user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L168
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D13(%rip),%r12
L58:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0xc(%rsp),%eax
  add    $0x4,%rbp
  cmp    %ebx,%eax
  jg     L58
  test   %eax,%eax
  jle    L168
  mov    $0x0,%ebx
  lea    0x10(%rsp),%rbp
  lea    D32(%rip),%r12
  lea    D51(%rip),%r13
  jmp    L53
L187:
  mov    %r13,%rdi
  call   <printf@plt>
L3:
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jle    L168
L53:
  cmpl   $0x1,0x0(%rbp,%rbx,4)
  je     L187
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L3
L168:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x3e8,%ebp
  lea    D32(%rip),%rbx
  lea    D51(%rip),%r12
  lea    D8(%rip),%r13
  jmp    L34
L96:
  mov    %r12,%rdi
  call   <printf@plt>
L69:
  add    $0x1,%r14
  cmp    %r14d,0xc(%rsp)
  jle    L31
L24:
  cmpl   $0x1,(%r15,%r14,4)
  je     L96
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L69
L31:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L160
  lea    0x10(%rsp),%rax
  lea    -0x1(%rdx),%edx
  lea    0x14(%rsp,%rdx,4),%rsi
  mov    $0x0,%edx
L103:
  cmpl   $0x1,(%rax)
  sete   %cl
  movzbl %cl,%ecx
  add    %ecx,%edx
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L103
  test   %edx,%edx
  je     L160
  sub    $0x1,%ebp
  je     L160
L34:
  lea    0x70(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    0xc(%rsp),%edx
  call   <next_generation@plt>
  mov    0xc(%rsp),%ecx
  test   %ecx,%ecx
  jle    L31
  lea    0x70(%rsp),%rax
  lea    0x10(%rsp),%rdx
  lea    -0x1(%rcx),%ecx
  lea    0x74(%rsp,%rcx,4),%rsi
L81:
  mov    (%rax),%ecx
  mov    %ecx,(%rdx)
  movl   $0x0,(%rax)
  add    $0x4,%rax
  add    $0x4,%rdx
  cmp    %rsi,%rax
  jne    L81
  mov    $0x0,%r14d
  lea    0x10(%rsp),%r15
  jmp    L24
L160:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  mov    $0x0,%eax
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D13
data_type: string
name: unpack_data
arguments: data_tag: D32
data_type: string
name: unpack_data
arguments: data_tag: D51
data_type: string
name: unpack_data
arguments: data_tag: D8
data_type: string

tool: D13: %d

tool: D32: .

tool: D51: *

tool: D8: 


assistant: ```c
int main() {
  int current[21];
  int next[21];
  int length;
  scanf("%d", &length);
  int i, i1, i2, i3, i4;
  for (i = 0; i < length; i++) {
    scanf("%d", &current[i]);
  }
  for (i3 = 0; i3 < length; i3++) {
    if (current[i3] == 1) {
      printf("*");
    } else {
      printf(".");
    }
  }
  printf("\n");
  for (i = 0; i < 1000; i++) {
    next_generation(current, next, length);
    for (i1 = 0; i1 < length; i1++) {
      current[i1] = next[i1];
      next[i1] = 0;
    }
    for (i3 = 0; i3 < length; i3++) {
      if (current[i3] == 1) {
        printf("*");
      } else {
        printf(".");
      }
    }
    printf("\n");
    int number = 0;
    for (i4 = 0; i4 < length; i4++) {
      if (current[i4] == 1) {
        number++;
      }
    }
    if (number == 0) {
      break;
    }
  }
}
```