user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse_cmd>:
  endbr64
  push   %r15
  xor    %edx,%edx
  mov    %edi,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x158,%rsp
  movdqa D12(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x148(%rsp)
  xor    %eax,%eax
  mov    %rdx,0x48(%rsp)
  lea    0x30(%rsp),%r13
  movabs $0x736f6300000000,%rax
  movaps %xmm0,0x50(%rsp)
  movdqa D24(%rip),%xmm0
  mov    %rax,0x40(%rsp)
  lea    0x7f(%rsp),%rax
  movaps %xmm0,0x60(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x70(%rsp)
  movdqa D29(%rip),%xmm0
  movq   $0x6e6973,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0x0,0xc(%rsp)
  mov    %rax,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
L46:
  mov    0x10(%rsp),%rcx
  movsbl 0x0(%r13,%rcx,1),%eax
  test   %al,%al
  je     L15
  movslq %ebp,%r14
  xor    %ebx,%ebx
  sub    %r14,%rcx
  mov    %rcx,%r12
  jmp    L5
  nopl   (%rax)
L83:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <getch@plt>
  mov    %al,0x80(%rsp,%r14,1)
  mov    %eax,%r15d
  mov    %eax,%esi
  add    $0x1,%r14
  lea    0x0(%r13,%r12,1),%rax
  movsbl (%rax,%r14,1),%eax
  test   %al,%al
  je     L75
L5:
  mov    %r14d,%ebp
  cmp    %r15d,%eax
  je     L83
  test   %ebx,%ebx
  je     L48
  movslq %r14d,%r12
  lea    -0x1(%rbx),%eax
  lea    0x80(%rsp,%r12,1),%r14
  add    0x18(%rsp),%r12
  sub    %rax,%r12
  nopw   0x0(%rax,%rax,1)
L58:
  movsbl -0x1(%r14),%edi
  xor    %eax,%eax
  sub    $0x1,%r14
  call   <ungetch@plt>
  cmp    %r12,%r14
  jne    L58
  sub    %ebx,%ebp
L48:
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%eax
  addq   $0x14,0x10(%rsp)
  cmp    $0x4,%eax
  jne    L46
  mov    $0x34,%eax
L78:
  mov    0x148(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L39
  add    $0x158,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  lea    -0x1(%rbp),%eax
  cltq
  movzbl 0x80(%rsp,%rax,1),%esi
  cs nopw 0x0(%rax,%rax,1)
L75:
  movsbl %sil,%edi
  xor    %eax,%eax
  call   <ungetch@plt>
  movslq 0xc(%rsp),%r15
  mov    0x20(%rsp,%r15,4),%eax
  jmp    L78
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D12
type_guessed: byte[16]
name: unpack_data
arguments: data_tag: D24
type_guessed: byte[16]
name: unpack_data
arguments: data_tag: D29
type_guessed: byte[16]

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x70,0x6f,0x77,0x00,0x00,0x00,0x00,0x00]}

tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x78,0x00,0x00,0x00]}

tool: {"D29": [0x31,0x00,0x00,0x00,0x32,0x00,0x00,0x00,0x33,0x00,0x00,0x00,0x35,0x00,0x00,0x00]}

assistant: ```c
int parse_cmd(int cstart) {
  char cmds[][20] = {"sin", "cos", "pow", "x"};
  int cmd_flag[] = {'1', '2', '3', '5'};
  int len = sizeof(cmds) / (sizeof(char) * 20);
  int i, j, top;
  char stack[200];
  top = 0;
  for (i = 0; i < len; i++) {
    j = 0;
    while (cmds[i][j] != '\0' && cstart == cmds[i][j]) {
      cstart = getch();
      stack[top++] = cstart;
      j++;
    }
    if (cmds[i][j] == '\0') {
      ungetch(stack[--top]);
      return cmd_flag[i];
    } else {
      while (j--) {
        ungetch(stack[--top]);
      }
    }
  }
  return '4';
}
```