ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<EndianSwap_4bytes>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  mov    %eax,(%rdi)
  ret
```

assistant: ```c
void EndianSwap_4bytes(unsigned char *data) {
  unsigned int *ival = (unsigned int *)data;
  unsigned int val = *ival;
  *ival = ((val >> 24) & 0x000000ff) | ((val >> 8) & 0x0000ff00) | ((val << 8) & 0x00ff0000) | ((val << 24) & 0xff000000);
}
```
user: Translate the below assembly code into C programming language:

```asm
<my_printf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x48,%rsp
  mov    %rdi,-0xb8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L3
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L3:
  mov    D8(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D14(%rip)
  nop
  leave
  ret
```

assistant: ```c
int __attribute__((no)) my_printf(const char *p, ...) {
  static volatile int x;
  ++x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<my_printf>:
  endbr64
  mov    D18(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D9(%rip)
  ret
```

assistant: ```c
int __attribute__((no)) my_printf(const char *p, ...) {
  static volatile int x;
  ++x;
}
```
user: What is the c source code of the assembly code below?

```asm
<my_printf>:
  endbr64
  mov    D19(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D14(%rip)
  ret
```

assistant: ```c
int __attribute__((no)) my_printf(const char *p, ...) {
  static volatile int x;
  ++x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_printf>:
  endbr64
  mov    D6(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D12(%rip)
  ret
```

assistant: ```c
int __attribute__((no)) my_printf(const char *p, ...) {
  static volatile int x;
  ++x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<print_fixed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0xf4240,-0x18(%rbp)
  jne    L34
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x431bde83,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x12,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x431bde83,%rax,%rax
  shr    $0x20,%rax
  sar    $0x12,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0xf4240,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L49
L34:
  cmpl   $0x3e8,-0x18(%rbp)
  jne    L7
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L49
L7:
  cmpl   $0x40,-0x18(%rbp)
  jne    L49
  mov    -0x14(%rbp),%eax
  lea    0x3f(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x6,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  cltd
  shr    $0x1a,%edx
  add    %edx,%eax
  and    $0x3f,%eax
  sub    %edx,%eax
  imul   $0x3e8,%eax,%eax
  lea    0x3f(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x6,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L49:
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
data_type: string

tool: D1: %d.%03d

assistant: ```c
void print_fixed(uint32_t value, int divider) {
  int i = 0, f = 0;
  if (divider == 1000000) {
    i = value / 1000000;
    f = (value % 1000000) / 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 1000) {
    i = value / 1000;
    f = value % 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 64) {
    i = value / 64;
    f = ((value % 64) * 1000) / 64;
    printf("%d.%03d", i, f);
  } else {
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_fixed>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  cmp    $0xf4240,%esi
  je     L44
  cmp    $0x3e8,%esi
  je     L3
  cmp    $0x40,%esi
  je     L43
L46:
  add    $0x8,%rsp
  ret
L44:
  movslq %edi,%rsi
  imul   $0x431bde83,%rsi,%rsi
  sar    $0x32,%rsi
  cltd
  sub    %edx,%esi
  imul   $0xf4240,%esi,%edx
  sub    %edx,%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L46
L3:
  movslq %edi,%rsi
  imul   $0x10624dd3,%rsi,%rsi
  sar    $0x26,%rsi
  cltd
  sub    %edx,%esi
  imul   $0x3e8,%esi,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L46
L43:
  cltd
  shr    $0x1a,%edx
  lea    (%rdi,%rdx,1),%ecx
  and    $0x3f,%ecx
  sub    %edx,%ecx
  imul   $0x3e8,%ecx,%ecx
  lea    0x3f(%rcx),%edx
  test   %ecx,%ecx
  cmovns %ecx,%edx
  sar    $0x6,%edx
  lea    0x3f(%rdi),%esi
  test   %edi,%edi
  cmovns %edi,%esi
  sar    $0x6,%esi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L46
```

tool_calls: name: extract_symbolic_data
arguments: tag: D22
type: string

tool: D22: %d.%03d

assistant: ```c
void print_fixed(uint32_t value, int divider) {
  int i = 0, f = 0;
  if (divider == 1000000) {
    i = value / 1000000;
    f = (value % 1000000) / 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 1000) {
    i = value / 1000;
    f = value % 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 64) {
    i = value / 64;
    f = ((value % 64) * 1000) / 64;
    printf("%d.%03d", i, f);
  } else {
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_fixed>:
  endbr64
  mov    %edi,%eax
  cmp    $0xf4240,%esi
  je     L29
  cmp    $0x3e8,%esi
  je     L15
  cmp    $0x40,%esi
  je     L14
  ret
  nopl   0x0(%rax)
L29:
  movslq %edi,%rsi
  cltd
  lea    D10(%rip),%rdi
  imul   $0x431bde83,%rsi,%rsi
  sar    $0x32,%rsi
  sub    %edx,%esi
  imul   $0xf4240,%esi,%edx
  sub    %edx,%eax
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sub    %eax,%edx
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  movslq %edi,%rsi
  cltd
  lea    D10(%rip),%rdi
  imul   $0x10624dd3,%rsi,%rsi
  sar    $0x26,%rsi
  sub    %edx,%esi
  imul   $0x3e8,%esi,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  jmp    1050 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L14:
  cltd
  lea    0x3f(%rdi),%esi
  shr    $0x1a,%edx
  lea    (%rdi,%rdx,1),%ecx
  and    $0x3f,%ecx
  sub    %edx,%ecx
  imul   $0x3e8,%ecx,%ecx
  test   %ecx,%ecx
  lea    0x3f(%rcx),%edx
  cmovns %ecx,%edx
  sar    $0x6,%edx
  test   %edi,%edi
  cmovns %edi,%esi
  lea    D10(%rip),%rdi
  sar    $0x6,%esi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D10
data_type: string

tool: D10: %d.%03d

assistant: ```c
void print_fixed(uint32_t value, int divider) {
  int i = 0, f = 0;
  if (divider == 1000000) {
    i = value / 1000000;
    f = (value % 1000000) / 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 1000) {
    i = value / 1000;
    f = value % 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 64) {
    i = value / 64;
    f = ((value % 64) * 1000) / 64;
    printf("%d.%03d", i, f);
  } else {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<print_fixed>:
  endbr64
  mov    %edi,%eax
  cmp    $0xf4240,%esi
  je     L10
  cmp    $0x3e8,%esi
  je     L4
  cmp    $0x40,%esi
  je     L29
  ret
  nopl   0x0(%rax)
L10:
  movslq %edi,%rsi
  cltd
  lea    D10(%rip),%rdi
  imul   $0x431bde83,%rsi,%rsi
  sar    $0x32,%rsi
  sub    %edx,%esi
  imul   $0xf4240,%esi,%edx
  sub    %edx,%eax
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sub    %eax,%edx
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L4:
  movslq %edi,%rsi
  cltd
  lea    D10(%rip),%rdi
  imul   $0x10624dd3,%rsi,%rsi
  sar    $0x26,%rsi
  sub    %edx,%esi
  imul   $0x3e8,%esi,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  jmp    1050 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L29:
  cltd
  lea    0x3f(%rdi),%esi
  shr    $0x1a,%edx
  lea    (%rdi,%rdx,1),%ecx
  and    $0x3f,%ecx
  sub    %edx,%ecx
  imul   $0x3e8,%ecx,%ecx
  test   %ecx,%ecx
  lea    0x3f(%rcx),%edx
  cmovns %ecx,%edx
  sar    $0x6,%edx
  test   %edi,%edi
  cmovns %edi,%esi
  lea    D10(%rip),%rdi
  sar    $0x6,%esi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D10
type_guessed: string

tool: D10: %d.%03d

assistant: ```c
void print_fixed(uint32_t value, int divider) {
  int i = 0, f = 0;
  if (divider == 1000000) {
    i = value / 1000000;
    f = (value % 1000000) / 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 1000) {
    i = value / 1000;
    f = value % 1000;
    printf("%d.%03d", i, f);
  } else if (divider == 64) {
    i = value / 64;
    f = ((value % 64) * 1000) / 64;
    printf("%d.%03d", i, f);
  } else {
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test3_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <MAIN__@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int test3_() {
  MAIN__();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<test3_>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <MAIN__@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int test3_() {
  MAIN__();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test3_>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <MAIN__@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int test3_() {
  MAIN__();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test3_>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <MAIN__@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int test3_() {
  MAIN__();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```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    D88(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x9,%eax
  ja     L35
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D36(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D36(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D100(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D122(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D94(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D118(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L35:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L95
  call   <__stack_chk_fail@plt>
L95:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D88
type_guessed: string
name: extract_symbolic_data
arguments: tag: D100
type_guessed: string
name: extract_symbolic_data
arguments: tag: D122
type_guessed: string
name: extract_symbolic_data
arguments: tag: D89
type_guessed: string
name: extract_symbolic_data
arguments: tag: D94
type_guessed: string
name: extract_symbolic_data
arguments: tag: D51
type_guessed: string
name: extract_symbolic_data
arguments: tag: D39
type_guessed: string
name: extract_symbolic_data
arguments: tag: D64
type_guessed: string
name: extract_symbolic_data
arguments: tag: D118
type_guessed: string
name: extract_symbolic_data
arguments: tag: D12
type_guessed: string
name: extract_symbolic_data
arguments: tag: D67
type_guessed: string

tool: D88: %d

tool: D100: zero


tool: D122: one


tool: D89: two


tool: D94: three


tool: D51: four


tool: D39: five


tool: D64: six


tool: D118: seven


tool: D12: eight


tool: D67: nine


assistant: ```c
int main() {
  int number;
  scanf("%d", &number);
  switch (number) {
  case 0:
    printf("zero\n");
    break;
  case 1:
    printf("one\n");
    break;
  case 2:
    printf("two\n");
    break;
  case 3:
    printf("three\n");
    break;
  case 4:
    printf("four\n");
    break;
  case 5:
    printf("five\n");
    break;
  case 6:
    printf("six\n");
    break;
  case 7:
    printf("seven\n");
    break;
  case 8:
    printf("eight\n");
    break;
  case 9:
    printf("nine\n");
    break;
  }
  return 0;
}
```
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    D111(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x9,0x4(%rsp)
  ja     L54
  mov    0x4(%rsp),%eax
  lea    D116(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D14(%rip),%rdi
  call   <printf@plt>
L54:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L75
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
  lea    D114(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D48(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D115(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D106(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D21(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D119(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D80(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L54
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L54
L75:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D111
datatype: string
name: extract_symbolic_data
arguments: data_tag: D14
datatype: string
name: extract_symbolic_data
arguments: data_tag: D114
datatype: string
name: extract_symbolic_data
arguments: data_tag: D48
datatype: string
name: extract_symbolic_data
arguments: data_tag: D115
datatype: string
name: extract_symbolic_data
arguments: data_tag: D106
datatype: string
name: extract_symbolic_data
arguments: data_tag: D21
datatype: string
name: extract_symbolic_data
arguments: data_tag: D119
datatype: string
name: extract_symbolic_data
arguments: data_tag: D80
datatype: string
name: extract_symbolic_data
arguments: data_tag: D24
datatype: string
name: extract_symbolic_data
arguments: data_tag: D18
datatype: string

tool: D111: %d

tool: D14: zero


tool: D114: one


tool: D48: two


tool: D115: three


tool: D106: four


tool: D21: five


tool: D119: six


tool: D80: seven


tool: D24: eight


tool: D18: nine


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

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D43(%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     L39
  mov    0x4(%rsp),%eax
  lea    D84(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D35(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L39:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D104(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D69(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D54(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D101(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D98(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D81(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D64(%rip),%rdi
  call   <printf@plt>
  jmp    L39
L67:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D43
type_guessed: string
name: retrieve_data
arguments: data_label: D35
type_guessed: string
name: retrieve_data
arguments: data_label: D104
type_guessed: string
name: retrieve_data
arguments: data_label: D69
type_guessed: string
name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D54
type_guessed: string
name: retrieve_data
arguments: data_label: D101
type_guessed: string
name: retrieve_data
arguments: data_label: D1
type_guessed: string
name: retrieve_data
arguments: data_label: D98
type_guessed: string
name: retrieve_data
arguments: data_label: D81
type_guessed: string
name: retrieve_data
arguments: data_label: D64
type_guessed: string

tool: D43: %d

tool: D35: nine


tool: D104: eight


tool: D69: zero


tool: D14: one


tool: D54: two


tool: D101: three


tool: D1: four


tool: D98: five


tool: D81: six


tool: D64: seven


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

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D119(%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     L39
  mov    0x4(%rsp),%eax
  lea    D102(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D15(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L39:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L68
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D25(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D115(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D56(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D33(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D60(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D104(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D45(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D103(%rip),%rdi
  call   <printf@plt>
  jmp    L39
  lea    D39(%rip),%rdi
  call   <printf@plt>
  jmp    L39
L68:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D119
type_infered: string
name: retrieve_data
arguments: tag: D15
type_infered: string
name: retrieve_data
arguments: tag: D25
type_infered: string
name: retrieve_data
arguments: tag: D115
type_infered: string
name: retrieve_data
arguments: tag: D56
type_infered: string
name: retrieve_data
arguments: tag: D33
type_infered: string
name: retrieve_data
arguments: tag: D60
type_infered: string
name: retrieve_data
arguments: tag: D104
type_infered: string
name: retrieve_data
arguments: tag: D45
type_infered: string
name: retrieve_data
arguments: tag: D103
type_infered: string
name: retrieve_data
arguments: tag: D39
type_infered: string

tool: D119: %d

tool: D15: nine


tool: D25: eight


tool: D115: zero


tool: D56: one


tool: D33: two


tool: D60: three


tool: D104: four


tool: D45: five


tool: D103: six


tool: D39: seven


assistant: ```c
int main() {
  int number;
  scanf("%d", &number);
  switch (number) {
  case 0:
    printf("zero\n");
    break;
  case 1:
    printf("one\n");
    break;
  case 2:
    printf("two\n");
    break;
  case 3:
    printf("three\n");
    break;
  case 4:
    printf("four\n");
    break;
  case 5:
    printf("five\n");
    break;
  case 6:
    printf("six\n");
    break;
  case 7:
    printf("seven\n");
    break;
  case 8:
    printf("eight\n");
    break;
  case 9:
    printf("nine\n");
    break;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strpbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L104
  mov    $0x0,%eax
  jmp    L52
L5:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L44
L34:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L32
  addq   $0x1,-0x8(%rbp)
L44:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L34
  jmp    L59
L32:
  nop
L59:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L22
  addq   $0x1,-0x18(%rbp)
L104:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
  jmp    L89
L22:
  nop
L89:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L36
  mov    -0x18(%rbp),%rax
  jmp    L52
L36:
  mov    $0x0,%eax
L52:
  pop    %rbp
  ret
```

assistant: ```c
char *strpbrk(const char *s, const char *accept) {
  const char *c = NULL;
  if (!*s)
    return NULL;
  while (*s) {
    for (c = accept; *c; c++) {
      if (*s == *c)
        break;
    }
    if (*c)
      break;
    s++;
  }
  return (*c == 0) ? NULL : (char *)s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strpbrk>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L55
  movzbl (%rsi),%r8d
  test   %r8b,%r8b
  sete   %r9b
  jmp    L28
L88:
  mov    %rsi,%rax
  mov    %r8d,%edx
L77:
  test   %dl,%dl
  jne    L17
  add    $0x1,%rdi
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L17
L28:
  cmp    %cl,%r8b
  je     L88
  test   %r9b,%r9b
  jne    L88
  mov    %rsi,%rax
L51:
  add    $0x1,%rax
  movzbl (%rax),%edx
  cmp    %cl,%dl
  je     L77
  test   %dl,%dl
  jne    L51
  jmp    L77
L17:
  cmpb   $0x0,(%rax)
  mov    $0x0,%eax
  cmovne %rdi,%rax
  ret
L55:
  mov    $0x0,%eax
  ret
```

assistant: ```c
char *strpbrk(const char *s, const char *accept) {
  const char *c = NULL;
  if (!*s)
    return NULL;
  while (*s) {
    for (c = accept; *c; c++) {
      if (*s == *c)
        break;
    }
    if (*c)
      break;
    s++;
  }
  return (*c == 0) ? NULL : (char *)s;
}
```