题目:

为了方便分析,我们将题目代码贴在下面:

int main()
{
  unsigned char puc[4];
  struct tagPIM
  {
    unsigned char ucPim1;
    unsigned char ucData0 : 1;
    unsigned char ucData1 : 2;
    unsigned char ucData2 : 3;
  }*pstPimData;
  pstPimData = (struct tagPIM*)puc;
  memset(puc,0,4);
  pstPimData->ucPim1 = 2; 
  pstPimData->ucData0 = 3;
  pstPimData->ucData1 = 4;
  pstPimData->ucData2 = 5;
  printf("%02x %02x %02x %02x\n",puc[0], puc[1], puc[2], puc[3]);
  return 0;
}

---

正确答案:B

接下来画图分析一下题目:

由图分析可知,最后puc[0]里存放的是00000010,puc[1]里存放的是00101001,puc[2]里存放的是00000000,puc[3]里存放的是00000000.用16进制打印出的结果即为:02 29 00 00.