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    $0x440,%rsp
  mov    %edi,-0x434(%rbp)
  mov    %rsi,-0x440(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x434(%rbp)
  jg     L103
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L90
L103:
  mov    -0x440(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x420(%rbp)
  mov    -0x440(%rbp),%rax
  mov    0x10(%rax),%rax
  mov    %rax,-0x418(%rbp)
  lea    -0x410(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x80,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,-0x428(%rbp)
L101:
  mov    -0x428(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x420(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L33
  mov    -0x428(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x418(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
L33:
  mov    -0x428(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x420(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x428(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x418(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L52
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x2,%eax
  jmp    L90
L58:
  mov    -0x428(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x420(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cltq
  mov    -0x410(%rbp,%rax,4),%edx
  mov    -0x428(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x420(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    $0x1,%edx
  cltq
  mov    %edx,-0x410(%rbp,%rax,4)
  mov    -0x428(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x418(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cltq
  mov    -0x410(%rbp,%rax,4),%edx
  mov    -0x428(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x418(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x1,%edx
  cltq
  mov    %edx,-0x410(%rbp,%rax,4)
  addl   $0x1,-0x428(%rbp)
  jmp    L101
L52:
  nop
  movl   $0x0,-0x424(%rbp)
  jmp    L113
L112:
  mov    -0x424(%rbp),%eax
  cltq
  mov    -0x410(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L31
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x2,%eax
  jmp    L90
L31:
  addl   $0x1,-0x424(%rbp)
L113:
  cmpl   $0xff,-0x424(%rbp)
  jle    L112
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L90:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L57
  call   <__stack_chk_fail@plt>
L57:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D21
data_type: string
name: retrieve_data
arguments: data_tag: D25
data_type: string
name: retrieve_data
arguments: data_tag: D20
data_type: string

tool: D21: Too few arguments.

tool: D25: The strings are not anagrams.


tool: D20: The strings are anagrams.


assistant: ```c
int main(int argc, char **argv) {
  if (argc < 3) {
    printf("Too few arguments.");
    return 1;
  }
  char *s1 = argv[1];
  char *s2 = argv[2];
  int chars[256] = {0};
  int i = 0;
  while (1) {
    if (!s1[i] || !s2[i]) {
      if (s1[i] != s2[i]) {
        printf("The strings are not anagrams.\n");
        return 2;
      }
      break;
    }
    chars[s1[i]] += 1;
    chars[s2[i]] -= 1;
    i += 1;
  }
  for (int i = 0; i < 256; ++i) {
    if (chars[i]) {
      printf("The strings are not anagrams.\n");
      return 2;
    }
  }
  printf("The strings are anagrams.\n");
  return 0;
}
```