user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  movq   $0x2a,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x2a,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x0,0x78(%rsp)
  movq   $0x0,0x80(%rsp)
  movq   $0x0,0x88(%rsp)
  movq   $0x0,0x90(%rsp)
  movq   $0x0,0x98(%rsp)
  movq   $0x0,0xa0(%rsp)
  movq   $0x0,0xa8(%rsp)
  movl   $0x0,0xc(%rsp)
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebp
  jmp    L98
L28:
  mov    %al,0x60(%rsp,%rbp,1)
  add    $0x1,%rbp
L98:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L28
  lea    D5(%rip),%rdi
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  lea    D37(%rip),%rdi
  call   <scanf@plt>
  test   %ebp,%ebp
  jle    L42
  mov    0xc(%rsp),%edi
  lea    -0x1(%rbp),%ebx
  mov    %ebp,%edx
  mov    $0x0,%ecx
  jmp    L34
L2:
  lea    -0x41(%rax),%esi
  cmp    $0x19,%sil
  ja     L79
  movsbl %al,%eax
  lea    -0x41(%rdi,%rax,1),%eax
  movslq %eax,%rsi
  imul   $0x4ec4ec4f,%rsi,%rsi
  sar    $0x23,%rsi
  mov    %eax,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%esi
  imul   $0x1a,%esi,%esi
  sub    %esi,%eax
  add    $0x41,%eax
L79:
  mov    %al,0x10(%rsp,%rcx,1)
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L58
L34:
  movzbl 0x60(%rsp,%rcx,1),%eax
  lea    -0x61(%rax),%esi
  cmp    $0x19,%sil
  ja     L2
  movsbl %al,%eax
  lea    -0x61(%rdi,%rax,1),%eax
  movslq %eax,%rsi
  imul   $0x4ec4ec4f,%rsi,%rsi
  sar    $0x23,%rsi
  mov    %eax,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%esi
  imul   $0x1a,%esi,%esi
  sub    %esi,%eax
  add    $0x61,%eax
  jmp    L79
L42:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  jmp    L63
L58:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rbp
  lea    0x11(%rsp,%rbx,1),%rbx
L80:
  movsbl 0x0(%rbp),%edi
  call   <putchar@plt>
  add    $0x1,%rbp
  cmp    %rbx,%rbp
  jne    L80
L63:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L106
  mov    $0x0,%eax
  add    $0xc8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L106:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D19
data_type: string
name: unpack_data
arguments: data_label: D5
data_type: string
name: unpack_data
arguments: data_label: D37
data_type: string
name: unpack_data
arguments: data_label: D21
data_type: string

tool: D19: enter message to be encrypted:

tool: D5: enter shift amount(1-25)

tool: D37: %d

tool: D21: encrypted message:

assistant: ```c
int main(void) {
  char encrypted_ms[80] = {'*'}, unencrypted_ms[80] = {'*'}, ch;
  int shift_amount = 0, index = 0;
  printf("enter message to be encrypted:\n");
  while ((ch = getchar()) != '\n') {
    unencrypted_ms[index] = ch;
    index++;
  }
  printf("enter shift amount(1-25)\n");
  scanf("%d", &shift_amount);
  for (int i = 0; i < index; i++) {
    if (unencrypted_ms[i] >= 'a' && unencrypted_ms[i] <= 'z') {
      encrypted_ms[i] = ((unencrypted_ms[i] - 'a') + shift_amount) % 26 + 'a';
    } else if (unencrypted_ms[i] >= 'A' && unencrypted_ms[i] <= 'Z') {
      encrypted_ms[i] = ((unencrypted_ms[i] - 'A') + shift_amount) % 26 + 'A';
    } else {
      encrypted_ms[i] = unencrypted_ms[i];
    }
  }
  printf("encrypted message:\n");
  for (int j = 0; j < index; j++) {
    printf("%c", encrypted_ms[j]);
  }
  printf("\n");
  return 0;
}
```