user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x3e8000(%rsp),%r11
L10:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L10
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8028(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L38
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  lea    D41(%rip),%r14
  lea    D11(%rip),%r13
  lea    D46(%rip),%r15
  jmp    L106
L156:
  lea    0x1c(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    0x1(%r12),%eax
  mov    %eax,0xc(%rsp)
  mov    0x1c(%rsp),%esi
  movslq %r12d,%r12
  mov    %esi,0x20(%rsp,%r12,4)
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%r12d
  jmp    L93
L158:
  test   %ebx,%ebx
  je     L40
L0:
  sub    $0x1,%ebx
  jmp    L93
L40:
  test   %r12d,%r12d
  jle    L0
  movslq %r12d,%rax
  lea    0x20(%rsp,%rax,4),%rdx
  lea    0x1f4020(%rsp),%rax
  lea    -0x1(%r12),%ecx
  lea    0x1f4024(%rsp,%rcx,4),%rsi
L117:
  mov    -0x4(%rdx),%ecx
  mov    %ecx,(%rax)
  sub    $0x4,%rdx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L117
  mov    %r12d,%eax
  mov    %ebx,%r12d
  mov    %eax,%ebx
  jmp    L0
L37:
  lea    -0x1(%rbx),%eax
  cltq
  mov    0x1f4020(%rsp,%rax,4),%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L93:
  add    $0x1,%ebp
  cmp    %ebp,0x14(%rsp)
  jle    L38
L106:
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x18(%rsp),%eax
  cmp    $0x1,%eax
  je     L156
  cmp    $0x2,%eax
  je     L158
  test   %ebx,%ebx
  jne    L37
  test   %r12d,%r12d
  jle    L37
  movslq %r12d,%rax
  lea    0x20(%rsp,%rax,4),%rdx
  lea    0x1f4020(%rsp),%rax
  lea    -0x1(%r12),%ecx
  lea    0x1f4024(%rsp,%rcx,4),%rsi
L127:
  mov    -0x4(%rdx),%ecx
  mov    %ecx,(%rax)
  sub    $0x4,%rdx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L127
  mov    %r12d,%eax
  mov    %ebx,%r12d
  mov    %eax,%ebx
  jmp    L37
L38:
  mov    0x3e8028(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L81
  mov    $0x0,%eax
  add    $0x3e8038,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L81:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D11
type_infered: string
name: retrieve_data
arguments: label: D41
type_infered: string
name: retrieve_data
arguments: label: D46
type_infered: string
name: retrieve_data
arguments: label: D29
type_infered: string

tool: D11: %d

tool: D41: INBOX_I = %d, OUTBOX_I = %d


tool: D46: %d


tool: D29: ADDED = %d


assistant: ```c
int main() {
  int q;
  scanf("%d", &q);
  int inbox[512000];
  int inbox_i = 0;
  int outbox[512000];
  int outbox_i = 0;
  for (int i = 0; i < q; i++) {
    printf("INBOX_I = %d, OUTBOX_I = %d\n", inbox_i, outbox_i);
    int action;
    scanf("%d", &action);
    if (action == 1) {
      int n;
      scanf("%d", &n);
      inbox[inbox_i++] = n;
      printf("ADDED = %d\n", n);
    } else if (action == 2) {
      if (outbox_i == 0)
        while (inbox_i > 0)
          outbox[outbox_i++] = inbox[--inbox_i];
      outbox_i--;
    } else {
      if (outbox_i == 0)
        while (inbox_i > 0)
          outbox[outbox_i++] = inbox[--inbox_i];
      printf("%d\n", outbox[outbox_i - 1]);
    }
  }
  return 0;
}
```