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C:带有const修饰的指针解读 (二)

2014-11-24 00:11:48 · 作者: · 浏览: 53
标签: 带有 const 修饰 指针 解读
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int main(int argc, char *argv[])
{
int i = 1;
int j = 100;

const int * const pTmp = &i;
printf("pTmp = %d\n", *pTmp);

i = 2; /*正确*/
printf("pTmp = %d\n", *pTmp);

pTmp = &j; /*错误：error: assignment of read-only variable `pTmp'*/
printf("pTmp = %d\n", *pTmp);


(*pTmp)++; /*错误：error: increment of read-only location*/
printf("pTmp = %d\n", *pTmp);

system("PAUSE");

return 0;
}

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