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
<so_ctor>:
  endbr64
  sub    $0x8,%rsp
  lea    D46(%rip),%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D29(%rip),%rsi
  mov    $0x0,%edi
  call   <dlsym@plt>
  test   %rax,%rax
  je     L14
  mov    (%rax),%esi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L4:
  add    $0x8,%rsp
  ret
L14:
  lea    D12(%rip),%rdi
  call   <puts@plt>
  jmp    L4
```

tool_calls: name: read_data
arguments: data_label: D46
type_guessed: string
name: read_data
arguments: data_label: D16
type_guessed: string
name: read_data
arguments: data_label: D29
type_guessed: string
name: read_data
arguments: data_label: D38
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string

tool: D46: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/frdeso/fun-with-so/refs/heads/master/aaa.c

tool: D16: %s: Constructor


tool: D29: my_variable

tool: D38: my_variable = %d



tool: D12: dlsym error

assistant: ```c
void __attribute__((constructor)) so_ctor(void) {
  int *my_variable_sym;
  printf("%s: Constructor\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/frdeso/fun-with-so/refs/heads/master/aaa.c");
  void *obj = dlsym(NULL, "my_variable");
  if (obj == NULL) {
    printf("dlsym error\n");
    return;
  }
  my_variable_sym = (int *)obj;
  printf("my_variable = %d\n\n", *my_variable_sym);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<so_ctor>:
  endbr64
  sub    $0x8,%rsp
  lea    D20(%rip),%rsi
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  lea    D9(%rip),%rsi
  call   <dlsym@plt>
  test   %rax,%rax
  je     L1
  mov    (%rax),%esi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1080 <printf@plt>
L1:
  lea    D45(%rip),%rdi
  pop    %rax
  jmp    1070 <puts@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D20
data_type: string
name: retrieve_data
arguments: tag: D23
data_type: string
name: retrieve_data
arguments: tag: D9
data_type: string
name: retrieve_data
arguments: tag: D30
data_type: string
name: retrieve_data
arguments: tag: D45
data_type: string

tool: D20: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/frdeso/fun-with-so/refs/heads/master/aaa.c

tool: D23: %s: Constructor


tool: D9: my_variable

tool: D30: my_variable = %d



tool: D45: dlsym error

assistant: ```c
void __attribute__((constructor)) so_ctor(void) {
  int *my_variable_sym;
  printf("%s: Constructor\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/frdeso/fun-with-so/refs/heads/master/aaa.c");
  void *obj = dlsym(NULL, "my_variable");
  if (obj == NULL) {
    printf("dlsym error\n");
    return;
  }
  my_variable_sym = (int *)obj;
  printf("my_variable = %d\n\n", *my_variable_sym);
}
```
user: What is the c source code of the assembly code below?

```asm
<so_ctor>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  lea    D8(%rip),%rsi
  call   <dlsym@plt>
  test   %rax,%rax
  je     L3
  mov    (%rax),%esi
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1080 <printf@plt>
L3:
  lea    D9(%rip),%rdi
  pop    %rax
  jmp    1070 <puts@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: string
name: unpack_data
arguments: data_tag: D16
datatype: string
name: unpack_data
arguments: data_tag: D8
datatype: string
name: unpack_data
arguments: data_tag: D43
datatype: string
name: unpack_data
arguments: data_tag: D9
datatype: string

tool: D0: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/frdeso/fun-with-so/refs/heads/master/aaa.c

tool: D16: %s: Constructor


tool: D8: my_variable

tool: D43: my_variable = %d



tool: D9: dlsym error

assistant: ```c
void __attribute__((constructor)) so_ctor(void) {
  int *my_variable_sym;
  printf("%s: Constructor\n", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/frdeso/fun-with-so/refs/heads/master/aaa.c");
  void *obj = dlsym(NULL, "my_variable");
  if (obj == NULL) {
    printf("dlsym error\n");
    return;
  }
  my_variable_sym = (int *)obj;
  printf("my_variable = %d\n\n", *my_variable_sym);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<modular_addition_size_t_unsafe>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jbe    L0
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  jmp    L19
L0:
  mov    -0x20(%rbp),%rax
  sub    -0x8(%rbp),%rax
L19:
  pop    %rbp
  ret
```

assistant: ```c
size_t modular_addition_size_t_unsafe(size_t x, size_t y, size_t m) {
  size_t z;
  z = m - x;
  if (z > y) {
    return x + y;
  } else {
    return y - z;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<modular_addition_size_t_unsafe>:
  endbr64
  mov    %rdx,%r8
  sub    %rdi,%r8
  lea    (%rdi,%rsi,1),%rcx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,%rdi
  cmp    %rsi,%r8
  mov    %rcx,%rax
  cmovbe %rdi,%rax
  ret
```

assistant: ```c
size_t modular_addition_size_t_unsafe(size_t x, size_t y, size_t m) {
  size_t z;
  z = m - x;
  if (z > y) {
    return x + y;
  } else {
    return y - z;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<modular_addition_size_t_unsafe>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  mov    %rdx,%r8
  sub    %rdi,%r8
  mov    %rax,%rcx
  sub    %rdx,%rcx
  cmp    %rsi,%r8
  cmovbe %rcx,%rax
  ret
```

assistant: ```c
size_t modular_addition_size_t_unsafe(size_t x, size_t y, size_t m) {
  size_t z;
  z = m - x;
  if (z > y) {
    return x + y;
  } else {
    return y - z;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<modular_addition_size_t_unsafe>:
  endbr64
  lea    (%rdi,%rsi,1),%rax
  mov    %rdx,%r8
  sub    %rdi,%r8
  mov    %rax,%rcx
  sub    %rdx,%rcx
  cmp    %rsi,%r8
  cmovbe %rcx,%rax
  ret
```

assistant: ```c
size_t modular_addition_size_t_unsafe(size_t x, size_t y, size_t m) {
  size_t z;
  z = m - x;
  if (z > y) {
    return x + y;
  } else {
    return y - z;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<F_FstC>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L11:
  addl   $0x1,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L3
  mov    $0x0,%eax
  jmp    L24
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jle    L11
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
L24:
  leave
  ret
```

assistant: ```c
char F_FstC(char *str, char **n) {
  int i = 0;
  while (str[i] <= 32) {
    i++;
    if (i >= strlen(str))
      return 0;
  }
  *n = str + i;
  return str[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<F_FstC>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%ebx
L5:
  lea    0x0(%rbp,%rbx,1),%rax
  cmpb   $0x20,0x0(%rbp,%rbx,1)
  jg     L14
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    $0x1,%rbx
  cmp    %ebx,%eax
  jg     L5
  mov    $0x0,%eax
  jmp    L4
L14:
  mov    %rax,(%r12)
  movzbl (%rax),%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char F_FstC(char *str, char **n) {
  int i = 0;
  while (str[i] <= 32) {
    i++;
    if (i >= strlen(str))
      return 0;
  }
  *n = str + i;
  return str[i];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<F_FstC>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L16:
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L13
L27:
  cmpb   $0x20,0x0(%rbp,%rbx,1)
  lea    0x0(%rbp,%rbx,1),%rax
  jle    L16
  mov    %rax,(%r12)
  pop    %rbx
  movzbl (%rax),%eax
  pop    %rbp
  pop    %r12
  ret
L13:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char F_FstC(char *str, char **n) {
  int i = 0;
  while (str[i] <= 32) {
    i++;
    if (i >= strlen(str))
      return 0;
  }
  *n = str + i;
  return str[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<F_FstC>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L4
L18:
  cmpb   $0x20,0x0(%rbp,%rbx,1)
  lea    0x0(%rbp,%rbx,1),%rax
  jle    L7
  mov    %rax,(%r12)
  pop    %rbx
  movzbl (%rax),%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char F_FstC(char *str, char **n) {
  int i = 0;
  while (str[i] <= 32) {
    i++;
    if (i >= strlen(str))
      return 0;
  }
  *n = str + i;
  return str[i];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<createIdPw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D13(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L19
  mov    $0xffffffff,%eax
  jmp    L9
L19:
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D4(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x1,%eax
L9:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D13
datatype: string
name: parse_data
arguments: data_tag: D4
datatype: string

tool: D13: w+

tool: D4: %s

assistant: ```c
int createIdPw(const char *id, const char *pw) {
  FILE *f;
  f = fopen(id, "w+");
  if (f == NULL)
    return -1;
  fprintf(f, "%s", pw);
  fclose(f);
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<createIdPw>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  lea    D13(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbx
  mov    $0xffffffff,%eax
  test   %rbx,%rbx
  je     L2
  mov    %rbp,%rdx
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <fprintf@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    $0x1,%eax
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D13
data_type: string
name: read_data
arguments: data_tag: D1
data_type: string

tool: D13: w+

tool: D1: %s

assistant: ```c
int createIdPw(const char *id, const char *pw) {
  FILE *f;
  f = fopen(id, "w+");
  if (f == NULL)
    return -1;
  fprintf(f, "%s", pw);
  fclose(f);
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<createIdPw>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    D3(%rip),%rsi
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    $0xffffffff,%eax
  test   %rbp,%rbp
  je     L9
  mov    %r12,%rdx
  lea    D16(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    $0x1,%eax
L9:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string
name: extract_symbolic_data
arguments: data_label: D16
data_type: string

tool: D3: w+

tool: D16: %s

assistant: ```c
int createIdPw(const char *id, const char *pw) {
  FILE *f;
  f = fopen(id, "w+");
  if (f == NULL)
    return -1;
  fprintf(f, "%s", pw);
  fclose(f);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<createIdPw>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    D3(%rip),%rsi
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    $0xffffffff,%eax
  test   %rbp,%rbp
  je     L9
  mov    %r12,%rdx
  lea    D6(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    $0x1,%eax
L9:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type: string
name: disassemble_data
arguments: data_label: D6
type: string

tool: D3: w+

tool: D6: %s

assistant: ```c
int createIdPw(const char *id, const char *pw) {
  FILE *f;
  f = fopen(id, "w+");
  if (f == NULL)
    return -1;
  fprintf(f, "%s", pw);
  fclose(f);
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D68(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x9,%eax
  jle    L22
  mov    %eax,%ecx
  mov    $0x0,%r12d
L63:
  mov    %ecx,%esi
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%edi
  sar    $0x1f,%edi
  mov    %edx,%ecx
  sub    %edi,%ecx
  add    $0x1,%r12d
  cmp    $0x63,%esi
  jg     L63
  cmp    $0x8,%r12d
  jle    L51
  lea    D18(%rip),%rdi
  call   <puts@plt>
  jmp    L70
L51:
  mov    $0xa,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L70
  mov    $0x64,%ebp
  mov    $0xa,%ebx
  mov    $0x0,%r13d
  lea    D1(%rip),%r14
L77:
  mov    0x4(%rsp),%ecx
  mov    %ecx,%eax
  cltd
  idiv   %ebp
  mov    %edx,%esi
  mov    %ecx,%eax
  cltd
  idiv   %ebx
  mov    %esi,%eax
  sub    %edx,%eax
  cltd
  idiv   %ebx
  mov    %eax,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x0(%rbp,%rbp,4),%ebp
  add    %ebp,%ebp
  lea    (%rbx,%rbx,4),%ebx
  add    %ebx,%ebx
  add    $0x1,%r13d
  cmp    %r12d,%r13d
  jne    L77
L70:
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L22:
  mov    $0x0,%r12d
  test   %eax,%eax
  jns    L51
  mov    $0xa,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  negl   0x4(%rsp)
  jmp    L70
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D68
type: string
name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D18
type: string
name: extract_symbolic_data
arguments: data_tag: D1
type: string
name: extract_symbolic_data
arguments: data_tag: D8
type: string

tool: D68: Enter an integer > 

tool: D14: %d

tool: D18: 
7
4
6
3
8
4
7
4
1
2

tool: D1: %d


tool: D8: That's all, have a nice day!

assistant: ```c
int main(void) {
  int x, y, i;
  int step1 = 10, step2 = 100;
  int num;
  int max = 0;
  printf("Enter an integer > ");
  scanf("%d", &x);
  y = x;
  while (y >= 10) {
    y /= 10;
    max++;
  }
  if (max >= 9)
    printf("\n7\n4\n6\n3\n8\n4\n7\n4\n1\n2\n");
  else {
    if (x < 0) {
      printf("%d\n", x % 10);
      x = x * -1;
    } else {
      printf("%d\n", x % 10);
    }
    for (i = 0; i < max; i++) {
      num = ((x % step2 - x % step1) / step1);
      printf("%d\n", num);
      step2 *= 10;
      step1 *= 10;
    }
  }
  printf("That's all, have a nice day!\n");
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D48(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D63(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  cmp    $0x9,%r8d
  jle    L35
  mov    %r8d,%edx
  xor    %ecx,%ecx
  mov    $0xcccccccd,%edi
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    %edx,%edx
  mov    %ecx,%r13d
  add    $0x1,%ecx
  mov    %rdx,%rsi
  imul   %rdi,%rdx
  shr    $0x23,%rdx
  cmp    $0x63,%esi
  jg     L16
  cmp    $0x8,%ecx
  jg     L79
  mov    %r8d,%eax
  mov    $0xa,%ecx
  mov    $0x64,%ebp
  xor    %r14d,%r14d
  cltd
  lea    D57(%rip),%r12
  mov    $0xa,%ebx
  idiv   %ecx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %edx,%esi
  call   <printf@plt>
  nopl   0x0(%rax)
L51:
  mov    0x4(%rsp),%ecx
  mov    %r12,%rdi
  mov    %ecx,%eax
  cltd
  idiv   %ebp
  mov    %ecx,%eax
  lea    0x0(%rbp,%rbp,4),%ebp
  add    %ebp,%ebp
  mov    %edx,%esi
  cltd
  idiv   %ebx
  sub    %edx,%esi
  mov    %esi,%eax
  cltd
  idiv   %ebx
  lea    (%rbx,%rbx,4),%ebx
  add    %ebx,%ebx
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14d,%eax
  add    $0x1,%r14d
  cmp    %r13d,%eax
  jne    L51
L52:
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L79:
  lea    D60(%rip),%rdi
  call   <puts@plt>
  jmp    L52
L35:
  mov    %r8d,%eax
  mov    $0xa,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%esi
  test   %r8d,%r8d
  jns    L42
  lea    D57(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  negl   0x4(%rsp)
  jmp    L52
L42:
  lea    D57(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L52
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D48
type_infered: string
name: extract_symbolic_data
arguments: data_label: D63
type_infered: string
name: extract_symbolic_data
arguments: data_label: D57
type_infered: string
name: extract_symbolic_data
arguments: data_label: D1
type_infered: string
name: extract_symbolic_data
arguments: data_label: D60
type_infered: string

tool: D48: Enter an integer > 

tool: D63: %d

tool: D57: %d


tool: D1: That's all, have a nice day!

tool: D60: 
7
4
6
3
8
4
7
4
1
2

assistant: ```c
int main(void) {
  int x, y, i;
  int step1 = 10, step2 = 100;
  int num;
  int max = 0;
  printf("Enter an integer > ");
  scanf("%d", &x);
  y = x;
  while (y >= 10) {
    y /= 10;
    max++;
  }
  if (max >= 9)
    printf("\n7\n4\n6\n3\n8\n4\n7\n4\n1\n2\n");
  else {
    if (x < 0) {
      printf("%d\n", x % 10);
      x = x * -1;
    } else {
      printf("%d\n", x % 10);
    }
    for (i = 0; i < max; i++) {
      num = ((x % step2 - x % step1) / step1);
      printf("%d\n", num);
      step2 *= 10;
      step1 *= 10;
    }
  }
  printf("That's all, have a nice day!\n");
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r14
  lea    D59(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  cmp    $0x9,%r8d
  jle    L51
  mov    %r8d,%edx
  xor    %ecx,%ecx
  mov    $0xcccccccd,%edi
  cs nopw 0x0(%rax,%rax,1)
L39:
  mov    %edx,%edx
  mov    %ecx,%r13d
  add    $0x1,%ecx
  mov    %rdx,%rsi
  imul   %rdi,%rdx
  shr    $0x23,%rdx
  cmp    $0x63,%esi
  jg     L39
  cmp    $0x8,%ecx
  jg     L35
  mov    %r8d,%eax
  mov    $0xa,%ecx
  mov    $0x64,%ebp
  xor    %r14d,%r14d
  cltd
  lea    D68(%rip),%r12
  mov    $0xa,%ebx
  idiv   %ecx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %edx,%esi
  call   <printf@plt>
  nopl   0x0(%rax)
L45:
  mov    0x4(%rsp),%ecx
  mov    %r12,%rdi
  mov    %ecx,%eax
  cltd
  idiv   %ebp
  mov    %ecx,%eax
  lea    0x0(%rbp,%rbp,4),%ebp
  add    %ebp,%ebp
  mov    %edx,%esi
  cltd
  idiv   %ebx
  sub    %edx,%esi
  mov    %esi,%eax
  cltd
  idiv   %ebx
  lea    (%rbx,%rbx,4),%ebx
  add    %ebx,%ebx
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14d,%eax
  add    $0x1,%r14d
  cmp    %r13d,%eax
  jne    L45
L56:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  lea    D64(%rip),%rdi
  call   <puts@plt>
  jmp    L56
L51:
  mov    %r8d,%eax
  mov    $0xa,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%esi
  test   %r8d,%r8d
  jns    L47
  lea    D68(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  negl   0x4(%rsp)
  jmp    L56
L47:
  lea    D68(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L56
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D59
datatype: string
name: parse_data
arguments: label: D24
datatype: string
name: parse_data
arguments: label: D68
datatype: string
name: parse_data
arguments: label: D16
datatype: string
name: parse_data
arguments: label: D64
datatype: string

tool: D59: Enter an integer > 

tool: D24: %d

tool: D68: %d


tool: D16: That's all, have a nice day!

tool: D64: 
7
4
6
3
8
4
7
4
1
2

assistant: ```c
int main(void) {
  int x, y, i;
  int step1 = 10, step2 = 100;
  int num;
  int max = 0;
  printf("Enter an integer > ");
  scanf("%d", &x);
  y = x;
  while (y >= 10) {
    y /= 10;
    max++;
  }
  if (max >= 9)
    printf("\n7\n4\n6\n3\n8\n4\n7\n4\n1\n2\n");
  else {
    if (x < 0) {
      printf("%d\n", x % 10);
      x = x * -1;
    } else {
      printf("%d\n", x % 10);
    }
    for (i = 0; i < max; i++) {
      num = ((x % step2 - x % step1) / step1);
      printf("%d\n", num);
      step2 *= 10;
      step1 *= 10;
    }
  }
  printf("That's all, have a nice day!\n");
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strxfrm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%r12
  mov    -0x30(%rbp),%rbx
  jmp    L68
L36:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L28
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  jmp    L68
L28:
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movb   $0x0,(%rax)
L68:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  test   %rax,%rax
  jne    L36
  jmp    L66
L23:
  add    $0x1,%rbx
L66:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L23
  mov    %rbx,%rax
  sub    -0x30(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jae    L78
  mov    -0x18(%rbp),%rax
  jmp    L76
L78:
  mov    -0x38(%rbp),%rax
L76:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
size_t strxfrm(char *dst, const char *src, size_t len) {
  size_t length;
  register char *ptr1, *ptr2;
  length = len;
  ptr1 = (char *)dst;
  ptr2 = (char *)src;
  while (length--) {
    if (*ptr2)
      *ptr1++ = *ptr2++;
    else
      *ptr1++ = '\0';
  }
  while (*ptr2)
    ptr2++;
  length = (ptr2 - src);
  if (length < len)
    return (length);
  return (len);
}
```