user: 这段汇编代码用C语言表示是什么样的？

```asm
<addbits2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x28(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jg     L25
  mov    -0x24(%rbp),%eax
  add    $0x2,%eax
  jmp    L52
L25:
  mov    -0x28(%rbp),%eax
L52:
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x30(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L49
L21:
  mov    -0x30(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x30(%rbp)
  movslq %eax,%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%esi
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x2c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  lea    -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    %ecx,%edi
  call   <add@plt>
  mov    %al,(%rbx)
L49:
  mov    -0x34(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jge    L93
  mov    -0x30(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L21
  jmp    L93
L60:
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x2c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  lea    -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    $0x30,%esi
  mov    %ecx,%edi
  call   <add@plt>
  mov    %al,(%rbx)
L93:
  mov    -0x34(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L60
  jmp    L19
L79:
  mov    -0x30(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x30(%rbp)
  movslq %eax,%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x2c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  lea    -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    %ecx,%esi
  mov    $0x30,%edi
  call   <add@plt>
  mov    %al,(%rbx)
L19:
  mov    -0x30(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L79
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *addbits2(char str1[], char str2[]) {
  int len1 = strlen(str1);
  int len2 = strlen(str2);
  char *res = (char *)malloc(sizeof(char) * ((len1 > len2) ? len1 : len2 + 2));
  int i, j, k;
  k = i = j = 0;
  int carry = 0;
  while (i < len1 && j < len2) {
    res[k++] = add(str1[i++], str2[j++], &carry);
  }
  while (i < len1) {
    res[k++] = add(str1[i++], '0', &carry);
  }
  while (j < len2) {
    res[k++] = add('0', str2[j++], &carry);
  }
  res[k] = '\0';
  return res;
}
```