user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movl   $0x1f,0x40(%rsp)
  movl   $0xc9,0x44(%rsp)
  movl   $0xfb,0x48(%rsp)
  movl   $0x82,0x4c(%rsp)
  movl   $0xaa,0x50(%rsp)
  movl   $0xd3,0x54(%rsp)
  movl   $0x5,0x58(%rsp)
  movl   $0x81,0x5c(%rsp)
  movl   $0x96,0x60(%rsp)
  movl   $0xe,0x64(%rsp)
  movl   $0x25,0x68(%rsp)
  movl   $0xf9,0x6c(%rsp)
  movl   $0xec,0x70(%rsp)
  movl   $0x12,0x74(%rsp)
  movl   $0x11,0x78(%rsp)
  movl   $0x81,0x7c(%rsp)
  mov    %rsp,%rsi
  lea    0x40(%rsp),%rdi
  mov    $0x0,%ecx
  mov    $0x0,%edx
  call   <forward4x4@plt>
  lea    0x10(%rsp),%rbp
  lea    0x50(%rsp),%r13
  mov    %rbp,%r12
  lea    D22(%rip),%r14
L41:
  lea    -0x10(%r12),%rbx
L28:
  mov    (%rbx),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L28
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x10,%r12
  cmp    %r13,%r12
  jne    L41
  movl   $0x7ba,0x80(%rsp)
  movl   $0x22,0x84(%rsp)
  movl   $0x1d6,0x88(%rsp)
  movl   $0xfffffec2,0x8c(%rsp)
  movl   $0x1eb,0x90(%rsp)
  movl   $0xfffffe5f,0x94(%rsp)
  movl   $0xfffffa1d,0x98(%rsp)
  movl   $0xfffffdf2,0x9c(%rsp)
  movl   $0x30,0xa0(%rsp)
  movl   $0xffffff9c,0xa4(%rsp)
  movl   $0xfffffe78,0xa8(%rsp)
  movl   $0x212,0xac(%rsp)
  movl   $0x53,0xb0(%rsp)
  movl   $0xfffffa37,0xb4(%rsp)
  movl   $0xffffffa5,0xb8(%rsp)
  movl   $0x214,0xbc(%rsp)
  mov    %rsp,%rsi
  lea    0x80(%rsp),%rdi
  mov    $0x0,%ecx
  mov    $0x0,%edx
  call   <inverse4x4@plt>
  lea    D9(%rip),%rdi
  call   <puts@plt>
  lea    D21(%rip),%r12
L49:
  lea    -0x10(%rbp),%rbx
L26:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L26
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x10,%rbp
  cmp    %r13,%rbp
  jne    L49
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  mov    $0x0,%eax
  add    $0xd0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D22
type_infered: string
name: retrieve_data
arguments: data_label: D9
type_infered: string
name: retrieve_data
arguments: data_label: D21
type_infered: string

tool: D22: %2d 

tool: D9: Inverse: 

tool: D21: %5d 

assistant: ```c
int main(void) {
  int out[4][4];
  int in3[4][4] = {{31, 201, 251, 130}, {170, 211, 5, 129}, {150, 14, 37, 249}, {236, 18, 17, 129}};
  int in[4][4] = {{31, 201, 251, 130}, {170, 211, 5, 129}, {150, 14, 37, 249}, {236, 18, 17, 129}};
  int in2[4][4] = {{0, 1, 1, 0}, {2, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}};
  forward4x4(in, out, 0, 0);
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      printf("%2d ", out[i][j]);
    }
    printf("\n");
  }
  int coeffs[4][4] = {{1978, 34, 470, -318}, {491, -417, -1507, -526}, {48, -100, -392, 530}, {83, -1481, -91, 532}};
  inverse4x4(coeffs, out, 0, 0);
  printf("Inverse: \n");
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < 4; j++) {
      printf("%5d ", out[i][j]);
    }
    printf("\n");
  }
  return 0;
}
```