user: 这段汇编代码用C语言表示是什么样的？

```asm
<getNum>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x108,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x38,%edx,%edx
  sub    %edx,%eax
  lea    0x1(%rax),%ebp
  lea    D11(%rip),%rsi
  lea    D22(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  lea    0x10(%rsp),%rdi
  mov    $0x1c,%ecx
  mov    $0x0,%eax
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  lea    D1(%rip),%r12
  jmp    L91
L78:
  lea    0xc(%rsp),%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <fscanf@plt>
  movslq 0xc(%rsp),%rax
  movl   $0x1,0x10(%rsp,%rax,4)
L91:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L78
  mov    %rbx,%rdi
  call   <fclose@plt>
  lea    0x14(%rsp),%rbx
  lea    0xf4(%rsp),%r12
  mov    %rbx,%rax
L144:
  cmpl   $0x0,(%rax)
  je     L24
  add    $0x4,%rax
  cmp    %rax,%r12
  jne    L144
  lea    D76(%rip),%rsi
  lea    D22(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rdi
  call   <fclose@plt>
L54:
  movl   $0x0,(%rbx)
  add    $0x4,%rbx
  cmp    %rbx,%r12
  jne    L54
L24:
  test   %ebp,%ebp
  jne    L108
  jmp    L123
L143:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x38,%edx,%edx
  sub    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%ebp
  je     L123
L108:
  movslq %ebp,%rax
  mov    0x10(%rsp,%rax,4),%edx
  mov    $0x38,%eax
L55:
  cmp    $0x1,%edx
  je     L143
  sub    $0x1,%eax
  jne    L55
  movslq %ebp,%rax
  movl   $0x1,0x10(%rsp,%rax,4)
L123:
  lea    D76(%rip),%rsi
  lea    D22(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%r12
  mov    $0x1,%ebx
  lea    D10(%rip),%r13
  jmp    L57
L49:
  add    $0x1,%rbx
  cmp    $0x39,%rbx
  je     L40
L57:
  mov    %ebx,%edx
  cmpl   $0x1,0x10(%rsp,%rbx,4)
  jne    L49
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  jmp    L49
L40:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0xf8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    %ebp,%eax
  add    $0x108,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D11
datatype: string
name: parse_data
arguments: label: D22
datatype: string
name: parse_data
arguments: label: D1
datatype: string
name: parse_data
arguments: label: D76
datatype: string
name: parse_data
arguments: label: D10
datatype: string

tool: D11: r

tool: D22: previous.dat

tool: D1: %d

tool: D76: w

tool: D10: %d


assistant: ```c
int getNum() {
  int i;
  srand(time(NULL));
  int temp = (rand() % 56) + 1;
  int finalNum = 0;
  FILE *fp3;
  fp3 = fopen("previous.dat", "r");
  int readNums[57] = {0};
  int arrayTemp;
  while (!feof(fp3)) {
    fscanf(fp3, "%d", &arrayTemp);
    readNums[arrayTemp] = 1;
  }
  fclose(fp3);
  int nothingLeft = 1;
  for (i = 1; i < 57; ++i) {
    if (readNums[i] == 0) {
      nothingLeft = 0;
      break;
    }
  }
  if (nothingLeft == 1) {
    fp3 = fopen("previous.dat", "w");
    fprintf(fp3, "%s", "");
    fclose(fp3);
    for (i = 1; i < 57; ++i) {
      readNums[i] = 0;
    }
  }
  int isFinal = 1;
  while (finalNum != temp) {
    isFinal = 1;
    for (i = 1; i < 57; ++i) {
      if (readNums[temp] == 1) {
        isFinal = 0;
        temp = (rand() % 56) + 1;
        break;
      }
    }
    if (isFinal == 1) {
      finalNum = temp;
      readNums[finalNum] = 1;
    }
  }
  fp3 = fopen("previous.dat", "w");
  for (i = 1; i < 57; ++i) {
    if (readNums[i] == 1) {
      fprintf(fp3, "%d\n", i);
    }
  }
  fclose(fp3);
  return finalNum;
}
```