user: 请从给定的汇编指令推导出C源代码：

```asm
<string_compare>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  movl   $0x736559,-0x10(%rbp)
  movl   $0x736579,-0xc(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L37
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L69
L32:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  addl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L15
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0x14(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L34
  lea    -0xc(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L15
L34:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  addl   $0x1,-0x18(%rbp)
L15:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L37:
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L32
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L69:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L60
  call   <__stack_chk_fail@plt>
L60:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D17
type: string
name: unpack_data
arguments: tag: D43
type: string
name: unpack_data
arguments: tag: D59
type: string
name: unpack_data
arguments: tag: D28
type: string
name: unpack_data
arguments: tag: D76
type: string
name: unpack_data
arguments: tag: D8
type: string
name: unpack_data
arguments: tag: D1
type: string

tool: D17: input : %s 


tool: D43: what should be : |%s| 


tool: D59: Correct! Your file(s) are %s


tool: D28: Do you want to get the answer? (yes or no) If no, you will continue: 

tool: D76: Lets try it now: 

tool: D8: 
input : |%s| 


tool: D1: Wrong! try again: 

assistant: ```c
void string_compare(char a[], char b[], char c[]) {
  int count = 1;
  char yes_no[4];
  char yes1[] = "Yes";
  char yes2[] = "yes";
  printf("input : %s \n", b);
  printf("what should be : |%s| \n", a);
  if ((strcmp(a, b)) == 0) {
    printf("Correct! Your file(s) are %s\n", c);
  } else {
    while (strcmp(a, b) != 0) {
      gets(b);
      count++;
      if ((count % 5) == 0) {
        printf("Do you want to get the answer? (yes or no) If no, you will continue: ");
        gets(yes_no);
        if ((strcmp(yes1, yes_no) == 0) || (strcmp(yes_no, yes2) == 0)) {
          printf("%s\n", a);
          printf("Lets try it now: ");
          gets(b);
          count++;
        }
      }
      printf("\ninput : |%s| \n", b);
      printf("Wrong! try again: ");
    }
    printf("Correct! Your file(s) are %s\n", c);
  }
}
```