12d
  nopl   0x0(%rax)
L15:
  mov    %r15,%rbx
  mov    %r13,%rdi
  lea    (%r12,%r15,1),%ebp
  xor    %eax,%eax
  call   <printf@plt>
  movsbl 0x1(%rbx),%esi
  lea    0x1(%r15),%r15
  test   %sil,%sil
  jne    L15
  movsbl -0x1(%r15),%esi
  lea    -0x1(%rbp),%r12d
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  not    %r12
  mov    %r14,%r15
  lea    D21(%rip),%r13
  mov    %esi,%edx
  add    %rbx,%r12
  call   <printf@plt>
  nop
L2:
  movsbl (%rbx),%esi
  sub    $0x1,%rbx
  xor    %eax,%eax
  mov    %r13,%rdi
  add    $0x1,%r15
  mov    %sil,-0x1(%r15)
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L2
  movslq %ebp,%rbp
  add    %rbp,%r14
L37:
  movb   $0x0,(%r14)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  movsbl -0x1(%rdi),%esi
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  mov    %esi,%edx
  call   <printf@plt>
  jmp    L37
```

tool_calls: name: unpack_data
arguments: data_tag: D11
data_type: string
name: unpack_data
arguments: data_tag: D28
data_type: string
name: unpack_data
arguments: data_tag: D21
data_type: string

tool: D11: int first while * source = %c;


tool: D28: before second while cycle * source = "%c" in char and %d in digit;


tool: D21: in second while * rev = %c;


assistant: ```c
void reverse_my(char *source, char *rev) {
  int i = 0;
  while (*source != '\0') {
    i++;
    printf("int first while * source = %c;\n", *source);
    *source++;
  };
  *source--;
  printf("before second while cycle * source = \"%c\" in char and %d in digit;\n", *source, *source);
  while (i > 0) {
    *rev++ = *source--;
    i--;
    *rev--;
    printf("in second while * rev = %c;\n", *rev);
    *rev++;
  };
  *rev = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<reverse_my>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L11
  mov    $0x1,%r12d
  lea    D31(%rip),%r13
  sub    %edi,%r12d
  nopl   0x0(%rax)
L3:
  mov    %r15,%rbx
  mov    %r13,%rdi
  lea    (%r12,%r15,1),%ebp
  xor    %eax,%eax
  call   <printf@plt>
  movsbl 0x1(%rbx),%esi
  lea    0x1(%r15),%r15
  test   %sil,%sil
  jne    L3
  movsbl -0x1(%r15),%esi
  lea    -0x1(%rbp),%r12d
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  not    %r12
  mov    %r14,%r15
  lea    D34(%rip),%r13
  mov    %esi,%edx
  add    %rbx,%r12
  call   <printf@plt>
  nop
L39:
  movsbl (%rbx),%esi
  sub    $0x1,%rbx
  xor    %eax,%eax
  mov    %r13,%rdi
  add    $0x1,%r15
  mov    %sil,-0x1(%r15)
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L39
  movslq %ebp,%rbp
  add    %rbp,%r14
L14:
  movb   $0x0,(%r14)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  movsbl -0x1(%rdi),%esi
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  mov    %esi,%edx
  call   <printf@plt>
  jmp    L14
```

tool_calls: name: retrieve_data
arguments: data_label: D31
datatype: string
name: retrieve_data
arguments: data_label: D24
datatype: string
name: retrieve_data
arguments: data_label: D34
datatype: string

tool: D31: int first while * source = %c;


tool: D24: before second while cycle * source = "%c" in char and %d in digit;


tool: D34: in second while * rev = %c;


assistant: ```c
void reverse_my(char *source, char *rev) {
  int i = 0;
  while (*source != '\0') {
    i++;
    printf("int first while * source = %c;\n", *source);
    *source++;
  };
  *source--;
  printf("before second while cycle * source = \"%c\" in char and %d in digit;\n", *source, *source);
  while (i > 0) {
    *rev++ = *source--;
    i--;
    *rev--;
    printf("in second while * rev = %c;\n", *rev);
    *rev++;
  };
  *rev = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  movl   $0x2,-0x14(%rbp)
L6:
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L80
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L79
L80:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  je     L6
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L6
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  jmp    L22
L57:
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  js     L22
  addl   $0x1,-0x10(%rbp)
L22:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jg     L57
  mov    -0x14(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L6
L79:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type: string
name: parse_data
arguments: data_tag: D9
type: string

tool: D7: %i %i

tool: D9: %i %i / %i


assistant: ```c
void main() {
  int n1 = 1;
  int n2 = 2;
  while (1) {
    int valorI = 0;
    int resto = 0;
    scanf("%i %i", &n1, &n2);
    if ((n1 == 0) && (n2 == 0))
      return;
    if (n1 != 0 && n2 != 0) {
      resto = n1 % n2;
      while (n1 > 0) {
        n1 -= n2;
        if (n1 >= 0)
          valorI++;
      }
      printf("%i %i / %i\n", valorI, resto, n2);
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```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
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  lea    D2(%rip),%rbx
  lea    D15(%rip),%r13
  jmp    L57
L69:
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jne    L51
L57:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jne    L69
  cmpl   $0x0,0x4(%rsp)
  jne    L57
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L51:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %ecx,%ecx
  jle    L46
  mov    $0x0,%eax
L79:
  sub    %esi,%ecx
  cmp    $0x80000000,%ecx
  adc    $0x0,%eax
  test   %ecx,%ecx
  jg     L79
  mov    %ecx,(%rsp)
L60:
  mov    %esi,%ecx
  mov    %eax,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L57
L46:
  mov    $0x0,%eax
  jmp    L60
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
type_guessed: string
name: disassemble_data
arguments: label: D15
type_guessed: string

tool: D2: %i %i

tool: D15: %i %i / %i


assistant: ```c
void main() {
  int n1 = 1;
  int n2 = 2;
  while (1) {
    int valorI = 0;
    int resto = 0;
    scanf("%i %i", &n1, &n2);
    if ((n1 == 0) && (n2 == 0))
      return;
    if (n1 != 0 && n2 != 0) {
      resto = n1 % n2;
      while (n1 > 0) {
        n1 -= n2;
        if (n1 >= 0)
          valorI++;
      }
      printf("%i %i / %i\n", valorI, resto, n2);
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D17(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  movl   $0x2,0x4(%rsp)
  nopl   0x0(%rax,%rax,1)
L33:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jne    L18
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L33
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L33
  mov    %ecx,%eax
  xor    %esi,%esi
  cltd
  idiv   %r8d
  test   %ecx,%ecx
  jg     L86
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L77:
  add    $0x1,%esi
  test   %ecx,%ecx
  jle    L20
L86:
  sub    %r8d,%ecx
  jns    L77
L20:
  mov    %ecx,(%rsp)
L28:
  mov    %r8d,%ecx
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L33
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
data_type: string
name: extract_symbolic_data
arguments: data_tag: D17
data_type: string

tool: D14: %i %i / %i


tool: D17: %i %i

assistant: ```c
void main() {
  int n1 = 1;
  int n2 = 2;
  while (1) {
    int valorI = 0;
    int resto = 0;
    scanf("%i %i", &n1, &n2);
    if ((n1 == 0) && (n2 == 0))
      return;
    if (n1 != 0 && n2 != 0) {
      resto = n1 % n2;
      while (n1 > 0) {
        n1 -= n2;
        if (n1 >= 0)
          valorI++;
      }
      printf("%i %i / %i\n", valorI, resto, n2);
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  movl   $0x2,0x4(%rsp)
  nopl   0x0(%rax,%rax,1)
L74:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jne    L18
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L74
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  mov    0x4(%rsp),%r8d
  test   %r8d,%r8d
  je     L74
  mov    %ecx,%eax
  xor    %esi,%esi
  cltd
  idiv   %r8d
  test   %ecx,%ecx
  jg     L89
  jmp    L61
  nopl   0x0(%rax,%rax,1)
L14:
  add    $0x1,%esi
  test   %ecx,%ecx
  jle    L29
L89:
  sub    %r8d,%ecx
  jns    L14
L29:
  mov    %ecx,(%rsp)
L61:
  mov    %r8d,%ecx
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L74
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string
name: parse_data
arguments: data_tag: D11
type: string

tool: D0: %i %i / %i


tool: D11: %i %i

assistant: ```c
void main() {
  int n1 = 1;
  int n2 = 2;
  while (1) {
    int valorI = 0;
    int resto = 0;
    scanf("%i %i", &n1, &n2);
    if ((n1 == 0) && (n2 == 0))
      return;
    if (n1 != 0 && n2 != 0) {
      resto = n1 % n2;
      while (n1 > 0) {
        n1 -= n2;
        if (n1 >= 0)
          valorI++;
      }
      printf("%i %i / %i\n", valorI, resto, n2);
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
L19:
  cmpl   $0x4,-0x4(%rbp)
  jg     L23
  addl   $0x1,-0x4(%rbp)
  jmp    L19
L23:
  cmpl   $0x4,-0x4(%rbp)
  jne    L21
  nop
L39:
  jmp    L39
L21:
  nop
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int counter = 0;
LOOPSTART:
  if (counter < 5) {
    counter++;
    goto LOOPSTART;
  }
  if (counter == 4) {
    goto ERROR;
  } else {
    goto END;
  }
ERROR:
  goto ERROR;
END:
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int counter = 0;
LOOPSTART:
  if (counter < 5) {
    counter++;
    goto LOOPSTART;
  }
  if (counter == 4) {
    goto ERROR;
  } else {
    goto END;
  }
ERROR:
  goto ERROR;
END:
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int counter = 0;
LOOPSTART:
  if (counter < 5) {
    counter++;
    goto LOOPSTART;
  }
  if (counter == 4) {
    goto ERROR;
  } else {
    goto END;
  }
ERROR:
  goto ERROR;
END:
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strndup_1_Bad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x6e49206f6c6c6548,%rax
  mov    %rax,-0x15(%rbp)
  movl   $0x21726566,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  movl   $0x64,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  lea    -0x15(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strndup@plt>
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L2
L18:
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x61,(%rax)
  addl   $0x1,-0x28(%rbp)
L2:
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L18
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
void strndup_1_Bad() {
  char s1[] = "Hello Infer!";
  int len = 100;
  char *s2 = strndup(s1, len);
  for (int i = 0; i < len; i++) {
    s2[i] = 'a';
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strndup_1_Bad>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x6e49206f6c6c6548,%rax
  mov    %rax,0xb(%rsp)
  movl   $0x21726566,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  lea    0xb(%rsp),%rdi
  mov    $0x64,%esi
  call   <strndup@plt>
  lea    0x64(%rax),%rdx
L11:
  movb   $0x61,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L11
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x28,%rsp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strndup_1_Bad() {
  char s1[] = "Hello Infer!";
  int len = 100;
  char *s2 = strndup(s1, len);
  for (int i = 0; i < len; i++) {
    s2[i] = 'a';
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strndup_1_Bad>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x64,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x6e49206f6c6c6548,%rax
  lea    0xb(%rsp),%rdi
  movl   $0x21726566,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    %rax,0xb(%rsp)
  call   <strndup@plt>
  movdqa D9(%rip),%xmm0
  movl   $0x61616161,0x60(%rax)
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  movups %xmm0,0x30(%rax)
  movups %xmm0,0x40(%rax)
  movups %xmm0,0x50(%rax)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
datatype: byte[16]

tool: {"D9": [0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61]}

assistant: ```c
void strndup_1_Bad() {
  char s1[] = "Hello Infer!";
  int len = 100;
  char *s2 = strndup(s1, len);
  for (int i = 0; i < len; i++) {
    s2[i] = 'a';
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<strndup_1_Bad>:
  endbr64
  sub    $0x28,%rsp
  mov    $0x64,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x6e49206f6c6c6548,%rax
  lea    0xb(%rsp),%rdi
  movl   $0x21726566,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    %rax,0xb(%rsp)
  call   <strndup@plt>
  movdqa D0(%rip),%xmm0
  movl   $0x61616161,0x60(%rax)
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  movups %xmm0,0x30(%rax)
  movups %xmm0,0x40(%rax)
  movups %xmm0,0x50(%rax)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D0
type: byte[16]

tool: {"D0": [0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61,0x61]}

assistant: ```c
void strndup_1_Bad() {
  char s1[] = "Hello Infer!";
  int len = 100;
  char *s2 = strndup(s1, len);
  for (int i = 0; i < len; i++) {
    s2[i] = 'a';
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_result_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movsbl -0x4(%rbp),%eax
  cmp    $0x78,%eax
  je     L12
  cmp    $0x78,%eax
  jg     L29
  cmp    $0x3d,%eax
  je     L34
  cmp    $0x6f,%eax
  jne    L29
  lea    D29(%rip),%rax
  jmp    L66
L12:
  lea    D7(%rip),%rax
  jmp    L66
L34:
  lea    D17(%rip),%rax
  jmp    L66
L29:
  nop
  lea    D0(%rip),%rax
L66:
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D29
type_infered: string
name: retrieve_data
arguments: tag: D7
type_infered: string
name: retrieve_data
arguments: tag: D17
type_infered: string
name: retrieve_data
arguments: tag: D0
type_infered: string

tool: D29: FRIEND

tool: D7: ENEMY

tool: D17: DRAW

tool: 

assistant: ```c
char *get_result_str(char result) {
  switch (result) {
  case 'o':
    return "FRIEND";
    break;
  case 'x':
    return "ENEMY";
    break;
  case '=':
    return "DRAW";
    break;
  default:
    break;
  }
  return "";
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_result_str>:
  endbr64
  lea    D13(%rip),%rax
  cmp    $0x6f,%dil
  je     L18
  lea    D25(%rip),%rax
  cmp    $0x78,%dil
  je     L18
  cmp    $0x3d,%dil
  lea    D33(%rip),%rax
  lea    D23(%rip),%rdx
  cmove  %rdx,%rax
L18:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
data_type: string
name: extract_symbolic_data
arguments: tag: D25
data_type: string
name: extract_symbolic_data
arguments: tag: D33
data_type: string
name: extract_symbolic_data
arguments: tag: D23
data_type: string

tool: D13: FRIEND

tool: D25: ENEMY

tool: 

tool: D23: DRAW

assistant: ```c
char *get_result_str(char result) {
  switch (result) {
  case 'o':
    return "FRIEND";
    break;
  case 'x':
    return "ENEMY";
    break;
  case '=':
    return "DRAW";
    break;
  default:
    break;
  }
  return "";
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_result_str>:
  endbr64
  lea    D12(%rip),%rax
  cmp    $0x6f,%dil
  je     L14
  lea    D9(%rip),%rax
  cmp    $0x78,%dil
  je     L14
  cmp    $0x3d,%dil
  lea    D11(%rip),%rax
  lea    D1(%rip),%rdx
  cmove  %rdx,%rax
L14:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
data_type: string
name: disassemble_data
arguments: data_tag: D9
data_type: string
name: disassemble_data
arguments: data_tag: D11
data_type: string
name: disassemble_data
arguments: data_tag: D1
data_type: string

tool: D12: FRIEND

tool: D9: ENEMY

tool: 

tool: D1: DRAW

assistant: ```c
char *get_result_str(char result) {
  switch (result) {
  case 'o':
    return "FRIEND";
    break;
  case 'x':
    return "ENEMY";
    break;
  case '=':
    return "DRAW";
    break;
  default:
    break;
  }
  return "";
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_result_str>:
  endbr64
  lea    D1(%rip),%rax
  cmp    $0x6f,%dil
  je     L10
  lea    D20(%rip),%rax
  cmp    $0x78,%dil
  je     L10
  cmp    $0x3d,%dil
  lea    D27(%rip),%rax
  lea    D31(%rip),%rdx
  cmove  %rdx,%rax
L10:
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
datatype: string
name: read_data
arguments: data_tag: D20
datatype: string
name: read_data
arguments: data_tag: D27
datatype: string
name: read_data
arguments: data_tag: D31
datatype: string

tool: D1: FRIEND

tool: D20: ENEMY

tool: 

tool: D31: DRAW

assistant: ```c
char *get_result_str(char result) {
  switch (result) {
  case 'o':
    return "FRIEND";
    break;
  case 'x':
    return "ENEMY";
    break;
  case '=':
    return "DRAW";
    break;
  default:
    break;
  }
  return "";
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_prog>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L31:
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <access@plt>
  test   %rax,%rax
  jne    L15
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  jmp    L32
L15:
  addl   $0x1,-0x4(%rbp)
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L31
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
L32:
  leave
  ret
```

assistant: ```c
char *get_prog(char *progs[]) {
  int i;
  for (i = 0; progs[i] != NULL; i++)
    if (access(progs[i], X_OK) == 0)
      return progs[i];
  return progs[i - 1];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<get_prog>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L43
  mov    %r13,%rbx
  mov    D4(%rip),%r12
  mov    $0x8,%ebp
  sub    %r13,%rbp
L21:
  mov    (%r12),%esi
  call   <access@plt>
  test   %rax,%rax
  je     L8
  lea    0x0(%rbp,%rbx,1),%rax
  add    $0x8,%rbx
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  jne    L21
L23:
  mov    -0x8(%r13,%rax,1),%rax
L40:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    (%rbx),%rax
  jmp    L40
L43:
  mov    $0x0,%eax
  jmp    L23
```

assistant: ```c
char *get_prog(char *progs[]) {
  int i;
  for (i = 0; progs[i] != NULL; i++)
    if (access(progs[i], X_OK) == 0)
      return progs[i];
  return progs[i - 1];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_prog>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L13
  mov    D7(%rip),%r12
  mov    %rbp,%rbx
  mov    $0x8,%r13d
  jmp    L43
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %r13,%rax
  sub    %rbp,%rax
  add    %rbx,%rax
  mov    0x8(%rbx),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  je     L36
L43:
  mov    (%r12),%esi
  call   <access@plt>
  test   %rax,%rax
  jne    L24
  mov    (%rbx),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L36:
  sub    $0x8,%rax
L11:
  mov    0x0(%rbp,%rax,1),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  mov    $0xfffffffffffffff8,%rax
  jmp    L11
```

assistant: ```c
char *get_prog(char *progs[]) {
  int i;
  for (i = 0; progs[i] != NULL; i++)
    if (access(progs[i], X_OK) == 0)
      return progs[i];
  return progs[i - 1];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_prog>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L46
  mov    D1(%rip),%r12
  mov    %rbp,%rbx
  mov    $0x8,%r13d
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L43:
  mov    %r13,%rax
  sub    %rbp,%rax
  add    %rbx,%rax
  mov    0x8(%rbx),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  je     L12
L11:
  mov    (%r12),%esi
  call   <access@plt>
  test   %rax,%rax
  jne    L43
  mov    (%rbx),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  sub    $0x8,%rax
L47:
  mov    0x0(%rbp,%rax,1),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  mov    $0xfffffffffffffff8,%rax
  jmp    L47
```

assistant: ```c
char *get_prog(char *progs[]) {
  int i;
  for (i = 0; progs[i] != NULL; i++)
    if (access(progs[i], X_OK) == 0)
      return progs[i];
  return progs[i - 1];
}
```