user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x60(%rbp)
  movl   $0x2,-0x5c(%rbp)
  movl   $0x3,-0x58(%rbp)
  movl   $0x4,-0x54(%rbp)
  movl   $0x5,-0x50(%rbp)
  movl   $0x6,-0x4c(%rbp)
  movl   $0x7,-0x48(%rbp)
  movl   $0x8,-0x44(%rbp)
  movl   $0x9,-0x40(%rbp)
  movl   $0xa,-0x3c(%rbp)
  lea    -0x60(%rbp),%rax
  mov    %rax,-0x70(%rbp)
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x80(%rbp)
  jmp    L34
L21:
  mov    -0x80(%rbp),%eax
  cltq
  mov    -0x60(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x80(%rbp)
L34:
  cmpl   $0x9,-0x80(%rbp)
  jle    L21
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x7c(%rbp)
  jmp    L61
L89:
  mov    -0x7c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L68
  mov    -0x7c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D73(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x7c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x70(%rbp),%rax
  add    %rax,%rdx
  mov    -0x7c(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  mov    (%rdx),%edx
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
L68:
  addl   $0x1,-0x7c(%rbp)
L61:
  cmpl   $0x9,-0x7c(%rbp)
  jle    L89
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x68(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x78(%rbp)
  jmp    L42
L80:
  mov    -0x78(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x78(%rbp)
L42:
  cmpl   $0x4,-0x78(%rbp)
  jle    L80
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x74(%rbp)
  jmp    L38
L11:
  mov    -0x74(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x74(%rbp)
L38:
  cmpl   $0x4,-0x74(%rbp)
  jle    L11
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D38
data_type: string
name: retrieve_data
arguments: data_label: D74
data_type: string
name: retrieve_data
arguments: data_label: D6
data_type: string
name: retrieve_data
arguments: data_label: D73
data_type: string
name: retrieve_data
arguments: data_label: D1
data_type: string
name: retrieve_data
arguments: data_label: D51
data_type: string
name: retrieve_data
arguments: data_label: D35
data_type: string
name: retrieve_data
arguments: data_label: D32
data_type: string

tool: D38: 
--- Vetor ---

tool: D74: 
valor : %d 

tool: D6: 

--- Ponteiro Vetor para Numero Pares ---

tool: D73: 
Numero Par : %d 

tool: D1: 


--- Novo Vetor ---

tool: D51: 
valor: %d 

tool: D35: 

--- Ponteiro Novo Vetor ---

tool: D32: 
Novo Vetor : %d 

assistant: ```c
int main(void) {
  int vetor[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
  int *ptr_vetor;
  int novo_vetor[10];
  int *ptr_novo_vetor;
  ptr_vetor = &vetor[0];
  printf("\n--- Vetor ---");
  for (int i = 0; i < 10; i++)
    printf("\nvalor : %d ", vetor[i]);
  printf("\n\n--- Ponteiro Vetor para Numero Pares ---");
  for (int i = 0; i < 10; i++) {
    if (ptr_vetor[i] % 2 == 0) {
      printf("\nNumero Par : %d ", ptr_vetor[i]);
      novo_vetor[i / 2] = ptr_vetor[i];
    }
  }
  ptr_novo_vetor = novo_vetor;
  printf("\n\n\n--- Novo Vetor ---");
  for (int i = 0; i < 10 / 2; i++)
    printf("\nvalor: %d ", novo_vetor[i]);
  printf("\n\n--- Ponteiro Novo Vetor ---");
  for (int i = 0; i < 10 / 2; i++)
    printf("\nNovo Vetor : %d ", *(ptr_novo_vetor + i));
}
```