在数值编程中，我经常想要将数字提升到表达式中的幂。这会导致代码出现数百次pow(x,2)或类似的事情。
这会使代码变得混乱并使其更难以解析，尤其是在将编写的方程与代码进行比较时。
我想在C ++中编写一个求幂运算符，如果可能的话。
a^b == pow(a, b)
我已尝试为operator()(double, double)或operator[](double, double)编写重载（选择因为它比（二进制）*更紧密。但是我无法使其工作。
这样的事情有可能吗？
也许是预处理器欺骗？
答案 0 :(得分：2)
这样的东西？
#include <iostream>
#include <cmath>
using namespace std;
class Exp {
public:
double val ;
Exp(double val) : val(val) {}
Exp operator^(const Exp &exp) {
return Exp(pow(val, exp.val));
}
operator double () {
return val;
}
};
ostream &operator << (ostream &out, const Exp &exp) {
out << exp.val;
return out;
}
int main(){
Exp a = 2;
Exp b = 3;
cout << (a + b) << ", " << (a ^ b) << endl;
return 0;
}
答案 1 :(得分：0)
您可以定义自己的用户定义文字，该文字返回带有调用操作符的对象以执行取幂，如
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这种方法的明显缺点是它只适用于文字值，即没有5.123_E(3)
（虽然a_E(b)可行）。
5.123_E(b)