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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
datatype: string
name: unpack_data
arguments: data_label: D8
datatype: string

tool: D5: a

tool: D8: b

assistant: ```c
int main(void) {
#pragma critTer : 4 :../ test / resources / switch.c:
  char x = 'a';
#pragma critTer : 5 :../ test / resources / switch.c:
#pragma critTer : 6 :../ test / resources / switch.c:
  switch (x) {
#pragma critTer : 7 :../ test / resources / switch.c:
  case 'a':
#pragma critTer : 8 :../ test / resources / switch.c:
    printf("a");
#pragma critTer : 9 :../ test / resources / switch.c:
  case 'b':
#pragma critTer : 10 :../ test / resources / switch.c:
    printf("b");
#pragma critTer : 11 :../ test / resources / switch.c:
    break;
#pragma critTer : 12 :../ test / resources / switch.c:
  }
#pragma critTer : 13 :../ test / resources / switch.c:
#pragma critTer : 14 :../ test / resources / switch.c:
  return 0;
#pragma critTer : 15 :../ test / resources / switch.c:
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x64,-0x3(%rbp)
  movb   $0x96,-0x2(%rbp)
  movzbl -0x3(%rbp),%edx
  movzbl -0x2(%rbp),%eax
  add    %edx,%eax
  mov    %al,-0x1(%rbp)
  movsbl -0x1(%rbp),%ecx
  movsbl -0x2(%rbp),%edx
  movsbl -0x3(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D1: %d + %d = %d

assistant: ```c
int main() {
  char first = 100;
  char second = 150;
  char result = first + second;
  printf("%d + %d = %d", first, second, result);
}
```
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
  sub    $0x8,%rsp
  mov    $0xfffffffa,%ecx
  mov    $0xffffff96,%edx
  mov    $0x64,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
data_type: string

tool: D3: %d + %d = %d

assistant: ```c
int main() {
  char first = 100;
  char second = 150;
  char result = first + second;
  printf("%d + %d = %d", first, second, result);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xfffffffa,%ecx
  mov    $0xffffff96,%edx
  mov    $0x64,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d + %d = %d

assistant: ```c
int main() {
  char first = 100;
  char second = 150;
  char result = first + second;
  printf("%d + %d = %d", first, second, result);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xfffffffa,%ecx
  mov    $0xffffff96,%edx
  mov    $0x64,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type: string

tool: D9: %d + %d = %d

assistant: ```c
int main() {
  char first = 100;
  char second = 150;
  char result = first + second;
  printf("%d + %d = %d", first, second, result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getX86XCR0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  mov    %eax,%ecx
  xgetbv
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int getX86XCR0(unsigned *rEAX, unsigned *rEDX) {
  __asm__(".byte 0x0f, 0x01, 0xd0" : "=a"(*rEAX), "=d"(*rEDX) : "c"(0));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getX86XCR0>:
  endbr64
  mov    $0x0,%ecx
  xgetbv
  mov    %eax,(%rdi)
  mov    %edx,(%rsi)
  mov    $0x0,%eax
  ret
```

assistant: ```c
int getX86XCR0(unsigned *rEAX, unsigned *rEDX) {
  __asm__(".byte 0x0f, 0x01, 0xd0" : "=a"(*rEAX), "=d"(*rEDX) : "c"(0));
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getX86XCR0>:
  endbr64
  xor    %ecx,%ecx
  xgetbv
  mov    %eax,(%rdi)
  xor    %eax,%eax
  mov    %edx,(%rsi)
  ret
```

assistant: ```c
int getX86XCR0(unsigned *rEAX, unsigned *rEDX) {
  __asm__(".byte 0x0f, 0x01, 0xd0" : "=a"(*rEAX), "=d"(*rEDX) : "c"(0));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<getX86XCR0>:
  endbr64
  xor    %ecx,%ecx
  xgetbv
  mov    %eax,(%rdi)
  xor    %eax,%eax
  mov    %edx,(%rsi)
  ret
```

assistant: ```c
int getX86XCR0(unsigned *rEAX, unsigned *rEDX) {
  __asm__(".byte 0x0f, 0x01, 0xd0" : "=a"(*rEAX), "=d"(*rEDX) : "c"(0));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<queue_trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L17
  mov    $0xffffffff,%eax
  jmp    L17
L17:
  pop    %rbp
  ret
```

assistant: ```c
int queue_trim(struct st_queue *qu) {
  if (NULL == qu) {
    return -1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<queue_trim>:
  endbr64
  test   %rdi,%rdi
  je     L0
  ret
L0:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int queue_trim(struct st_queue *qu) {
  if (NULL == qu) {
    return -1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<queue_trim>:
  endbr64
  test   %rdi,%rdi
  je     L7
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int queue_trim(struct st_queue *qu) {
  if (NULL == qu) {
    return -1;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<queue_trim>:
  endbr64
  test   %rdi,%rdi
  je     L4
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int queue_trim(struct st_queue *qu) {
  if (NULL == qu) {
    return -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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  lea    D17(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D38(%rip),%rax
  mov    %rax,-0x18(%rbp)
  lea    D23(%rip),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x6,-0x3c(%rbp)
  jmp    L16
L28:
  mov    -0x28(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x3c(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L14
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  jmp    L10
L14:
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x28(%rbp)
L10:
  subl   $0x1,-0x3c(%rbp)
L16:
  cmpl   $0x0,-0x3c(%rbp)
  jg     L28
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D17
type_guessed: string
name: disassemble_data
arguments: data_tag: D38
type_guessed: string
name: disassemble_data
arguments: data_tag: D23
type_guessed: string
name: disassemble_data
arguments: data_tag: D25
type_guessed: string

tool: D17: one

tool: D38: two

tool: D23: three

tool: D25: %s %s %s


assistant: ```c
int main() {
  char *one = "one";
  char *two = "two";
  char *three = "three";
  char *s1 = one;
  char *s2 = two;
  char *s3 = three;
  char *swap = NULL;
  for (int i = 6; i > 0; i--) {
    printf("%s %s %s\n", s1, s2, s3);
    if (i % 2) {
      swap = s1;
      s1 = s2;
      s2 = swap;
    } else {
      swap = s2;
      s2 = s3;
      s3 = swap;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x6,%ebx
  lea    D35(%rip),%r13
  lea    D16(%rip),%rbp
  lea    D34(%rip),%r12
  lea    D4(%rip),%r14
  jmp    L20
L6:
  mov    %r12,%rax
  mov    %rbp,%r12
  mov    %rax,%rbp
L31:
  sub    $0x1,%ebx
  je     L35
L20:
  mov    %r13,%rcx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <printf@plt>
  test   $0x1,%bl
  jne    L6
  mov    %rbp,%rax
  mov    %r13,%rbp
  mov    %rax,%r13
  jmp    L31
L35:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: tag: D35
type: string
name: unpack_data
arguments: tag: D16
type: string
name: unpack_data
arguments: tag: D34
type: string
name: unpack_data
arguments: tag: D4
type: string

tool: D35: three

tool: D16: two

tool: D34: one

tool: D4: %s %s %s


assistant: ```c
int main() {
  char *one = "one";
  char *two = "two";
  char *three = "three";
  char *s1 = one;
  char *s2 = two;
  char *s3 = three;
  char *swap = NULL;
  for (int i = 6; i > 0; i--) {
    printf("%s %s %s\n", s1, s2, s3);
    if (i % 2) {
      swap = s1;
      s1 = s2;
      s2 = swap;
    } else {
      swap = s2;
      s2 = s3;
      s3 = swap;
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  lea    D39(%rip),%r14
  push   %r13
  lea    D14(%rip),%r13
  mov    %r14,%rdx
  push   %r12
  lea    D19(%rip),%r12
  mov    %r13,%rdi
  push   %rbp
  lea    D29(%rip),%rbp
  mov    %r12,%rsi
  push   %rbx
  mov    %rbp,%rcx
  mov    $0x6,%ebx
  call   <printf@plt>
L32:
  sub    $0x1,%ebx
  je     L25
L24:
  mov    %r14,%rcx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  test   $0x1,%bl
  je     L7
  mov    %r12,%rax
  mov    %rbp,%r12
  mov    %rax,%rbp
  sub    $0x1,%ebx
  jne    L24
L25:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rbp,%rax
  mov    %r14,%rbp
  mov    %rax,%r14
  jmp    L32
```

tool_calls: name: disassemble_data
arguments: tag: D39
type_guessed: string
name: disassemble_data
arguments: tag: D14
type_guessed: string
name: disassemble_data
arguments: tag: D19
type_guessed: string
name: disassemble_data
arguments: tag: D29
type_guessed: string

tool: D39: two

tool: D14: %s %s %s


tool: D19: one

tool: D29: three

assistant: ```c
int main() {
  char *one = "one";
  char *two = "two";
  char *three = "three";
  char *s1 = one;
  char *s2 = two;
  char *s3 = three;
  char *swap = NULL;
  for (int i = 6; i > 0; i--) {
    printf("%s %s %s\n", s1, s2, s3);
    if (i % 2) {
      swap = s1;
      s1 = s2;
      s2 = swap;
    } else {
      swap = s2;
      s2 = s3;
      s3 = swap;
    }
  }
}
```
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    D26(%rip),%r14
  push   %r13
  lea    D20(%rip),%r13
  mov    %r14,%rdx
  push   %r12
  lea    D15(%rip),%r12
  mov    %r13,%rdi
  push   %rbp
  lea    D6(%rip),%rbp
  mov    %r12,%rsi
  push   %rbx
  mov    %rbp,%rcx
  mov    $0x5,%ebx
  call   <printf@plt>
  jmp    L27
L31:
  mov    %r12,%rax
  mov    %rbp,%r12
  mov    %rax,%rbp
  sub    $0x1,%ebx
  je     L38
L27:
  mov    %r14,%rcx
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  test   $0x1,%bl
  jne    L31
  mov    %rbp,%rax
  mov    %r14,%rbp
  mov    %rax,%r14
  sub    $0x1,%ebx
  jne    L27
L38:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: label: D26
datatype: string
name: read_data
arguments: label: D20
datatype: string
name: read_data
arguments: label: D15
datatype: string
name: read_data
arguments: label: D6
datatype: string

tool: D26: two

tool: D20: %s %s %s


tool: D15: one

tool: D6: three

assistant: ```c
int main() {
  char *one = "one";
  char *two = "two";
  char *three = "three";
  char *s1 = one;
  char *s2 = two;
  char *s3 = three;
  char *swap = NULL;
  for (int i = 6; i > 0; i--) {
    printf("%s %s %s\n", s1, s2, s3);
    if (i % 2) {
      swap = s1;
      s1 = s2;
      s2 = swap;
    } else {
      swap = s2;
      s2 = s3;
      s3 = swap;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  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    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x9,%eax
  ja     L55
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D110(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D110(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D112(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L55
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L55:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L44
  call   <__stack_chk_fail@plt>
L44:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D13
type_infered: string
name: parse_data
arguments: label: D38
type_infered: string
name: parse_data
arguments: label: D112
type_infered: string
name: parse_data
arguments: label: D30
type_infered: string
name: parse_data
arguments: label: D39
type_infered: string
name: parse_data
arguments: label: D1
type_infered: string
name: parse_data
arguments: label: D55
type_infered: string
name: parse_data
arguments: label: D53
type_infered: string
name: parse_data
arguments: label: D7
type_infered: string
name: parse_data
arguments: label: D60
type_infered: string
name: parse_data
arguments: label: D71
type_infered: string

tool: D13: %d

tool: D38: zero

tool: D112: one

tool: D30: two

tool: D39: three

tool: D1: four

tool: D55: five

tool: D53: six

tool: D7: seven

tool: D60: eight

tool: D71: nine

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 0:
    printf("zero");
    break;
  case 1:
    printf("one");
    break;
  case 2:
    printf("two");
    break;
  case 3:
    printf("three");
    break;
  case 4:
    printf("four");
    break;
  case 5:
    printf("five");
    break;
  case 6:
    printf("six");
    break;
  case 7:
    printf("seven");
    break;
  case 8:
    printf("eight");
    break;
  case 9:
    printf("nine");
    break;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D48(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x9,0x4(%rsp)
  ja     L9
  mov    0x4(%rsp),%eax
  lea    D11(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D78(%rip),%rdi
  call   <printf@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L71
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
  lea    D74(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D28(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D58(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D35(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D44(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D55(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D114(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D94(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D80(%rip),%rdi
  call   <printf@plt>
  jmp    L9
L71:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D48
type_guessed: string
name: retrieve_data
arguments: tag: D78
type_guessed: string
name: retrieve_data
arguments: tag: D74
type_guessed: string
name: retrieve_data
arguments: tag: D28
type_guessed: string
name: retrieve_data
arguments: tag: D58
type_guessed: string
name: retrieve_data
arguments: tag: D35
type_guessed: string
name: retrieve_data
arguments: tag: D44
type_guessed: string
name: retrieve_data
arguments: tag: D55
type_guessed: string
name: retrieve_data
arguments: tag: D114
type_guessed: string
name: retrieve_data
arguments: tag: D94
type_guessed: string
name: retrieve_data
arguments: tag: D80
type_guessed: string

tool: D48: %d

tool: D78: zero

tool: D74: one

tool: D28: two

tool: D58: three

tool: D35: four

tool: D44: five

tool: D55: six

tool: D114: seven

tool: D94: eight

tool: D80: nine

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 0:
    printf("zero");
    break;
  case 1:
    printf("one");
    break;
  case 2:
    printf("two");
    break;
  case 3:
    printf("three");
    break;
  case 4:
    printf("four");
    break;
  case 5:
    printf("five");
    break;
  case 6:
    printf("six");
    break;
  case 7:
    printf("seven");
    break;
  case 8:
    printf("eight");
    break;
  case 9:
    printf("nine");
    break;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D11(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x9,0x4(%rsp)
  ja     L22
  mov    0x4(%rsp),%eax
  lea    D24(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D79(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D98(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D33(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D96(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D75(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D73(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D31(%rip),%rdi
  call   <printf@plt>
  jmp    L22
  lea    D49(%rip),%rdi
  call   <printf@plt>
  jmp    L22
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D11
type_infered: string
name: read_data
arguments: tag: D79
type_infered: string
name: read_data
arguments: tag: D98
type_infered: string
name: read_data
arguments: tag: D3
type_infered: string
name: read_data
arguments: tag: D33
type_infered: string
name: read_data
arguments: tag: D96
type_infered: string
name: read_data
arguments: tag: D75
type_infered: string
name: read_data
arguments: tag: D73
type_infered: string
name: read_data
arguments: tag: D6
type_infered: string
name: read_data
arguments: tag: D31
type_infered: string
name: read_data
arguments: tag: D49
type_infered: string

tool: D11: %d

tool: D79: nine

tool: D98: eight

tool: D3: zero

tool: D33: one

tool: D96: two

tool: D75: three

tool: D73: four

tool: D6: five

tool: D31: six

tool: D49: seven

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 0:
    printf("zero");
    break;
  case 1:
    printf("one");
    break;
  case 2:
    printf("two");
    break;
  case 3:
    printf("three");
    break;
  case 4:
    printf("four");
    break;
  case 5:
    printf("five");
    break;
  case 6:
    printf("six");
    break;
  case 7:
    printf("seven");
    break;
  case 8:
    printf("eight");
    break;
  case 9:
    printf("nine");
    break;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D96(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x9,0x4(%rsp)
  ja     L19
  mov    0x4(%rsp),%eax
  lea    D103(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D113(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D110(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D34(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D105(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D44(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  lea    D52(%rip),%rdi
  call   <printf@plt>
  jmp    L19
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D96
type_infered: string
name: read_data
arguments: data_label: D113
type_infered: string
name: read_data
arguments: data_label: D110
type_infered: string
name: read_data
arguments: data_label: D17
type_infered: string
name: read_data
arguments: data_label: D34
type_infered: string
name: read_data
arguments: data_label: D6
type_infered: string
name: read_data
arguments: data_label: D105
type_infered: string
name: read_data
arguments: data_label: D44
type_infered: string
name: read_data
arguments: data_label: D14
type_infered: string
name: read_data
arguments: data_label: D15
type_infered: string
name: read_data
arguments: data_label: D52
type_infered: string

tool: D96: %d

tool: D113: nine

tool: D110: eight

tool: D17: zero

tool: D34: one

tool: D6: two

tool: D105: three

tool: D44: four

tool: D14: five

tool: D15: six

tool: D52: seven

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 0:
    printf("zero");
    break;
  case 1:
    printf("one");
    break;
  case 2:
    printf("two");
    break;
  case 3:
    printf("three");
    break;
  case 4:
    printf("four");
    break;
  case 5:
    printf("five");
    break;
  case 6:
    printf("six");
    break;
  case 7:
    printf("seven");
    break;
  case 8:
    printf("eight");
    break;
  case 9:
    printf("nine");
    break;
  }
}
```