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解决不用sizeof求出int大小的方法
来源: 互联网 发布时间:2014-10-17
本文导语: 代码如下所示: 代码如下:#include int main(int argc, char *argv[]) { int a[2]; unsigned int add1 = &a[0]; unsigned int add2 = &a[1]; printf("The address of a[0] is %u/n",add1); printf("The address of a[1] is %u/n",add2); printf("...
代码如下所示:
代码如下:
#include
int main(int argc, char *argv[])
{
int a[2];
unsigned int add1 = &a[0];
unsigned int add2 = &a[1];
printf("The address of a[0] is %u/n",add1);
printf("The address of a[1] is %u/n",add2);
printf("The size of int is %u/n", add2 - add1);
}
输出结果是:
The address of a[0] is 3218821936
The address of a[1] is 3218821940
The size of int is 4