利用递归实现X的Y次方运算

[potato6517]

//递归调用实现x exp y
#include<iostream.h>
using std::cout;
using std::cin;
using std::endl;

int power(int base,int exponent);

int main()
{
int x,y;
cout<<"请输入底数x:";
cin>>x;
cout<<"请输入指数y:";
cin>>y;
cout<<"x的y次方为:"<<power(x,y)<<endl;
return 0;
}
int power(int base,int exponent)
{
if(exponent==1)//递归出口
return base;
else
return base*power(base,exponent-1);
}

[icerlion]

exponent=0呢？

lz思考问题要全面

[potato6517]

哦，感谢，忽略了，呵呵