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    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x15bc7,-0xc0(%rbp)
  movabs $0x6e694c20694c,%rax
  mov    $0x0,%edx
  mov    %rax,-0xb8(%rbp)
  mov    %rdx,-0xb0(%rbp)
  movl   $0x0,-0xa8(%rbp)
  movb   $0x4d,-0xa4(%rbp)
  movabs $0x6a69654220333231,%rax
  movabs $0x64616f5220676e69,%rdx
  mov    %rax,-0xa3(%rbp)
  mov    %rdx,-0x9b(%rbp)
  movl   $0x0,-0x93(%rbp)
  movq   $0x15bc8,-0x80(%rbp)
  movabs $0x69654c20694c,%rax
  mov    $0x0,%edx
  mov    %rax,-0x78(%rbp)
  mov    %rdx,-0x70(%rbp)
  movl   $0x0,-0x68(%rbp)
  movb   $0x46,-0x64(%rbp)
  movabs $0x616873676e6f685a,%rax
  movabs $0x64616f52206e,%rdx
  mov    %rax,-0x63(%rbp)
  mov    %rdx,-0x5b(%rbp)
  movl   $0x0,-0x53(%rbp)
  movq   $0x15bc9,-0x48(%rbp)
  movabs $0x676e61695820694c,%rax
  mov    $0x0,%edx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movl   $0x0,-0x30(%rbp)
  movb   $0x46,-0x2c(%rbp)
  movabs $0x6f7567676e6f685a,%rax
  movabs $0x64616f5220,%rdx
  mov    %rax,-0x2b(%rbp)
  mov    %rdx,-0x23(%rbp)
  movl   $0x0,-0x1b(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movzbl -0xa4(%rbp),%eax
  movsbl %al,%edx
  mov    -0xc0(%rbp),%rax
  lea    -0xc0(%rbp),%rcx
  lea    0x1d(%rcx),%rdi
  lea    -0xc0(%rbp),%rcx
  lea    0x8(%rcx),%rsi
  mov    %rdi,%r8
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0xc4(%rbp)
  jmp    L6
L25:
  mov    -0xc4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x80(%rbp),%rcx
  mov    -0xc4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  shl    $0x3,%rax
  add    $0x10,%rax
  add    %rcx,%rax
  lea    0xd(%rax),%rdi
  mov    -0xc4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  shl    $0x3,%rax
  add    %rbp,%rax
  sub    $0x64,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  lea    -0x80(%rbp),%rsi
  mov    -0xc4(%rbp),%eax
  movslq %eax,%rcx
  mov    %rcx,%rax
  shl    $0x3,%rax
  sub    %rcx,%rax
  shl    $0x3,%rax
  add    %rsi,%rax
  lea    0x8(%rax),%rsi
  mov    -0xc4(%rbp),%eax
  movslq %eax,%rcx
  mov    %rcx,%rax
  shl    $0x3,%rax
  sub    %rcx,%rax
  shl    $0x3,%rax
  add    %rbp,%rax
  add    $0xffffffffffffff80,%rax
  mov    (%rax),%rax
  mov    %rdi,%r8
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc4(%rbp)
L6:
  cmpl   $0x1,-0xc4(%rbp)
  jle    L25
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D27
datatype: string
name: extract_symbolic_data
arguments: tag: D12
datatype: string
name: extract_symbolic_data
arguments: tag: D1
datatype: string

tool: D27: Student a:

tool: D12: NO.:%ld
name:%s
sex:%c
address:%s


tool: D1: 
Student %d:


assistant: ```c
void main() {
  struct student {
    long int num;
    char name[20];
    char sex;
    char addr[20];
  } a = {89031, "Li Lin", 'M', "123 Beijing Road"};
  int i;
  struct student s[] = {{89032, "Li Lei", 'F', "Zhongshan Road"}, {89033, "Li Xiang", 'F', "Zhongguo Road"}};
  printf("Student a:\n");
  printf("NO.:%ld\nname:%s\nsex:%c\naddress:%s\n", a.num, a.name, a.sex, a.addr);
  for (i = 0; i < 2; i++) {
    printf("\nStudent %d:\n", i);
    printf("NO.:%ld\nname:%s\nsex:%c\naddress:%s\n", s[i].num, s[i].name, s[i].sex, s[i].addr);
  }
}
```