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
<get_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L28
L25:
  addl   $0x1,-0x4(%rbp)
  addq   $0x1,-0x18(%rbp)
L28:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    -0x18(%rbp),%rdx
  mov    %al,(%rdx)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L1
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L25
L1:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int get_line(char *line, int max_line_length) {
  int length = 0;
  while ((*line = getchar()) != '\n' && length <= max_line_length) {
    ++length;
    ++line;
  }
  *line = '\0';
  return length;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_line>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    $0x0,%ebp
L11:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  cmp    %r12d,%ebp
  jg     L13
  cmp    $0xa,%al
  je     L13
  add    $0x1,%ebp
  add    $0x1,%rbx
  jmp    L11
L13:
  movb   $0x0,(%rbx)
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_line(char *line, int max_line_length) {
  int length = 0;
  while ((*line = getchar()) != '\n' && length <= max_line_length) {
    ++length;
    ++line;
  }
  *line = '\0';
  return length;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_line>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L8:
  cmp    $0xa,%al
  je     L21
  add    $0x1,%r12d
  add    $0x1,%rbx
L23:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  cmp    %ebp,%r12d
  jle    L8
L21:
  mov    %r12d,%eax
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_line(char *line, int max_line_length) {
  int length = 0;
  while ((*line = getchar()) != '\n' && length <= max_line_length) {
    ++length;
    ++line;
  }
  *line = '\0';
  return length;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_line>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L22:
  cmp    $0xa,%al
  je     L13
  add    $0x1,%r12d
  add    $0x1,%rbx
L15:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  cmp    %ebp,%r12d
  jle    L22
L13:
  mov    %r12d,%eax
  movb   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_line(char *line, int max_line_length) {
  int length = 0;
  while ((*line = getchar()) != '\n' && length <= max_line_length) {
    ++length;
    ++line;
  }
  *line = '\0';
  return length;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x31,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_guessed: string
name: retrieve_data
arguments: tag: D14
type_guessed: string
name: retrieve_data
arguments: tag: D77
type_guessed: string

tool: D2: 123456789

tool: D14: 123123

tool: D77: ===============

assistant: ```c
int main() {
  printf("%s\n", ho_strrchr("123456789", '1'));
  printf("%s\n", ho_strrchr("123456789", '2'));
  printf("%s\n", ho_strrchr("123456789", '3'));
  printf("%s\n", ho_strrchr("123456789", '9'));
  printf("%s\n", ho_strrchr("123456789", 'a'));
  printf("%s\n", ho_strrchr("123123", '1'));
  printf("===============\n");
  printf("%s\n", ho_strrchr2("123456789", '1'));
  printf("%s\n", ho_strrchr2("123456789", '2'));
  printf("%s\n", ho_strrchr2("123456789", '3'));
  printf("%s\n", ho_strrchr2("123456789", '9'));
  printf("%s\n", ho_strrchr2("123456789", 'a'));
  printf("%s\n", ho_strrchr2("123123", '1'));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x31,%esi
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  lea    D10(%rip),%rbp
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D21(%rip),%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  mov    %rbx,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D10
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D21
type_guessed: string

tool: D7: 123456789

tool: D10: 123123

tool: D21: ===============

assistant: ```c
int main() {
  printf("%s\n", ho_strrchr("123456789", '1'));
  printf("%s\n", ho_strrchr("123456789", '2'));
  printf("%s\n", ho_strrchr("123456789", '3'));
  printf("%s\n", ho_strrchr("123456789", '9'));
  printf("%s\n", ho_strrchr("123456789", 'a'));
  printf("%s\n", ho_strrchr("123123", '1'));
  printf("===============\n");
  printf("%s\n", ho_strrchr2("123456789", '1'));
  printf("%s\n", ho_strrchr2("123456789", '2'));
  printf("%s\n", ho_strrchr2("123456789", '3'));
  printf("%s\n", ho_strrchr2("123456789", '9'));
  printf("%s\n", ho_strrchr2("123456789", 'a'));
  printf("%s\n", ho_strrchr2("123123", '1'));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x31,%esi
  lea    D3(%rip),%r12
  push   %rbp
  lea    D23(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  mov    %r12,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  mov    $0x31,%esi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_guessed: string
name: parse_data
arguments: data_tag: D23
type_guessed: string
name: parse_data
arguments: data_tag: D11
type_guessed: string

tool: D3: 123123

tool: D23: 123456789

tool: D11: ===============

assistant: ```c
int main() {
  printf("%s\n", ho_strrchr("123456789", '1'));
  printf("%s\n", ho_strrchr("123456789", '2'));
  printf("%s\n", ho_strrchr("123456789", '3'));
  printf("%s\n", ho_strrchr("123456789", '9'));
  printf("%s\n", ho_strrchr("123456789", 'a'));
  printf("%s\n", ho_strrchr("123123", '1'));
  printf("===============\n");
  printf("%s\n", ho_strrchr2("123456789", '1'));
  printf("%s\n", ho_strrchr2("123456789", '2'));
  printf("%s\n", ho_strrchr2("123456789", '3'));
  printf("%s\n", ho_strrchr2("123456789", '9'));
  printf("%s\n", ho_strrchr2("123456789", 'a'));
  printf("%s\n", ho_strrchr2("123123", '1'));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x31,%esi
  lea    D20(%rip),%r12
  push   %rbp
  lea    D29(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  mov    %r12,%rdi
  call   <ho_strrchr@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x31,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x32,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x33,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x39,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x61,%esi
  mov    %rbp,%rdi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  mov    $0x31,%esi
  call   <ho_strrchr2@plt>
  mov    %rax,%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D20
type_infered: string
name: parse_data
arguments: tag: D29
type_infered: string
name: parse_data
arguments: tag: D11
type_infered: string

tool: D20: 123123

tool: D29: 123456789

tool: D11: ===============

assistant: ```c
int main() {
  printf("%s\n", ho_strrchr("123456789", '1'));
  printf("%s\n", ho_strrchr("123456789", '2'));
  printf("%s\n", ho_strrchr("123456789", '3'));
  printf("%s\n", ho_strrchr("123456789", '9'));
  printf("%s\n", ho_strrchr("123456789", 'a'));
  printf("%s\n", ho_strrchr("123123", '1'));
  printf("===============\n");
  printf("%s\n", ho_strrchr2("123456789", '1'));
  printf("%s\n", ho_strrchr2("123456789", '2'));
  printf("%s\n", ho_strrchr2("123456789", '3'));
  printf("%s\n", ho_strrchr2("123456789", '9'));
  printf("%s\n", ho_strrchr2("123456789", 'a'));
  printf("%s\n", ho_strrchr2("123123", '1'));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_executable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  je     L66
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <warn@plt>
  mov    $0x0,%eax
  jmp    L41
L66:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <S_ISREG@plt>
  test   %eax,%eax
  jne    L54
  mov    $0x0,%eax
  jmp    L41
L54:
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <access@plt>
  test   %rax,%rax
  je     L26
  mov    $0x0,%eax
  jmp    L41
L26:
  mov    $0x1,%eax
L41:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int is_executable(const char *path) {
  struct stat stbuf;
  if (stat(path, &stbuf)) {
    warn(path);
    return 0;
  }
  if (!S_ISREG(stbuf.st_mode)) {
    return 0;
  }
  if (access(path, X_OK)) {
    return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_executable>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %rax,%rax
  jne    L11
  mov    0x4(%rsp),%edi
  call   <S_ISREG@plt>
  test   %eax,%eax
  je     L34
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  mov    %rbx,%rdi
  call   <access@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L34
L11:
  mov    %rbx,%rdi
  call   <warn@plt>
  mov    $0x0,%eax
L34:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x10,%rsp
  pop    %rbx
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_executable(const char *path) {
  struct stat stbuf;
  if (stat(path, &stbuf)) {
    warn(path);
    return 0;
  }
  if (!S_ISREG(stbuf.st_mode)) {
    return 0;
  }
  if (access(path, X_OK)) {
    return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_executable>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %rax,%rax
  jne    L24
  mov    0x4(%rsp),%edi
  call   <S_ISREG@plt>
  test   %eax,%eax
  jne    L29
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L49
  add    $0x10,%rsp
  pop    %rbp
  ret
  xchg   %ax,%ax
L29:
  mov    D1(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <access@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L7
  nopl   0x0(%rax)
L24:
  mov    %rbp,%rdi
  call   <warn@plt>
  xor    %eax,%eax
  jmp    L7
L49:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_executable(const char *path) {
  struct stat stbuf;
  if (stat(path, &stbuf)) {
    warn(path);
    return 0;
  }
  if (!S_ISREG(stbuf.st_mode)) {
    return 0;
  }
  if (access(path, X_OK)) {
    return 0;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<is_executable>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %rax,%rax
  jne    L4
  mov    0x4(%rsp),%edi
  call   <S_ISREG@plt>
  test   %eax,%eax
  jne    L18
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbp
  ret
  xchg   %ax,%ax
L18:
  mov    D5(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <access@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L7
  nopl   0x0(%rax)
L4:
  mov    %rbp,%rdi
  call   <warn@plt>
  xor    %eax,%eax
  jmp    L7
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_executable(const char *path) {
  struct stat stbuf;
  if (stat(path, &stbuf)) {
    warn(path);
    return 0;
  }
  if (!S_ISREG(stbuf.st_mode)) {
    return 0;
  }
  if (access(path, X_OK)) {
    return 0;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x7,%edi
  call   <bitcount@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type: string

tool: D9: Bitcount -> %d

assistant: ```c
int main(void) {
  printf("Bitcount -> %d", bitcount(7));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%edi
  call   <bitcount@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D9: Bitcount -> %d

assistant: ```c
int main(void) {
  printf("Bitcount -> %d", bitcount(7));
  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
  mov    $0x7,%edi
  call   <bitcount@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
data_type: string

tool: D8: Bitcount -> %d

assistant: ```c
int main(void) {
  printf("Bitcount -> %d", bitcount(7));
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%edi
  call   <bitcount@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
type_guessed: string

tool: D8: Bitcount -> %d

assistant: ```c
int main(void) {
  printf("Bitcount -> %d", bitcount(7));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<delay>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L5:
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jb     L5
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void delay(unsigned int delta) {
  unsigned int t = 0;
  while (t < delta) {
    t++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<delay>:
  endbr64
  test   %edi,%edi
  je     L18
  mov    $0x0,%eax
L1:
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L1
L18:
  ret
```

assistant: ```c
void delay(unsigned int delta) {
  unsigned int t = 0;
  while (t < delta) {
    t++;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<delay>:
  endbr64
  ret
```

assistant: ```c
void delay(unsigned int delta) {
  unsigned int t = 0;
  while (t < delta) {
    t++;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<delay>:
  endbr64
  ret
```

assistant: ```c
void delay(unsigned int delta) {
  unsigned int t = 0;
  while (t < delta) {
    t++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multiply189>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L34
L11:
  movl   $0x0,-0x3c(%rbp)
  jmp    L14
L12:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L14:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L12
  addl   $0x1,-0x40(%rbp)
L34:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L11
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply189(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<multiply189>:
  endbr64
  test   %edi,%edi
  jle    L2
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L9:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L27:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L27
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  ret
```

assistant: ```c
void multiply189(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<multiply189>:
  endbr64
  test   %edi,%edi
  jle    L1
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L27:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L27
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L4
  pop    %rbx
  ret
L1:
  ret
```

assistant: ```c
void multiply189(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```