Question

我在RPN calc中实现cmath函数时遇到问题。我注意到通话的不同行为。 如果我在POW语句中有pow（y，x），我将从表达式中得到0.5作为答案： 2 1 arctan cos pow（正确）。但是，如果输入以下内容，则使用相同的代码： 3配方8战俘，我收到0（不正确）。.

但是，如果我在pow（）调用（pow（x，y））中切换y和x 并输入： 2 1 arctan cos pow我收到0.9（不正确）如果我输入：我收到了3则8磅的战利品（正确）。.

我尝试过atan(stack.peek());，但是返回pow(y,x)时返回0，而返回pow(x,y);时返回1。

在我不确定的地方，我的逻辑肯定是错的。谢谢

stack.cpp

int main(int argc, char *argv[]){
    HPStack stack;
    string line;

    while(getline(cin,line)) {
        stringstream ss(line);
        string token;
        while(ss >> token){
            if(isdigit(token[0])){
                stack.push(atof(token.data()));
            } else if (token == "+"){
                double x = stack.pop();
                double y = stack.pop();
                stack.push(y+x);
            } else if (token == "-"){
                double x = stack.pop();
                double y = stack.pop();
                stack.push(y-x);
            } else if (token == "/"){
                double x = stack.pop();
                double y = stack.pop();
                stack.push(y/x);
            } else if (token == "*"){
                double x = stack.pop();
                double y = stack.pop();
                stack.push(y*x);
            } else if (token == "COS"){
                stack.push(cos(stack.peek()));
            } else if (token == "ARCTAN"){
                double x = stack.pop();
                double y = stack.pop();
                stack.push(atan2(x,y));
            else if (token == "RECIP"){
                double x = stack.peek();
                double y = stack.peek();
                double recip = 1/y;
                stack.push(recip);
            } else if (token == "POW"){
                double x = stack.pop();
                double y = stack.pop();
                double sum = pow(y,x);
                double answer = (int)(sum*10.0)/10.0;
                stack.push(answer);
            } 
        }
        cout << stack.peek() << endl;
    }
    return 0;
}

HPStack.h

#ifndef HPSTACK_H_
#define HPSTACK_H_

class HPStack {
    public:
        void push(double value);
        double pop();
        double peek();
    private:
        double x=0.0,y=0.0,z=0.0,t=0.0;
};

#endif

HPStack.cpp

#include "HPStack.h"

void HPStack::push(double value){
    t = z;
    z = y;
    y = x;
    x = value;
}

double HPStack::pop(){
    double out; 
    out = x;
    x = y;
    y = z;
    z = t;
    return out;
}

double HPStack::peek(){
    return x;
}

使用RPN计算器的cmath函数实现

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