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
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x50(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x50(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L5:
  cmp    %rdx,%rsp
  je     L58
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L5
L58:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L49
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L49:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x40(%rbp),%rax
  movl   $0x1,0x4(%rax)
  movl   $0x2,-0x4c(%rbp)
  jmp    L40
L21:
  mov    -0x4c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x40(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%ecx
  mov    -0x4c(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x40(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  add    %eax,%ecx
  mov    -0x40(%rbp),%rax
  mov    -0x4c(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x4c(%rbp)
L40:
  mov    -0x50(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jl     L21
  mov    -0x50(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4c(%rbp)
  jmp    L50
L33:
  mov    -0x40(%rbp),%rax
  mov    -0x4c(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4c(%rbp)
L50:
  mov    -0x50(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jl     L33
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L71
  call   <__stack_chk_fail@plt>
L71:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D35
type_infered: string
name: retrieve_data
arguments: tag: D0
type_infered: string
name: retrieve_data
arguments: tag: D3
type_infered: string
name: retrieve_data
arguments: tag: D8
type_infered: string

tool: D35: Enter the length of the sequence: 

tool: D0: %d

tool: D3: 
Fibonacci sequence of length %d:


tool: D8: %d 

assistant: ```c
int main() {
  int length, i;
  printf("Enter the length of the sequence: ");
  scanf("%d", &length);
  int arr[length];
  arr[0] = 0;
  arr[1] = 1;
  for (i = 2; i < length; i++) {
    arr[i] = arr[i - 1] + arr[i - 2];
  }
  printf("\nFibonacci sequence of length %d:\n", length);
  for (i = 0; i < length; i++)
    printf("%d ", arr[i]);
  printf("\n");
}
```