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
  sub    $0xb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xa4(%rbp)
  movl   $0x0,-0xa0(%rbp)
  movl   $0x0,-0x9c(%rbp)
  lea    D33(%rip),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x98(%rbp)
  cmpq   $0x0,-0x98(%rbp)
  jne    L92
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L98
L89:
  movl   $0x0,-0xa4(%rbp)
  mov    -0xa4(%rbp),%eax
  mov    %eax,-0xa8(%rbp)
  movl   $0x0,-0xac(%rbp)
  jmp    L7
L44:
  mov    -0xac(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L47
  cmpl   $0x0,-0xa8(%rbp)
  jne    L61
  movl   $0x1,-0xa8(%rbp)
L61:
  mov    -0xa4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  mov    -0xac(%rbp),%edx
  movslq %edx,%rdx
  movzbl -0x70(%rbp,%rdx,1),%edx
  cltq
  mov    %dl,-0x90(%rbp,%rax,1)
  jmp    L18
L47:
  cmpl   $0x0,-0xa8(%rbp)
  je     L18
  movl   $0x0,-0xa8(%rbp)
  mov    -0xa4(%rbp),%eax
  cltq
  movb   $0x0,-0x90(%rbp,%rax,1)
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <isAllowedWord@plt>
  test   %rax,%rax
  je     L106
  lea    -0x90(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xa0(%rbp)
L106:
  addl   $0x1,-0x9c(%rbp)
  movl   $0x0,-0xa4(%rbp)
L18:
  addl   $0x1,-0xac(%rbp)
L7:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xac(%rbp)
  jl     L44
L92:
  mov    -0x98(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  mov    $0x64,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L89
  mov    -0xa0(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x9c(%rbp),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x98(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L98:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D33
type: string
name: extract_symbolic_data
arguments: label: D14
type: string
name: extract_symbolic_data
arguments: label: D24
type: string
name: extract_symbolic_data
arguments: label: D22
type: string
name: extract_symbolic_data
arguments: label: D5
type: string
name: extract_symbolic_data
arguments: label: D38
type: string

tool: D33: r

tool: D14: D:\_strukturno\vlez.txt

tool: D24: Vlez.txt can not be opened!

tool: D22: %s 

tool: D5: Total allowed words: %d


tool: D38: Total words: %d


assistant: ```c
int main() {
  int i, inWord;
  int count = 0, totalallowed = 0, totalwords = 0;
  char row[100], zbor[20];
  FILE *ptr;
  if ((ptr = fopen("D:\\_strukturno\\vlez.txt", "r")) == NULL) {
    printf("Vlez.txt can not be opened!");
    return -1;
  }
  while (fgets(row, 100, ptr) != NULL) {
    inWord = count = 0;
    for (i = 0; i < strlen(row); i++) {
      if (isalnum(row[i])) {
        if (!inWord) {
          inWord = 1;
        }
        zbor[count++] = row[i];
      } else if (inWord) {
        inWord = 0;
        zbor[count] = '\0';
        if (isAllowedWord(zbor)) {
          printf("%s ", zbor);
          totalallowed++;
        }
        totalwords++;
        count = 0;
      }
    }
  }
  printf("Total allowed words: %d\n", totalallowed);
  printf("Total words: %d\n", totalwords);
  fclose(ptr);
  return 0;
}
```