user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<intersect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf80,%rsp
  mov    %rdi,-0x1f68(%rbp)
  mov    %esi,-0x1f6c(%rbp)
  mov    %rdx,-0x1f78(%rbp)
  mov    %ecx,-0x1f70(%rbp)
  mov    %r8,-0x1f80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1f50(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x1f4,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0xfb0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x1f4,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    -0x1f6c(%rbp),%edx
  mov    -0x1f70(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x1f5c(%rbp)
  mov    -0x1f5c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x1f58(%rbp)
  movl   $0x0,-0x1f60(%rbp)
  jmp    L72
L73:
  mov    -0x1f60(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x1f68(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x1f50(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x1f50(%rbp,%rax,4)
  addl   $0x1,-0x1f60(%rbp)
L72:
  mov    -0x1f60(%rbp),%eax
  cmp    -0x1f6c(%rbp),%eax
  jl     L73
  movl   $0x0,-0x1f60(%rbp)
  jmp    L46
L42:
  mov    -0x1f60(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x1f78(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0xfb0(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0x1f60(%rbp)
L46:
  mov    -0x1f60(%rbp),%eax
  cmp    -0x1f70(%rbp),%eax
  jl     L42
  movl   $0x0,-0x1f60(%rbp)
  jmp    L14
L51:
  mov    -0x1f80(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%ecx
  mov    -0x1f80(%rbp),%rdx
  mov    %ecx,(%rdx)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x1f58(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1f60(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x1f60(%rbp),%eax
  cltq
  mov    -0x1f50(%rbp,%rax,4),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1f60(%rbp),%eax
  cltq
  mov    %edx,-0x1f50(%rbp,%rax,4)
  mov    -0x1f60(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1f60(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
L55:
  mov    -0x1f60(%rbp),%eax
  cltq
  mov    -0x1f50(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L60
  mov    -0x1f60(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L51
L60:
  addl   $0x1,-0x1f60(%rbp)
L14:
  cmpl   $0x3e7,-0x1f60(%rbp)
  jle    L55
  mov    -0x1f58(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int *intersect(int *nums1, int nums1Size, int *nums2, int nums2Size, int *returnSize) {
  int arr1[1000] = {0};
  int arr2[1000] = {0};
  int min = nums1Size > nums2Size ? nums2Size : nums1Size;
  int i;
  int *ret = (int *)malloc(sizeof(int) * min);
  for (i = 0; i < nums1Size; i++) {
    arr1[nums1[i]]++;
  }
  for (i = 0; i < nums2Size; i++) {
    arr2[nums2[i]]++;
  }
  for (i = 0; i < 1000; i++) {
    while (arr1[i] != 0 && arr2[i] != 0) {
      ret[(*returnSize)++] = i;
      arr1[i]--;
      arr2[i]--;
    }
  }
  return ret;
}
```