user: 这段汇编代码用C语言表示是什么样的？

```asm
<juxtaposition_hv>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x0,%eax
L17:
  mov    0x0(%rbp,%rax,1),%edx
  mov    %edx,(%r15,%rax,1)
  mov    %edx,0x0(%r13,%rax,1)
  mov    0x80(%rbp,%rax,1),%edx
  mov    %edx,(%r14,%rax,1)
  mov    %edx,(%r12,%rax,1)
  add    $0x4,%rax
  cmp    $0x80,%rax
  jne    L17
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%edx
  mov    $0x4,%esi
  mov    %r15,%rdi
  call   <affiche_tab@plt>
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%edx
  mov    $0x4,%esi
  mov    %r14,%rdi
  call   <affiche_tab@plt>
  mov    %r13,%rsi
  mov    %r15,%rdi
  call   <juxtaposition_horizontale@plt>
  mov    %rax,%rbp
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <juxtaposition_horizontale@plt>
  mov    %rax,%r14
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%edx
  mov    $0x8,%esi
  mov    %rbp,%rdi
  call   <affiche_tab@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%edx
  mov    $0x8,%esi
  mov    %r14,%rdi
  call   <affiche_tab@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L14:
  mov    0x0(%rbp,%rax,1),%edx
  mov    %edx,(%rbx,%rax,1)
  mov    (%r14,%rax,1),%edx
  mov    %edx,0x80(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    $0x80,%rax
  jne    L14
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D29
data_type: string
name: disassemble_data
arguments: tag: D33
data_type: string
name: disassemble_data
arguments: tag: D32
data_type: string
name: disassemble_data
arguments: tag: D2
data_type: string
name: disassemble_data
arguments: tag: D1
data_type: string

tool: D29: temp1 


tool: D33:  
 temp2 
 

tool: D32: temp1 juxtap 


tool: D2:  
 temp2 juxtap 
 

tool: D1: fin temp 


assistant: ```c
uint8_t *juxtaposition_hv(uint8_t *bloc, uint8_t *out) {
  uint8_t *temp1 = malloc(32 * sizeof(uint8_t));
  uint8_t *temp1bis = malloc(32 * sizeof(uint8_t));
  uint8_t *temp2 = malloc(32 * sizeof(uint8_t));
  uint8_t *temp2bis = malloc(32 * sizeof(uint8_t));
  for (uint8_t i = 0; i < 32; i++) {
    temp1[i] = bloc[i];
    temp1bis[i] = bloc[i];
    temp2[i] = bloc[i + 32];
    temp2bis[i] = bloc[i + 32];
  }
  printf("temp1 \n");
  affiche_tab(temp1, 4, 8);
  printf(" \n temp2 \n ");
  affiche_tab(temp2, 4, 8);
  temp1 = juxtaposition_horizontale(temp1, temp1bis);
  temp2 = juxtaposition_horizontale(temp2, temp2bis);
  printf("temp1 juxtap \n");
  affiche_tab(temp1, 8, 4);
  printf(" \n temp2 juxtap \n ");
  affiche_tab(temp2, 8, 4);
  printf("fin temp \n");
  for (uint8_t i = 0; i < 32; i++) {
    out[i] = temp1[i];
    out[i + 32] = temp2[i];
  }
  free(temp1);
  free(temp2);
  free(temp1bis);
  free(temp2bis);
  return out;
}
```