ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<vid_to_reg>:
  endbr64
  sub    $0x5a,%esi
  cmp    $0x1,%esi
  ja     L14
  lea    -0x44c(%rdi),%eax
  cmp    $0x2ee,%eax
  ja     L19
  imul   $0xfffffff6,%edi,%edi
  add    $0x4843,%edi
  movslq %edi,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x23,%rdx
  sar    $0x1f,%edi
  sub    %edi,%edx
  add    $0x5,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  ret
L14:
  mov    $0xffffffff,%eax
  ret
L19:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int vid_to_reg(int val, int vrm) {
  switch (vrm) {
  case 91:
  case 90:
    return ((val >= 1100) && (val <= 1850) ? ((18499 - val * 10) / 25 + 5) / 10 : -1);
  default:
    return -1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<vid_to_reg>:
  endbr64
  sub    $0x5a,%esi
  cmp    $0x1,%esi
  ja     L15
  lea    -0x44c(%rdi),%eax
  cmp    $0x2ee,%eax
  ja     L15
  imul   $0xfffffff6,%edi,%edi
  mov    $0xcccccccd,%edx
  add    $0x4843,%edi
  movslq %edi,%rax
  sar    $0x1f,%edi
  imul   $0x51eb851f,%rax,%rax
  sar    $0x23,%rax
  sub    %edi,%eax
  add    $0x5,%eax
  imul   %rdx,%rax
  shr    $0x23,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int vid_to_reg(int val, int vrm) {
  switch (vrm) {
  case 91:
  case 90:
    return ((val >= 1100) && (val <= 1850) ? ((18499 - val * 10) / 25 + 5) / 10 : -1);
  default:
    return -1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vid_to_reg>:
  endbr64
  sub    $0x5a,%esi
  cmp    $0x1,%esi
  ja     L1
  lea    -0x44c(%rdi),%eax
  cmp    $0x2ee,%eax
  ja     L1
  imul   $0xfffffff6,%edi,%edi
  mov    $0xcccccccd,%edx
  add    $0x4843,%edi
  movslq %edi,%rax
  sar    $0x1f,%edi
  imul   $0x51eb851f,%rax,%rax
  sar    $0x23,%rax
  sub    %edi,%eax
  add    $0x5,%eax
  imul   %rdx,%rax
  shr    $0x23,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int vid_to_reg(int val, int vrm) {
  switch (vrm) {
  case 91:
  case 90:
    return ((val >= 1100) && (val <= 1850) ? ((18499 - val * 10) / 25 + 5) / 10 : -1);
  default:
    return -1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ch_str_cmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  jmp    L34
L20:
  addq   $0x1,-0x10(%rbp)
L34:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x4(%rbp)
  je     L25
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L20
L25:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x4(%rbp)
  jne    L14
  mov    -0x10(%rbp),%rax
  jmp    L16
L14:
  mov    $0x0,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
char *ch_str_cmp(char c, const char *str) {
  while ((c != *str) && (*str != '\0'))
    ++str;
  if (c == *str) {
    return (char *)str;
  } else
    return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ch_str_cmp>:
  endbr64
  mov    %rsi,%rax
  mov    %edi,%ecx
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L10
  cmp    %dil,%dl
  je     L10
L20:
  add    $0x1,%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L10
  cmp    %cl,%dl
  jne    L20
L10:
  cmp    %dil,%dl
  mov    $0x0,%edx
  cmovne %rdx,%rax
  ret
```

assistant: ```c
char *ch_str_cmp(char c, const char *str) {
  while ((c != *str) && (*str != '\0'))
    ++str;
  if (c == *str) {
    return (char *)str;
  } else
    return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ch_str_cmp>:
  endbr64
  movzbl (%rsi),%eax
  mov    %edi,%edx
  cmp    %dil,%al
  jne    L27
  jmp    L39
L19:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  cmp    %dl,%al
  je     L39
L27:
  test   %al,%al
  jne    L19
  xor    %eax,%eax
  test   %dil,%dil
  cmove  %rsi,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %rsi,%rax
  ret
```

assistant: ```c
char *ch_str_cmp(char c, const char *str) {
  while ((c != *str) && (*str != '\0'))
    ++str;
  if (c == *str) {
    return (char *)str;
  } else
    return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ch_str_cmp>:
  endbr64
  movzbl (%rsi),%eax
  mov    %edi,%edx
  cmp    %dil,%al
  jne    L20
  jmp    L0
L35:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  cmp    %dl,%al
  je     L0
L20:
  test   %al,%al
  jne    L35
  xor    %eax,%eax
  test   %dil,%dil
  cmove  %rsi,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %rsi,%rax
  ret
```

assistant: ```c
char *ch_str_cmp(char c, const char *str) {
  while ((c != *str) && (*str != '\0'))
    ++str;
  if (c == *str) {
    return (char *)str;
  } else
    return NULL;
}
```
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
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <test@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
datatype: string

tool: D8: C Programming Language

assistant: ```c
int main() {
  test("C Programming Language");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D4
datatype: string

tool: D4: C Programming Language

assistant: ```c
int main() {
  test("C Programming Language");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type: string

tool: D2: C Programming Language

assistant: ```c
int main() {
  test("C Programming Language");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string

tool: D2: C Programming Language

assistant: ```c
int main() {
  test("C Programming Language");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<help>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D21
type_infered: string
name: disassemble_data
arguments: data_tag: D35
type_infered: string
name: disassemble_data
arguments: data_tag: D29
type_infered: string
name: disassemble_data
arguments: data_tag: D38
type_infered: string
name: disassemble_data
arguments: data_tag: D41
type_infered: string

tool: D21: u up     d down   m mix


tool: D35: f front  b back   * reset


tool: D29: l left   r right


tool: D38: [ load   ] save


tool: D41: q quit



assistant: ```c
void help() {
  printf("u up     d down   m mix\n");
  printf("f front  b back   * reset\n");
  printf("l left   r right\n");
  printf("[ load   ] save\n");
  printf("q quit\n\n");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D45(%rip),%rdi
  call   <printf@plt>
  lea    D47(%rip),%rdi
  call   <printf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D14
type_guessed: string
name: parse_data
arguments: label: D42
type_guessed: string
name: parse_data
arguments: label: D45
type_guessed: string
name: parse_data
arguments: label: D47
type_guessed: string
name: parse_data
arguments: label: D32
type_guessed: string

tool: D14: u up     d down   m mix


tool: D42: f front  b back   * reset


tool: D45: l left   r right


tool: D47: [ load   ] save


tool: D32: q quit



assistant: ```c
void help() {
  printf("u up     d down   m mix\n");
  printf("f front  b back   * reset\n");
  printf("l left   r right\n");
  printf("[ load   ] save\n");
  printf("q quit\n\n");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D38(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D33
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D16
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D35
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D38
type_guessed: string

tool: D33: u up     d down   m mix


tool: D8: f front  b back   * reset


tool: D16: l left   r right


tool: D35: [ load   ] save


tool: D38: q quit



assistant: ```c
void help() {
  printf("u up     d down   m mix\n");
  printf("f front  b back   * reset\n");
  printf("l left   r right\n");
  printf("[ load   ] save\n");
  printf("q quit\n\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D24
type_infered: string
name: extract_symbolic_data
arguments: data_label: D35
type_infered: string
name: extract_symbolic_data
arguments: data_label: D11
type_infered: string
name: extract_symbolic_data
arguments: data_label: D18
type_infered: string
name: extract_symbolic_data
arguments: data_label: D0
type_infered: string

tool: D24: u up     d down   m mix


tool: D35: f front  b back   * reset


tool: D11: l left   r right


tool: D18: [ load   ] save


tool: D0: q quit



assistant: ```c
void help() {
  printf("u up     d down   m mix\n");
  printf("f front  b back   * reset\n");
  printf("l left   r right\n");
  printf("[ load   ] save\n");
  printf("q quit\n\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  lea    D12(%rip),%rdx
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D13
type_guessed: string

tool: D13: &d=0x%x &e=0x%x


assistant: ```c
int foo() {
  int d;
  static int e = 5;
  printf("&d=0x%x &e=0x%x\n", &d, &e);
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdx
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
data_type: string

tool: D1: &d=0x%x &e=0x%x


assistant: ```c
int foo() {
  int d;
  static int e = 5;
  printf("&d=0x%x &e=0x%x\n", &d, &e);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  sub    $0x18,%rsp
  lea    D11(%rip),%rdx
  lea    D1(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D1
type_infered: string

tool: D1: &d=0x%x &e=0x%x


assistant: ```c
int foo() {
  int d;
  static int e = 5;
  printf("&d=0x%x &e=0x%x\n", &d, &e);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo>:
  endbr64
  sub    $0x18,%rsp
  lea    D8(%rip),%rdx
  lea    D10(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D10
data_type: string

tool: D10: &d=0x%x &e=0x%x


assistant: ```c
int foo() {
  int d;
  static int e = 5;
  printf("&d=0x%x &e=0x%x\n", &d, &e);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<is_exist_event_list>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <get_task_by_event_list@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int is_exist_event_list(void *event_list) {
  struct tag_osn_task *task = get_task_by_event_list(event_list);
  return task ? 1 : 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<is_exist_event_list>:
  endbr64
  sub    $0x8,%rsp
  call   <get_task_by_event_list@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int is_exist_event_list(void *event_list) {
  struct tag_osn_task *task = get_task_by_event_list(event_list);
  return task ? 1 : 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
<is_exist_event_list>:
  endbr64
  sub    $0x8,%rsp
  call   <get_task_by_event_list@plt>
  test   %rax,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_exist_event_list(void *event_list) {
  struct tag_osn_task *task = get_task_by_event_list(event_list);
  return task ? 1 : 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<is_exist_event_list>:
  endbr64
  sub    $0x8,%rsp
  call   <get_task_by_event_list@plt>
  test   %rax,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_exist_event_list(void *event_list) {
  struct tag_osn_task *task = get_task_by_event_list(event_list);
  return task ? 1 : 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Q_CleanStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L10
L28:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5e,%al
  jne    L6
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L6
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x5e,%al
  je     L6
  addq   $0x1,-0x8(%rbp)
  jmp    L37
L6:
  cmpl   $0x1f,-0x14(%rbp)
  jle    L37
  cmpl   $0x7e,-0x14(%rbp)
  jg     L37
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    %dl,(%rax)
L37:
  addq   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L28
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *Q_CleanStr(char *string) {
  char *d;
  char *s;
  int c;
  s = string;
  d = string;
  while ((c = *s) != 0) {
    if ((*(s) == '^' && *((s) + 1) && *((s) + 1) != '^')) {
      s++;
    } else if (c >= 0x20 && c <= 0x7E) {
      *d++ = c;
    }
    s++;
  }
  *d = '\0';
  return string;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Q_CleanStr>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L66
  mov    %rdi,%rsi
  mov    %rdi,%r9
  jmp    L80
L20:
  movzbl 0x1(%rsi),%ecx
  cmp    $0x5e,%cl
  je     L77
  test   %cl,%cl
  jne    L64
L77:
  mov    %dl,(%r9)
  mov    %rsi,%rcx
  lea    0x1(%r9),%r9
L82:
  lea    0x1(%rcx),%rsi
  movzbl 0x1(%rcx),%edx
  test   %dl,%dl
  je     L89
L80:
  cmp    $0x5e,%dl
  je     L20
  movsbl %dl,%r8d
  sub    $0x20,%r8d
  mov    %rsi,%rcx
  cmp    $0x5e,%r8d
  ja     L82
  jmp    L77
L64:
  lea    0x1(%rsi),%rcx
  jmp    L82
L66:
  mov    %rdi,%r9
L89:
  movb   $0x0,(%r9)
  ret
```

assistant: ```c
char *Q_CleanStr(char *string) {
  char *d;
  char *s;
  int c;
  s = string;
  d = string;
  while ((c = *s) != 0) {
    if ((*(s) == '^' && *((s) + 1) && *((s) + 1) != '^')) {
      s++;
    } else if (c >= 0x20 && c <= 0x7E) {
      *d++ = c;
    }
    s++;
  }
  *d = '\0';
  return string;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Q_CleanStr>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L6
  mov    %rdi,%rsi
  mov    %rdi,%r9
  jmp    L15
  cs nopw 0x0(%rax,%rax,1)
L46:
  movsbl %dl,%r8d
  mov    %rsi,%rcx
  sub    $0x20,%r8d
  cmp    $0x5e,%r8d
  jbe    L52
L64:
  movzbl 0x1(%rcx),%edx
  lea    0x1(%rcx),%rsi
  test   %dl,%dl
  je     L54
L15:
  cmp    $0x5e,%dl
  jne    L46
  movzbl 0x1(%rsi),%ecx
  cmp    $0x5e,%cl
  je     L52
  test   %cl,%cl
  je     L52
  movzbl 0x2(%rsi),%edx
  lea    0x1(%rsi),%rcx
  lea    0x1(%rcx),%rsi
  test   %dl,%dl
  jne    L15
L54:
  movb   $0x0,(%r9)
  ret
  nopl   0x0(%rax)
L52:
  mov    %dl,(%r9)
  mov    %rsi,%rcx
  add    $0x1,%r9
  jmp    L64
  nopl   0x0(%rax)
L6:
  mov    %rdi,%r9
  movb   $0x0,(%r9)
  ret
```

assistant: ```c
char *Q_CleanStr(char *string) {
  char *d;
  char *s;
  int c;
  s = string;
  d = string;
  while ((c = *s) != 0) {
    if ((*(s) == '^' && *((s) + 1) && *((s) + 1) != '^')) {
      s++;
    } else if (c >= 0x20 && c <= 0x7E) {
      *d++ = c;
    }
    s++;
  }
  *d = '\0';
  return string;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Q_CleanStr>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L27
  mov    %rdi,%rsi
  mov    %rdi,%r9
  jmp    L80
  cs nopw 0x0(%rax,%rax,1)
L42:
  movsbl %dl,%r8d
  mov    %rsi,%rcx
  sub    $0x20,%r8d
  cmp    $0x5e,%r8d
  jbe    L85
L52:
  movzbl 0x1(%rcx),%edx
  lea    0x1(%rcx),%rsi
  test   %dl,%dl
  je     L89
L80:
  cmp    $0x5e,%dl
  jne    L42
  movzbl 0x1(%rsi),%ecx
  cmp    $0x5e,%cl
  je     L85
  test   %cl,%cl
  je     L85
  movzbl 0x2(%rsi),%edx
  lea    0x1(%rsi),%rcx
  lea    0x1(%rcx),%rsi
  test   %dl,%dl
  jne    L80
L89:
  movb   $0x0,(%r9)
  ret
  nopl   0x0(%rax)
L85:
  mov    %dl,(%r9)
  mov    %rsi,%rcx
  add    $0x1,%r9
  jmp    L52
  nopl   0x0(%rax)
L27:
  mov    %rdi,%r9
  movb   $0x0,(%r9)
  ret
```

assistant: ```c
char *Q_CleanStr(char *string) {
  char *d;
  char *s;
  int c;
  s = string;
  d = string;
  while ((c = *s) != 0) {
    if ((*(s) == '^' && *((s) + 1) && *((s) + 1) != '^')) {
      s++;
    } else if (c >= 0x20 && c <= 0x7E) {
      *d++ = c;
    }
    s++;
  }
  *d = '\0';
  return string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <init@plt>
  movl   $0x1,-0x4(%rbp)
  jmp    L9
L32:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <fib@plt>
  cmp    $0x3e6,%eax
  jle    L8
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L15
L8:
  addl   $0x1,-0x4(%rbp)
L9:
  cmpl   $0x1387,-0x4(%rbp)
  jle    L32
L15:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string

tool: D9: Fibonacci number %d is the first to have 1000 digits


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  init();
  for (i = 1; i < 5000; i++) {
    if (fib(i) >= 999) {
      printf("Fibonacci number %d is the first to have 1000 digits\n", i);
      break;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <init@plt>
  mov    $0x1,%ebx
L18:
  mov    %ebx,%edi
  call   <fib@plt>
  cmp    $0x3e6,%eax
  jg     L15
  add    $0x1,%ebx
  cmp    $0x1388,%ebx
  jne    L18
  jmp    L24
L15:
  mov    %ebx,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
L24:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_infered: string

tool: D6: Fibonacci number %d is the first to have 1000 digits


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  init();
  for (i = 1; i < 5000; i++) {
    if (fib(i) >= 999) {
      printf("Fibonacci number %d is the first to have 1000 digits\n", i);
      break;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    $0x1,%ebp
  call   <init@plt>
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L7:
  add    $0x1,%ebp
  cmp    $0x1388,%ebp
  je     L1
L18:
  mov    %ebp,%edi
  call   <fib@plt>
  cmp    $0x3e6,%eax
  jle    L7
  mov    %ebp,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
L1:
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type: string

tool: D8: Fibonacci number %d is the first to have 1000 digits


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  init();
  for (i = 1; i < 5000; i++) {
    if (fib(i) >= 999) {
      printf("Fibonacci number %d is the first to have 1000 digits\n", i);
      break;
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    $0x1,%ebp
  call   <init@plt>
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L14:
  add    $0x1,%ebp
  cmp    $0x1388,%ebp
  je     L9
L28:
  mov    %ebp,%edi
  call   <fib@plt>
  cmp    $0x3e6,%eax
  jle    L14
  mov    %ebp,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
L9:
  xor    %eax,%eax
  pop    %rbp
  ret
```

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

tool: D3: Fibonacci number %d is the first to have 1000 digits


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  init();
  for (i = 1; i < 5000; i++) {
    if (fib(i) >= 999) {
      printf("Fibonacci number %d is the first to have 1000 digits\n", i);
      break;
    }
  }
  return 0;
}
```