user: 这段汇编代码用C语言表示是什么样的?

```asm
<bubble_sort>:
  endbr64
  push   %r14
  lea    D5(%rip),%rdi
  push   %r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  movdqa D19(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D35(%rip),%rax
  mov    %rsp,%r14
  lea    0x18(%rsp),%rbp
  movaps %xmm0,(%rsp)
  mov    %r14,%rbx
  mov    %rax,0x10(%rsp)
  call   <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L14
  mov    $0xa,%edi
  xor    %r13d,%r13d
  call   <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L34:
  cmp    $0x5,%r13d
  je     L81
  movq   0x10(%rsp),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%eax
  movd   %xmm2,%edx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%edx
  jge    L57
  movq   %xmm1,0x10(%rsp)
L57:
  cmp    $0x4,%r13d
  je     L81
  movq   0xc(%rsp),%xmm0
  pshufd $0xe5,%xmm0,%xmm3
  movd   %xmm0,%eax
  movd   %xmm3,%edx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%edx
  jge    L89
  movq   %xmm1,0xc(%rsp)
L89:
  cmp    $0x3,%r13d
  je     L81
  movq   0x8(%rsp),%xmm0
  pshufd $0xe5,%xmm0,%xmm4
  movd   %xmm0,%eax
  movd   %xmm4,%edx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%edx
  jge    L118
  movq   %xmm1,0x8(%rsp)
L118:
  cmp    $0x2,%r13d
  je     L81
  movq   0x4(%rsp),%xmm0
  pshufd $0xe5,%xmm0,%xmm5
  movd   %xmm0,%eax
  movd   %xmm5,%edx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%edx
  jge    L37
  movq   %xmm1,0x4(%rsp)
L37:
  test   %r13d,%r13d
  jne    L81
  movq   (%rsp),%xmm0
  pshufd $0xe5,%xmm0,%xmm6
  movd   %xmm0,%eax
  movd   %xmm6,%edx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%edx
  jge    L81
  movq   %xmm1,(%rsp)
L81:
  mov    %r14,%rbx
  nopl   0x0(%rax,%rax,1)
L36:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L36
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <putchar@plt>
  cmp    $0x6,%r13d
  jne    L34
  nopl   (%rax)
L47:
  mov    (%r14),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%r14
  call   <printf@plt>
  cmp    %rbp,%r14
  jne    L47
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L116
  add    $0x20,%rsp
  lea    D27(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <puts@plt>
L116:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: string
name: read_data
arguments: data_label: D15
type_guessed: string
name: read_data
arguments: data_label: D19
type_guessed: byte[16]
name: read_data
arguments: data_label: D35
type_guessed: byte[8]
name: read_data
arguments: data_label: D27
type_guessed: string

tool: D5: start 

tool: D15: %d 

tool: {"D19": [0x50,0x00,0x00,0x00,0x29,0x00,0x00,0x00,0x23,0x00,0x00,0x00,0x5a,0x00,0x00,0x00]}

tool: {"D35": [0x28,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: D27: end  

assistant: ```c
void bubble_sort() {
  int a[] = {80, 41, 35, 90, 40, 20};
  int t, i, j;
  printf("start \n");
  for (j = 0; j < 6; j++) {
    printf("%d ", a[j]);
  }
  printf("\n");
  for (i = 0; i < 6; i++) {
    for (j = 6 - 1; j > i; j--) {
      if (a[j] < a[j - 1]) {
        t = a[j];
        a[j] = a[j - 1];
        a[j - 1] = t;
      }
    }
    for (j = 0; j < 6; j++) {
      printf("%d ", a[j]);
    }
    printf("\n");
  }
  for (i = 0; i < 6; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  printf("end  \n");
}
```