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我将如何添加一个允许用户输入类似 2 + 2 或 10 / 5 的函数,然后简单地运行对象来计算它,就像他们使用我的“输入第一个输入”语句手动输入它时一样。因此,作为作业的一部分,我需要允许用户在控制台中输入类似 10 / 5 + 1 / 2 的内容。我还需要能够允许运算符重载,我不确定我的程序当前是否允许这样做。任何帮助,将不胜感激。谢谢!

#include <iostream>
#include <conio.h>

using namespace std;

class Rational
{
    private:
        float numInput;

    public:
        Rational(): numInput(0)
        {}

        void getValues()
        {           
            cout << "Enter number: ";
            cin >> numInput;
        }

        void showValues()
        {
            cout << numInput << endl;
        }

        Rational operator + (Rational) const;
        Rational operator - (Rational) const;
        Rational operator * (Rational) const;
        Rational operator / (Rational) const;
};

Rational Rational::operator + (Rational arg2) const
{
    Rational temp;
    temp.numInput = numInput + arg2.numInput;
    return temp;
}

Rational Rational::operator - (Rational arg2) const
{
    Rational temp;
    temp.numInput = numInput - arg2.numInput;
    return temp;
}

Rational Rational::operator * (Rational arg2) const
{
    Rational temp;
    temp.numInput = numInput * arg2.numInput;
    return temp;
}

Rational Rational::operator / (Rational arg2) const
{
    Rational temp;
    temp.numInput = numInput / arg2.numInput;
    return temp;
}

int main()
{
    Rational mathOb1, mathOb2, outputOb;
    int choice;
    mathOb1.getValues();
    cout << "First number entered: ";
    mathOb1.showValues();
    cout << endl;
    cout << "Enter operator: + = 1, - = 2, * = 3, / = 4  ";
    cin >> choice;
    cout << endl;
    mathOb2.getValues();
    cout << "Second number entered: ";
    mathOb2.showValues();    cout << endl;

    switch (choice)
    {
        case 1:
            outputOb = mathOb1 + mathOb2;
            break;
        case 2:
            outputOb = mathOb1 - mathOb2;
            break;
        case 3:
            outputOb = mathOb1 * mathOb2;
            break;
        case 4:
            outputOb = mathOb1 / mathOb2;
            break;
        default:
            cout << "Invalid choice! " << endl;
    }
    cout << "Answer: ";
    outputOb.showValues();
    cout << endl;

    return 0;
}
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1 回答 1

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你不能使用cin >> {int},如果你提供 a ,那只会失败char,你会从那里卡住。

只需使用std::getline并从那里解析出令牌:

std::string expression;
std::getline(std::cin, expression);

然后,您可以使用多种方法中的任何一种将 a 拆分为该问题string中表达的标记,并循环遍历标记并根据它们是运算符还是数字来解释它们。

于 2015-01-31T22:20:11.810 回答