ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L10:
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  je     L5
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  je     L5
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <gcd@plt>
  mov    %eax,%ecx
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L10
L5:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D15
data_type: string
name: read_data
arguments: tag: D21
data_type: string
name: read_data
arguments: tag: D36
data_type: string
name: read_data
arguments: tag: D13
data_type: string

tool: D15: Example of Greatest Common Divisor Recursive Function


tool: D21: Enter two values (0 to stop): 

tool: D36: %ld %ld

tool: D13: The greatest common divisor of %ld and %ld is %ld.



assistant: ```c
int main() {
  long num1, num2;
  printf("Example of Greatest Common Divisor "
         "Recursive Function\n\n");
  do {
    printf("Enter two values (0 to stop): ");
    scanf("%ld %ld", &num1, &num2);
    if (num1 == 0L || num2 == 0L)
      break;
    printf("The greatest common divisor of "
           "%ld and %ld is %ld.\n\n",
           num1, num2, gcd(num1, num2));
  } while (1);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <puts@plt>
  lea    D24(%rip),%r12
  lea    0x10(%rsp),%rbp
  lea    D39(%rip),%rbx
  lea    D38(%rip),%r13
  jmp    L2
L25:
  call   <gcd@plt>
  mov    %eax,%ecx
  mov    0x10(%rsp),%rdx
  mov    0x8(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L2:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rdi
  test   %rdi,%rdi
  je     L36
  mov    0x10(%rsp),%rsi
  test   %rsi,%rsi
  jne    L25
L36:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D30
type: string
name: extract_symbolic_data
arguments: data_tag: D24
type: string
name: extract_symbolic_data
arguments: data_tag: D39
type: string
name: extract_symbolic_data
arguments: data_tag: D38
type: string

tool: D30: Example of Greatest Common Divisor Recursive Function


tool: D24: Enter two values (0 to stop): 

tool: D39: %ld %ld

tool: D38: The greatest common divisor of %ld and %ld is %ld.



assistant: ```c
int main() {
  long num1, num2;
  printf("Example of Greatest Common Divisor "
         "Recursive Function\n\n");
  do {
    printf("Enter two values (0 to stop): ");
    scanf("%ld %ld", &num1, &num2);
    if (num1 == 0L || num2 == 0L)
      break;
    printf("The greatest common divisor of "
           "%ld and %ld is %ld.\n\n",
           num1, num2, gcd(num1, num2));
  } while (1);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r14
  lea    D13(%rip),%rdi
  lea    D7(%rip),%r14
  push   %r13
  lea    D34(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%rbp
  call   <puts@plt>
  jmp    L25
  nopl   (%rax)
L15:
  mov    0x10(%rsp),%rsi
  test   %rsi,%rsi
  je     L14
  call   <gcd@plt>
  mov    0x10(%rsp),%rdx
  mov    0x8(%rsp),%rsi
  mov    %r14,%rdi
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
L25:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rdi
  test   %rdi,%rdi
  jne    L15
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D13
data_type: string
name: disassemble_data
arguments: tag: D7
data_type: string
name: disassemble_data
arguments: tag: D34
data_type: string
name: disassemble_data
arguments: tag: D1
data_type: string

tool: D13: Example of Greatest Common Divisor Recursive Function


tool: D7: The greatest common divisor of %ld and %ld is %ld.



tool: D34: Enter two values (0 to stop): 

tool: D1: %ld %ld

assistant: ```c
int main() {
  long num1, num2;
  printf("Example of Greatest Common Divisor "
         "Recursive Function\n\n");
  do {
    printf("Enter two values (0 to stop): ");
    scanf("%ld %ld", &num1, &num2);
    if (num1 == 0L || num2 == 0L)
      break;
    printf("The greatest common divisor of "
           "%ld and %ld is %ld.\n\n",
           num1, num2, gcd(num1, num2));
  } while (1);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  lea    D22(%rip),%rdi
  lea    D30(%rip),%r14
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D26(%rip),%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%rbp
  call   <puts@plt>
  jmp    L21
  nopl   (%rax)
L12:
  mov    0x10(%rsp),%rsi
  test   %rsi,%rsi
  je     L19
  call   <gcd@plt>
  mov    0x10(%rsp),%rdx
  mov    0x8(%rsp),%rsi
  mov    %r14,%rdi
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
L21:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rdi
  test   %rdi,%rdi
  jne    L12
L19:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D22
data_type: string
name: extract_symbolic_data
arguments: label: D30
data_type: string
name: extract_symbolic_data
arguments: label: D2
data_type: string
name: extract_symbolic_data
arguments: label: D26
data_type: string

tool: D22: Example of Greatest Common Divisor Recursive Function


tool: D30: The greatest common divisor of %ld and %ld is %ld.



tool: D2: Enter two values (0 to stop): 

tool: D26: %ld %ld

assistant: ```c
int main() {
  long num1, num2;
  printf("Example of Greatest Common Divisor "
         "Recursive Function\n\n");
  do {
    printf("Enter two values (0 to stop): ");
    scanf("%ld %ld", &num1, &num2);
    if (num1 == 0L || num2 == 0L)
      break;
    printf("The greatest common divisor of "
           "%ld and %ld is %ld.\n\n",
           num1, num2, gcd(num1, num2));
  } while (1);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<vaasprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <va_copy@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  mov    %rax,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  cmpl   $0x0,-0xc(%rbp)
  jns    L19
  mov    $0x0,%eax
  jmp    L7
L19:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L42
  mov    $0x0,%eax
  jmp    L7
L42:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L43
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L7
L43:
  mov    -0x8(%rbp),%rax
L7:
  leave
  ret
```

assistant: ```c
char *vaasprintf(char const *const fmt, va_list ap) {
  va_list ap2;
  va_copy(ap2, ap);
  int rc = vsnprintf(NULL, 0, fmt, ap2);
  va_end(ap2);
  if (rc < 0)
    return NULL;
  char *str = malloc(rc + 1);
  if (!str)
    return NULL;
  rc = vsnprintf(str, rc + 1, fmt, ap);
  if (rc < 0) {
    free(str);
    return NULL;
  }
  return str;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vaasprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_copy@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0x0,%ebp
  test   %ebx,%ebx
  js     L6
  add    $0x1,%ebx
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L6
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    %ebx,%esi
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  test   %eax,%eax
  js     L26
L6:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x0,%ebp
  jmp    L6
```

assistant: ```c
char *vaasprintf(char const *const fmt, va_list ap) {
  va_list ap2;
  va_copy(ap2, ap);
  int rc = vsnprintf(NULL, 0, fmt, ap2);
  va_end(ap2);
  if (rc < 0)
    return NULL;
  char *str = malloc(rc + 1);
  if (!str)
    return NULL;
  rc = vsnprintf(str, rc + 1, fmt, ap);
  if (rc < 0) {
    free(str);
    return NULL;
  }
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vaasprintf>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  call   <va_copy@plt>
  xor    %edi,%edi
  xor    %esi,%esi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <va_end@plt>
  test   %ebx,%ebx
  js     L5
  lea    0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L5
  mov    %r14d,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  test   %eax,%eax
  js     L19
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %r12,%rdi
  call   <free@plt>
L5:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *vaasprintf(char const *const fmt, va_list ap) {
  va_list ap2;
  va_copy(ap2, ap);
  int rc = vsnprintf(NULL, 0, fmt, ap2);
  va_end(ap2);
  if (rc < 0)
    return NULL;
  char *str = malloc(rc + 1);
  if (!str)
    return NULL;
  rc = vsnprintf(str, rc + 1, fmt, ap);
  if (rc < 0) {
    free(str);
    return NULL;
  }
  return str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<vaasprintf>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %ebp,%edi
  call   <va_copy@plt>
  xor    %edi,%edi
  xor    %esi,%esi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <va_end@plt>
  test   %ebx,%ebx
  js     L24
  lea    0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L24
  mov    %r14d,%ecx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  test   %eax,%eax
  js     L14
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %r12,%rdi
  call   <free@plt>
L24:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *vaasprintf(char const *const fmt, va_list ap) {
  va_list ap2;
  va_copy(ap2, ap);
  int rc = vsnprintf(NULL, 0, fmt, ap2);
  va_end(ap2);
  if (rc < 0)
    return NULL;
  char *str = malloc(rc + 1);
  if (!str)
    return NULL;
  rc = vsnprintf(str, rc + 1, fmt, ap);
  if (rc < 0) {
    free(str);
    return NULL;
  }
  return str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type: string

tool: D4: %d

assistant: ```c
int getint() {
  int tmp;
  scanf("%d", &tmp);
  return tmp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getint>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D9
data_type: string

tool: D9: %d

assistant: ```c
int getint() {
  int tmp;
  scanf("%d", &tmp);
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getint>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
datatype: string

tool: D5: %d

assistant: ```c
int getint() {
  int tmp;
  scanf("%d", &tmp);
  return tmp;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getint>:
  endbr64
  sub    $0x18,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: string

tool: D9: %d

assistant: ```c
int getint() {
  int tmp;
  scanf("%d", &tmp);
  return tmp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<drawFace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <glBegin@plt>
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <glNormal3i@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  mov    $0x0,%eax
  call   <glEnd@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void drawFace(void) {
  glBegin(GL_QUADS);
  glNormal3i(0, 0, 1);
  glVertex3f(1, 1, 1);
  glVertex3f(-1, 1, 1);
  glVertex3f(-1, -1, 1);
  glVertex3f(1, -1, 1);
  glEnd();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<drawFace>:
  endbr64
  sub    $0x8,%rsp
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <glBegin@plt>
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <glNormal3i@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  mov    $0x0,%eax
  call   <glEnd@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void drawFace(void) {
  glBegin(GL_QUADS);
  glNormal3i(0, 0, 1);
  glVertex3f(1, 1, 1);
  glVertex3f(-1, 1, 1);
  glVertex3f(-1, -1, 1);
  glVertex3f(1, -1, 1);
  glEnd();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<drawFace>:
  endbr64
  sub    $0x8,%rsp
  mov    D2(%rip),%rax
  mov    (%rax),%edi
  call   <glBegin@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <glNormal3i@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <glEnd@plt>
```

assistant: ```c
void drawFace(void) {
  glBegin(GL_QUADS);
  glNormal3i(0, 0, 1);
  glVertex3f(1, 1, 1);
  glVertex3f(-1, 1, 1);
  glVertex3f(-1, -1, 1);
  glVertex3f(1, -1, 1);
  glEnd();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<drawFace>:
  endbr64
  sub    $0x8,%rsp
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <glBegin@plt>
  mov    $0x1,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <glNormal3i@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0xffffffff,%edi
  call   <glVertex3f@plt>
  mov    $0x1,%edx
  mov    $0xffffffff,%esi
  mov    $0x1,%edi
  call   <glVertex3f@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <glEnd@plt>
```

assistant: ```c
void drawFace(void) {
  glBegin(GL_QUADS);
  glNormal3i(0, 0, 1);
  glVertex3f(1, 1, 1);
  glVertex3f(-1, 1, 1);
  glVertex3f(-1, -1, 1);
  glVertex3f(1, -1, 1);
  glEnd();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<degfcorr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %edi,-0x94(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x90(%rbp)
  movsd  D105(%rip),%xmm0
  movsd  %xmm0,-0x88(%rbp)
  movsd  D68(%rip),%xmm0
  movsd  %xmm0,-0x80(%rbp)
  movsd  D31(%rip),%xmm0
  movsd  %xmm0,-0x78(%rbp)
  movsd  D71(%rip),%xmm0
  movsd  %xmm0,-0x70(%rbp)
  movsd  D154(%rip),%xmm0
  movsd  %xmm0,-0x68(%rbp)
  movsd  D0(%rip),%xmm0
  movsd  %xmm0,-0x60(%rbp)
  movsd  D136(%rip),%xmm0
  movsd  %xmm0,-0x58(%rbp)
  movsd  D11(%rip),%xmm0
  movsd  %xmm0,-0x50(%rbp)
  movsd  D34(%rip),%xmm0
  movsd  %xmm0,-0x48(%rbp)
  movsd  D39(%rip),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  D54(%rip),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  D72(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  D35(%rip),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  D4(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D144(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  cmpl   $0x0,-0x94(%rbp)
  jle    L18
  cmpl   $0xf,-0x94(%rbp)
  jg     L18
  mov    -0x94(%rbp),%eax
  cltq
  movsd  -0x90(%rbp,%rax,8),%xmm0
  jmp    L20
L18:
  movsd  D144(%rip),%xmm0
L20:
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D105
data_type: f64
name: unpack_data
arguments: data_tag: D68
data_type: f64
name: unpack_data
arguments: data_tag: D31
data_type: f64
name: unpack_data
arguments: data_tag: D71
data_type: f64
name: unpack_data
arguments: data_tag: D154
data_type: f64
name: unpack_data
arguments: data_tag: D0
data_type: f64
name: unpack_data
arguments: data_tag: D136
data_type: f64
name: unpack_data
arguments: data_tag: D11
data_type: f64
name: unpack_data
arguments: data_tag: D34
data_type: f64
name: unpack_data
arguments: data_tag: D39
data_type: f64
name: unpack_data
arguments: data_tag: D54
data_type: f64
name: unpack_data
arguments: data_tag: D72
data_type: f64
name: unpack_data
arguments: data_tag: D35
data_type: f64
name: unpack_data
arguments: data_tag: D4
data_type: f64
name: unpack_data
arguments: data_tag: D144
data_type: f64

tool: D105: 5e-05

tool: D68: 9e-05

tool: D31: 0.00012

tool: D71: 0.00015

tool: D154: 0.0002

tool: D0: 0.0003

tool: D136: 0.00045

tool: D11: 0.0006

tool: D34: 0.0009

tool: D39: 0.0015

tool: D54: 0.0021

tool: D72: 0.0027

tool: D35: 0.0033

tool: D4: 0.0046

tool: D144: 0.0058

assistant: ```c
double degfcorr(int ai) {
  const double degf[16] = {0.00000, 0.00005, 0.00009, 0.00012, 0.00015, 0.00020, 0.00030, 0.00045, 0.00060, 0.00090, 0.00150, 0.00210, 0.00270, 0.00330, 0.00460, 0.00580};
  return 0 < ai && ai <= 15 ? degf[ai] : 0.0058;
}
```