user: 这段汇编代码用C语言表示是什么样的？

```asm
<caesar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %esi,-0x12c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x118(%rbp)
  jmp    L72
L2:
  mov    -0x118(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x118(%rbp),%eax
  cltq
  mov    %dl,-0x110(%rbp,%rax,1)
  addl   $0x1,-0x118(%rbp)
L72:
  mov    -0x128(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x118(%rbp)
  jl     L2
  movl   $0x0,-0x114(%rbp)
  jmp    L31
L1:
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L86
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L86
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x61(%rax),%edx
  mov    -0x12c(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    $0x61,%eax
  mov    %eax,%edx
  mov    -0x114(%rbp),%eax
  cltq
  mov    %dl,-0x110(%rbp,%rax,1)
  jmp    L52
L86:
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L52
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L52
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x41(%rax),%edx
  mov    -0x12c(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    $0x41,%eax
  mov    %eax,%edx
  mov    -0x114(%rbp),%eax
  cltq
  mov    %dl,-0x110(%rbp,%rax,1)
L52:
  addl   $0x1,-0x114(%rbp)
L31:
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x128(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
  lea    -0x110(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
type: string

tool: D8: %s 
 

assistant: ```c
void caesar(char message[], int key) {
  char messageOut[264];
  for (int j = 0; j < (int)strlen(message); j++) {
    messageOut[j] = message[j];
  }
  int i = 0;
  while (message[i] != '\0') {
    if (message[i] >= 'a' && message[i] <= 'z') {
      messageOut[i] = (message[i] - 'a' + key) % 26 + 'a';
    } else if (message[i] >= 'A' && message[i] <= 'Z') {
      messageOut[i] = (message[i] - 'A' + key) % 26 + 'A';
    }
    i++;
  }
  printf("%s \n ", messageOut);
}
```