数值的整数次方

题目：实现函数double Power（double base,int exponent），求base的exponent次方。

要点：定义域解析、错误处理、效率

代码：

#include<iostream>
using namespace std;
bool g_InvalidInput=false;
double powerWithUnsignedExponent(double base,unsigned int exponent);
bool equal(double num1,double num2)
{
if((num1-num2>-0.0000001)&&(num1-num2<0.0000001)) return true;
else return false;
}
double power(double base,int exponent)
{
g_InvalidInput=false;
if(equal(base,0.0)&&exponent<0)
{
g_InvalidInput=true;
return 0.0;
}
unsigned int absExponent=(unsigned int)(exponent);
if(exponent<0) absExponent=(unsigned int)(-exponent);

double result=powerWithUnsignedExponent(base,absExponent);
if(exponent<0) result=1.0/result;
return result;
}
double powerWithUnsignedExponent(double base,unsigned int exponent)
{
if(exponent==0) return 1;
if(exponent==1) return base;

double result=powerWithUnsignedExponent(base,exponent>>1);
result*=result;
if(exponent&0x1==1) result*=base;
return result;
}
int main()
{
double base;
int exponent;
cout<<"please enter base:";
cin>>base;
cout<<"please enter exponent:";
cin>>exponent;
cout<<base<<"^"<<exponent<<"="<<power(base,exponent)<<endl;
return 0;
}

reference：剑指Offer（何海涛著）