user: 这段汇编代码用C语言表示是什么样的：

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L75
L79:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L23
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  jmp    L75
L23:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L75:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L77
  mov    -0x10(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L79
  jmp    L77
L85:
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L77:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L85
  jmp    L14
L10:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L14:
  mov    -0x10(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L10
  mov    -0x2c(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void merge(int *nums1, int m, int *nums2, int n) {
  int *nums3 = (int *)malloc(sizeof(int) * (m + n));
  int i = 0;
  int j = 0;
  int k = 0;
  while (i < m && j < n)
    if (nums1[i] < nums2[j])
      nums3[k++] = nums1[i++];
    else
      nums3[k++] = nums2[j++];
  while (i < m)
    nums3[k++] = nums1[i++];
  while (j < n)
    nums3[k++] = nums2[j++];
  memcpy(nums1, nums3, sizeof(int) * (m + n));
  free(nums3);
}
```