user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<addbits2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%r15d
  mov    %eax,0xc(%rsp)
  call   <strlen@plt>
  cmp    %eax,%r15d
  lea    0x2(%rax),%edi
  mov    %eax,%ebx
  mov    %eax,0x14(%rsp)
  cmovg  %r15d,%edi
  call   <malloc@plt>
  cmp    %ebx,%r15d
  movl   $0x0,0x24(%rsp)
  cmovle %r15d,%ebx
  mov    %rax,%r12
  mov    %ebx,0x10(%rsp)
  test   %ebx,%ebx
  jle    L8
  movslq %ebx,%rax
  mov    %r12,%r14
  lea    0x24(%rsp),%rbx
  mov    %r12,%r15
  mov    %rax,0x18(%rsp)
  add    %r12,%rax
  neg    %r14
  mov    %rax,(%rsp)
  nopl   0x0(%rax)
L70:
  lea    0x0(%r13,%r14,1),%rax
  mov    %rbx,%rdx
  movsbl (%rax,%r15,1),%esi
  lea    0x0(%rbp,%r14,1),%rax
  movsbl (%rax,%r15,1),%edi
  add    $0x1,%r15
  call   <add@plt>
  mov    %al,-0x1(%r15)
  cmp    (%rsp),%r15
  jne    L70
  mov    0x18(%rsp),%rax
  mov    0xc(%rsp),%ebx
  mov    0x10(%rsp),%ecx
  add    %r12,%rax
  cmp    %ecx,%ebx
  jle    L28
L9:
  movslq %ecx,%rax
  sub    %r12,%rbp
  lea    0x1(%r12,%rax,1),%rdx
  lea    (%rax,%r12,1),%r14
  lea    -0x1(%rbx),%eax
  lea    0x24(%rsp),%rbx
  sub    %ecx,%eax
  lea    (%rdx,%rax,1),%r15
  nop
L71:
  movsbl 0x0(%rbp,%r14,1),%edi
  mov    %rbx,%rdx
  mov    $0x30,%esi
  add    $0x1,%r14
  call   <add@plt>
  mov    %al,-0x1(%r14)
  cmp    %r15,%r14
  jne    L71
  movslq 0xc(%rsp),%rax
  add    %r12,%rax
L31:
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%ecx
  cmp    %ecx,%esi
  jle    L61
  lea    -0x1(%rsi),%eax
  movslq 0xc(%rsp),%rdx
  lea    0x24(%rsp),%rbx
  sub    %ecx,%eax
  lea    0x1(%r12,%rax,1),%rbp
  movslq %ecx,%rax
  lea    (%r12,%rdx,1),%r14
  sub    %r12,%rax
  add    %rdx,%rbp
  sub    %rdx,%rax
  add    %rax,%r13
  nopl   0x0(%rax,%rax,1)
L6:
  movsbl 0x0(%r13,%r14,1),%esi
  mov    %rbx,%rdx
  mov    $0x30,%edi
  add    $0x1,%r14
  call   <add@plt>
  mov    %al,-0x1(%r14)
  cmp    %rbp,%r14
  jne    L6
  mov    0x14(%rsp),%eax
  sub    0x10(%rsp),%eax
  add    0xc(%rsp),%eax
  cltq
  add    %r12,%rax
L61:
  movb   $0x0,(%rax)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x38,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  movl   $0x0,0x10(%rsp)
  mov    0xc(%rsp),%ebx
  mov    0x10(%rsp),%ecx
  cmp    %ecx,%ebx
  jg     L9
  nopl   0x0(%rax,%rax,1)
L28:
  mov    0x10(%rsp),%ecx
  mov    %ecx,0xc(%rsp)
  jmp    L31
L24:
  call   <__stack_chk_fail@plt>
```

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;
}
```