ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strlcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L13
  subq   $0x1,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L13
L40:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  je     L71
  subq   $0x1,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L40
  jmp    L13
L71:
  nop
L13:
  cmpq   $0x0,-0x8(%rbp)
  jne    L70
  cmpq   $0x0,-0x38(%rbp)
  je     L14
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
L14:
  nop
L50:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L50
L70:
  mov    -0x10(%rbp),%rax
  sub    -0x30(%rbp),%rax
  sub    $0x1,%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlcpy(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  if (n != 0 && --n != 0) {
    do {
      if ((*d++ = *s++) == 0)
        break;
    } while (--n != 0);
  }
  if (n == 0) {
    if (siz != 0)
      *d = '\0';
    while (*s++)
      ;
  }
  return (size_t)(s - src - 1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlcpy>:
  endbr64
  test   %rdx,%rdx
  je     L7
  sub    $0x1,%rdx
  je     L22
  mov    %rsi,%rax
L60:
  add    $0x1,%rax
  add    $0x1,%rdi
  movzbl -0x1(%rax),%ecx
  mov    %cl,-0x1(%rdi)
  test   %cl,%cl
  je     L83
  sub    $0x1,%rdx
  jne    L60
  jmp    L56
L7:
  mov    %rsi,%rax
L16:
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L16
L74:
  sub    %rsi,%rax
  sub    $0x1,%rax
  ret
L22:
  mov    %rsi,%rax
L56:
  movb   $0x0,(%rdi)
  jmp    L16
L83:
  test   %rdx,%rdx
  jne    L74
  jmp    L56
```

assistant: ```c
size_t strlcpy(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  if (n != 0 && --n != 0) {
    do {
      if ((*d++ = *s++) == 0)
        break;
    } while (--n != 0);
  }
  if (n == 0) {
    if (siz != 0)
      *d = '\0';
    while (*s++)
      ;
  }
  return (size_t)(s - src - 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strlcpy>:
  endbr64
  mov    %rsi,%rax
  test   %rdx,%rdx
  je     L39
  cmp    $0x1,%rdx
  je     L75
  mov    %rdi,%rcx
  jmp    L38
  nopw   0x0(%rax,%rax,1)
L64:
  mov    %rdx,%r8
  sub    %rcx,%r8
  lea    -0x1(%rdi,%r8,1),%r8
  test   %r8,%r8
  je     L53
L38:
  movzbl (%rax),%r8d
  add    $0x1,%rax
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  test   %r8b,%r8b
  jne    L64
L24:
  sub    %rsi,%rax
  sub    $0x1,%rax
  ret
  nopl   (%rax)
L53:
  mov    %rcx,%rdi
L75:
  movb   $0x0,(%rdi)
  cs nopw 0x0(%rax,%rax,1)
L39:
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  je     L24
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L39
  jmp    L24
```

assistant: ```c
size_t strlcpy(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  if (n != 0 && --n != 0) {
    do {
      if ((*d++ = *s++) == 0)
        break;
    } while (--n != 0);
  }
  if (n == 0) {
    if (siz != 0)
      *d = '\0';
    while (*s++)
      ;
  }
  return (size_t)(s - src - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strlcpy>:
  endbr64
  mov    %rsi,%rax
  test   %rdx,%rdx
  je     L37
  cmp    $0x1,%rdx
  je     L65
  mov    %rdi,%rcx
  jmp    L35
  nopw   0x0(%rax,%rax,1)
L62:
  mov    %rdx,%r8
  sub    %rcx,%r8
  lea    -0x1(%rdi,%r8,1),%r8
  test   %r8,%r8
  je     L33
L35:
  movzbl (%rax),%r8d
  add    $0x1,%rax
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  test   %r8b,%r8b
  jne    L62
L55:
  sub    %rsi,%rax
  sub    $0x1,%rax
  ret
  nopl   (%rax)
L33:
  mov    %rcx,%rdi
L65:
  movb   $0x0,(%rdi)
  cs nopw 0x0(%rax,%rax,1)
L37:
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  je     L55
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L37
  jmp    L55
```

assistant: ```c
size_t strlcpy(char *dst, const char *src, size_t siz) {
  char *d = dst;
  const char *s = src;
  size_t n = siz;
  if (n != 0 && --n != 0) {
    do {
      if ((*d++ = *s++) == 0)
        break;
    } while (--n != 0);
  }
  if (n == 0) {
    if (siz != 0)
      *d = '\0';
    while (*s++)
      ;
  }
  return (size_t)(s - src - 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x94(%rbp)
  lea    -0x90(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <itos@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <Reverse@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
type: string

tool: D12: %s


assistant: ```c
int main(void) {
  int n = -00;
  char s[128];
  itos(n, s);
  printf("%s\n", s);
  Reverse(s);
  printf("%s\n", s);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <itos@plt>
  mov    %rbx,%rsi
  lea    D6(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <Reverse@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: %s


assistant: ```c
int main(void) {
  int n = -00;
  char s[128];
  itos(n, s);
  printf("%s\n", s);
  Reverse(s);
  printf("%s\n", s);
  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
  push   %r12
  xor    %edi,%edi
  lea    D3(%rip),%r12
  push   %rbp
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <itos@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <Reverse@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_guessed: string

tool: D3: %s


assistant: ```c
int main(void) {
  int n = -00;
  char s[128];
  itos(n, s);
  printf("%s\n", s);
  Reverse(s);
  printf("%s\n", s);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r12
  xor    %edi,%edi
  lea    D5(%rip),%r12
  push   %rbp
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <itos@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <Reverse@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D5
data_type: string

tool: D5: %s


assistant: ```c
int main(void) {
  int n = -00;
  char s[128];
  itos(n, s);
  printf("%s\n", s);
  Reverse(s);
  printf("%s\n", s);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<altera_a10_cdr_lookup_lf_resistor_pd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x6acfbf,-0x4(%rbp)
  ja     L45
  cmpl   $0x4,-0x8(%rbp)
  je     L84
  cmpl   $0x4,-0x8(%rbp)
  ja     L70
  cmpl   $0x1,-0x8(%rbp)
  je     L28
  cmpl   $0x2,-0x8(%rbp)
  je     L84
  jmp    L70
L28:
  mov    $0x2,%eax
  jmp    L85
L84:
  mov    $0x3,%eax
  jmp    L85
L70:
  mov    $0x0,%eax
  jmp    L85
L45:
  cmpl   $0x2,-0x8(%rbp)
  ja     L134
  cmpl   $0x0,-0x8(%rbp)
  jne    L133
  jmp    L87
L134:
  cmpl   $0x4,-0x8(%rbp)
  jne    L87
L133:
  mov    $0x3,%eax
  jmp    L85
L87:
  mov    $0x0,%eax
L85:
  pop    %rbp
  ret
```

assistant: ```c
unsigned int altera_a10_cdr_lookup_lf_resistor_pd(unsigned int fvco, unsigned int lpd) {
  if (fvco < 7000000) {
    switch (lpd) {
    case 1:
      return 2;
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  } else {
    switch (lpd) {
    case 1:
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<altera_a10_cdr_lookup_lf_resistor_pd>:
  endbr64
  cmp    $0x6acfbf,%edi
  ja     L23
  lea    -0x2(%rsi),%eax
  test   $0xfffffffd,%eax
  je     L11
  cmp    $0x1,%esi
  sete   %al
  movzbl %al,%eax
  add    %eax,%eax
  ret
L23:
  cmp    $0x2,%esi
  jbe    L9
  cmp    $0x4,%esi
  sete   %al
  movzbl %al,%eax
  lea    (%rax,%rax,2),%eax
  ret
L9:
  test   %esi,%esi
  mov    $0x3,%eax
  cmove  %esi,%eax
  ret
L11:
  mov    $0x3,%eax
  ret
```

assistant: ```c
unsigned int altera_a10_cdr_lookup_lf_resistor_pd(unsigned int fvco, unsigned int lpd) {
  if (fvco < 7000000) {
    switch (lpd) {
    case 1:
      return 2;
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  } else {
    switch (lpd) {
    case 1:
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<altera_a10_cdr_lookup_lf_resistor_pd>:
  endbr64
  cmp    $0x6acfbf,%edi
  ja     L25
  lea    -0x2(%rsi),%eax
  and    $0xfffffffd,%eax
  je     L14
  xor    %eax,%eax
  cmp    $0x1,%esi
  sete   %al
  add    %eax,%eax
  ret
  nop
L25:
  cmp    $0x2,%esi
  ja     L18
  neg    %esi
  sbb    %eax,%eax
  and    $0x3,%eax
  ret
  nopl   (%rax)
L18:
  xor    %eax,%eax
  cmp    $0x4,%esi
  sete   %al
  lea    (%rax,%rax,2),%eax
  ret
  nopl   0x0(%rax)
L14:
  mov    $0x3,%eax
  ret
```

assistant: ```c
unsigned int altera_a10_cdr_lookup_lf_resistor_pd(unsigned int fvco, unsigned int lpd) {
  if (fvco < 7000000) {
    switch (lpd) {
    case 1:
      return 2;
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  } else {
    switch (lpd) {
    case 1:
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<altera_a10_cdr_lookup_lf_resistor_pd>:
  endbr64
  cmp    $0x6acfbf,%edi
  ja     L12
  lea    -0x2(%rsi),%eax
  and    $0xfffffffd,%eax
  je     L17
  xor    %eax,%eax
  cmp    $0x1,%esi
  sete   %al
  add    %eax,%eax
  ret
  nop
L12:
  cmp    $0x2,%esi
  ja     L24
  neg    %esi
  sbb    %eax,%eax
  and    $0x3,%eax
  ret
  nopl   (%rax)
L24:
  xor    %eax,%eax
  cmp    $0x4,%esi
  sete   %al
  lea    (%rax,%rax,2),%eax
  ret
  nopl   0x0(%rax)
L17:
  mov    $0x3,%eax
  ret
```

assistant: ```c
unsigned int altera_a10_cdr_lookup_lf_resistor_pd(unsigned int fvco, unsigned int lpd) {
  if (fvco < 7000000) {
    switch (lpd) {
    case 1:
      return 2;
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  } else {
    switch (lpd) {
    case 1:
    case 2:
    case 4:
      return 3;
    default:
      return 0;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x0,-0x14(%rbp)
  lea    D23(%rip),%rax
  mov    %rax,-0x10(%rbp)
  lea    D15(%rip),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L26
L28:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <processRoom@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L26
  movl   $0x1,-0x14(%rbp)
L26:
  cmpl   $0x0,-0x14(%rbp)
  je     L28
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D23
data_type: string
name: retrieve_data
arguments: label: D15
data_type: string
name: retrieve_data
arguments: label: D0
data_type: string
name: retrieve_data
arguments: label: D33
data_type: string
name: retrieve_data
arguments: label: D28
data_type: string

tool: D23: 0001

tool: D15: r

tool: D0: adventure.csv

tool: D33: WELCOME TO JORK, the hyperinteractive text game


tool: D28: THE GAME IS OVER. GO HOME.


assistant: ```c
int main() {
  int game_over = 0;
  char *current_room = "0001";
  FILE *adventure_file = fopen("adventure.csv", "r");
  printf("WELCOME TO JORK, the hyperinteractive text game\n");
  while (!game_over) {
    current_room = processRoom(adventure_file, current_room);
    if (!current_room) {
      game_over = 1;
    }
  }
  printf("THE GAME IS OVER. GO HOME.\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  lea    D26(%rip),%rsi
  lea    D48(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rsi
L8:
  mov    %rbx,%rdi
  call   <processRoom@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L8
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: label: D26
type: string
name: retrieve_data
arguments: label: D48
type: string
name: retrieve_data
arguments: label: D32
type: string
name: retrieve_data
arguments: label: D14
type: string
name: retrieve_data
arguments: label: D13
type: string

tool: D26: r

tool: D48: adventure.csv

tool: D32: WELCOME TO JORK, the hyperinteractive text game


tool: D14: 0001

tool: D13: THE GAME IS OVER. GO HOME.


assistant: ```c
int main() {
  int game_over = 0;
  char *current_room = "0001";
  FILE *adventure_file = fopen("adventure.csv", "r");
  printf("WELCOME TO JORK, the hyperinteractive text game\n");
  while (!game_over) {
    current_room = processRoom(adventure_file, current_room);
    if (!current_room) {
      game_over = 1;
    }
  }
  printf("THE GAME IS OVER. GO HOME.\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D34(%rip),%rsi
  lea    D5(%rip),%rdi
  call   <fopen@plt>
  lea    D1(%rip),%rdi
  mov    %rax,%rbx
  call   <printf@plt>
  lea    D28(%rip),%rsi
  xchg   %ax,%ax
L6:
  mov    %rbx,%rdi
  call   <processRoom@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L6
  lea    D36(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D34
type_infered: string
name: extract_symbolic_data
arguments: label: D5
type_infered: string
name: extract_symbolic_data
arguments: label: D1
type_infered: string
name: extract_symbolic_data
arguments: label: D28
type_infered: string
name: extract_symbolic_data
arguments: label: D36
type_infered: string

tool: D34: r

tool: D5: adventure.csv

tool: D1: WELCOME TO JORK, the hyperinteractive text game


tool: D28: 0001

tool: D36: THE GAME IS OVER. GO HOME.


assistant: ```c
int main() {
  int game_over = 0;
  char *current_room = "0001";
  FILE *adventure_file = fopen("adventure.csv", "r");
  printf("WELCOME TO JORK, the hyperinteractive text game\n");
  while (!game_over) {
    current_room = processRoom(adventure_file, current_room);
    if (!current_room) {
      game_over = 1;
    }
  }
  printf("THE GAME IS OVER. GO HOME.\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D27(%rip),%rsi
  lea    D26(%rip),%rdi
  call   <fopen@plt>
  lea    D43(%rip),%rdi
  mov    %rax,%rbx
  call   <printf@plt>
  lea    D35(%rip),%rsi
  xchg   %ax,%ax
L5:
  mov    %rbx,%rdi
  call   <processRoom@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L5
  lea    D16(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: tag: D27
type_guessed: string
name: parse_data
arguments: tag: D26
type_guessed: string
name: parse_data
arguments: tag: D43
type_guessed: string
name: parse_data
arguments: tag: D35
type_guessed: string
name: parse_data
arguments: tag: D16
type_guessed: string

tool: D27: r

tool: D26: adventure.csv

tool: D43: WELCOME TO JORK, the hyperinteractive text game


tool: D35: 0001

tool: D16: THE GAME IS OVER. GO HOME.


assistant: ```c
int main() {
  int game_over = 0;
  char *current_room = "0001";
  FILE *adventure_file = fopen("adventure.csv", "r");
  printf("WELCOME TO JORK, the hyperinteractive text game\n");
  while (!game_over) {
    current_room = processRoom(adventure_file, current_room);
    if (!current_room) {
      game_over = 1;
    }
  }
  printf("THE GAME IS OVER. GO HOME.\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
data_type: string

tool: D0: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_guessed: string

tool: D2: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
datatype: string

tool: D6: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_infered: string

tool: D3: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sunday_search>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x240,%rsp
  mov    %rdi,-0x238(%rbp)
  mov    %rsi,-0x240(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x210(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x40,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,-0x220(%rbp)
  movl   $0x0,-0x21c(%rbp)
  movl   $0x0,-0x218(%rbp)
  mov    -0x240(%rbp),%rdx
  lea    -0x210(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <set_offset2tail@plt>
  mov    %eax,-0x214(%rbp)
L26:
  mov    -0x218(%rbp),%eax
  mov    %eax,-0x220(%rbp)
  movl   $0x0,-0x21c(%rbp)
  jmp    L67
L66:
  mov    -0x21c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x240(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x221(%rbp)
  jne    L79
  addl   $0x1,-0x220(%rbp)
  addl   $0x1,-0x21c(%rbp)
L67:
  mov    -0x220(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x238(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x221(%rbp)
  cmpb   $0x0,-0x221(%rbp)
  jne    L66
  jmp    L43
L79:
  nop
L43:
  mov    -0x21c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x240(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L117
  mov    -0x218(%rbp),%eax
  jmp    L95
L117:
  cmpb   $0x0,-0x221(%rbp)
  jne    L82
  mov    $0xffffffff,%eax
  jmp    L95
L82:
  mov    -0x218(%rbp),%edx
  mov    -0x214(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x238(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x221(%rbp)
  movsbl -0x221(%rbp),%eax
  cltq
  mov    -0x210(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L46
  movsbl -0x221(%rbp),%eax
  cltq
  mov    -0x210(%rbp,%rax,4),%eax
  jmp    L68
L46:
  mov    -0x214(%rbp),%eax
L68:
  add    %eax,-0x218(%rbp)
  jmp    L26
L95:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
int sunday_search(const char *str1, const char *str2) {
  int pos_in_str2[128] = {0};
  int idx1 = 0;
  int idx2 = 0;
  int base_idx = 0;
  char nextchar;
  int len2 = set_offset2tail(pos_in_str2, str2);
  while (1) {
    for (idx1 = base_idx, idx2 = 0; (nextchar = str1[idx1]) != '\0'; idx1++, idx2++) {
      if (nextchar != str2[idx2])
        break;
    }
    if (str2[idx2] == '\0') {
      return base_idx;
    }
    if (nextchar == '\0') {
      return -1;
    }
    nextchar = str1[base_idx + len2];
    base_idx += (pos_in_str2[nextchar] == 0) ? len2 : pos_in_str2[nextchar];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sunday_search>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x218,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x40,%ecx
  rep stos %rax,%es:(%rdi)
  mov    %rsp,%rdi
  call   <set_offset2tail@plt>
  mov    %eax,%r8d
  mov    $0x0,%eax
  mov    $0x0,%r9d
  jmp    L66
L74:
  mov    %r9d,%esi
L1:
  movslq %esi,%rsi
  cmpb   $0x0,(%rbx,%rsi,1)
  je     L67
  test   %cl,%cl
  je     L29
  lea    (%rax,%r8,1),%edx
  movslq %edx,%rdx
  movsbq 0x0(%rbp,%rdx,1),%rdx
  mov    (%rsp,%rdx,4),%edx
  test   %edx,%edx
  cmove  %r8d,%edx
  add    %edx,%eax
L66:
  movslq %eax,%rdi
  add    %rbp,%rdi
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L74
  mov    $0x1,%edx
L50:
  lea    -0x1(%rdx),%esi
  cmp    %cl,-0x1(%rbx,%rdx,1)
  jne    L1
  mov    %edx,%esi
  add    $0x1,%rdx
  movzbl -0x1(%rdi,%rdx,1),%ecx
  test   %cl,%cl
  jne    L50
  jmp    L1
L29:
  mov    $0xffffffff,%eax
L67:
  mov    0x208(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L61
  add    $0x218,%rsp
  pop    %rbx
  pop    %rbp
  ret
L61:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sunday_search(const char *str1, const char *str2) {
  int pos_in_str2[128] = {0};
  int idx1 = 0;
  int idx2 = 0;
  int base_idx = 0;
  char nextchar;
  int len2 = set_offset2tail(pos_in_str2, str2);
  while (1) {
    for (idx1 = base_idx, idx2 = 0; (nextchar = str1[idx1]) != '\0'; idx1++, idx2++) {
      if (nextchar != str2[idx2])
        break;
    }
    if (str2[idx2] == '\0') {
      return base_idx;
    }
    if (nextchar == '\0') {
      return -1;
    }
    nextchar = str1[base_idx + len2];
    base_idx += (pos_in_str2[nextchar] == 0) ? len2 : pos_in_str2[nextchar];
  }
}
```