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c++ - 为什么这个重载运算符给出错误的输出？

问题描述

加减整数的输出给了我正确的解决方案，但是当尝试将字符串添加到 int 时应该会产生错误，但事实并非如此。不知道我哪里出错了。

ERROR + 3 is ERROR
ERROR - 4 is ERROR
ERROR + hello is ERROR
ERROR - world is ERROR
3 + ERROR is ERROR
3 - ERROR is ERROR
3 + 4 is 7
3 - 4 is -1
**3 + hello is 7  
3 - world is -1
4 + ERROR is 7** 
4 + 3 is 7
4 - 3 is 1
**4 + hello is 7
4 - world is 1
**hello + ERROR is 7
hello - 3 is 1
hello + 4 is 7**
hello + world is helloworld
**hello - world is 1**
world + ERROR is helloworld
world - ERROR is 1
world + 3 is helloworld
world - 3 is 1
world + 4 is helloworld
world - 4 is 1
world + hello is worldhello
world - hello is 1

下面是带有 getter 的头文件，用于返回一个 in 或一个字符串。

#ifndef VALUE_H
#define VALUE_H

#include <iostream>
#include <string>
using namespace std;

enum ValType
{
    VINT,
    VSTR,
    VERR
};

class Value
{
    ValType T;
    int I;
    string S;

public:
    Value() : T(VERR), I(0) {}
    Value(int vi) : T(VINT), I(vi) {}
    Value(string vs) : T(VSTR), I(0), S(vs) {}

    ValType GetType() const { return T; }
    bool IsErr() const { return T == VERR; }
    bool IsInt() const { return T == VINT; }
    bool IsStr() const { return T == VSTR; }

    int GetInt() const
    {
        if (IsInt())
            return I;
        throw "RUNTIME ERROR: Value not an integer";
    }
    string GetStr() const
    {
        if (IsStr())
            return S;
        throw "RUNTIME ERROR: Value not a string";
    }

    // add op to this
    Value operator+(const Value &op) const
    {
        try
        {
            if (IsInt() && op.IsInt()) // integer addition
                return GetInt() + op.GetInt();
            if (IsStr() && op.IsStr()) // string addition
                return GetStr() + op.GetStr();
        }
        catch (std::string &e)
        {
            cout << "RUNTIME ERROR " << e << endl;
        }

        //throw "RUNTIME ERROR...";
    }

    // subtract op from this
    Value operator-(const Value &op) const
    {
        try
        {
            if (IsInt() && op.IsInt()) // integer addition
                return GetInt() - op.GetInt();
        }
        catch (std::string &e)
        {
            cout << "RUNTIME ERROR " << e << endl;
        }
        //throw "RUNTIME ERROR...";
    }

    // multiply this by op
    Value operator*(const Value &op) const;

    // divide this by op
    Value operator/(const Value &op) const;

    friend ostream &operator<<(ostream &out, const Value &op)
    {
        if (op.IsInt())
            out << op.I;
        else if (op.IsStr())
            out << op.S;
        else
            out << "ERROR";
        return out;
    }
};

#endif

下面是主要驱动，将整数和字符串作为值，循环遍历它们

#include <iostream>
#include <string>
#include <vector>
using namespace std;
#include "val.h"

void doOp(const Value &a, const Value &b)
{
    cout << a << " + " << b << " is " << flush << a + b << endl;
    cout << a << " - " << b << " is " << flush << a - b << endl;
}



int main()
{

    Value a;
    Value b(3);
    Value c(4);
    Value d("hello");
    Value e("world");

    vector<Value> vals({a, b, c, d, e});

    cout << "a=" << a << endl;
    cout << "b=" << b << endl;
    cout << "c=" << c << endl;
    cout << "d=" << d << endl;
    cout << "e=" << e << endl;

    for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
            if (i != j)
                doOp(vals[i], vals[j]);

    return 0;
}

标签： c++operator-overloading

解决方案

当被比较的两个对象不是同一类型时，您的operator+和的返回值operator-是不确定的。Value当发生不匹配时，您需要返回一个空的，因为您的默认构造函数将成员Value初始化为：ValType TVERR

Value operator+(const Value &op) const
{
    if (IsInt() && op.IsInt()) // integer addition
        return GetInt() + op.GetInt();
    if (IsStr() && op.IsStr()) // string addition
        return GetStr() + op.GetStr();
    return Value();
}

// subtract op from this
Value operator-(const Value &op) const
{
    if (IsInt() && op.IsInt()) // integer subtraction
        return GetInt() - op.GetInt();

    // in comments, you said you want "hello - world" to be "hello"
    if (IsStr() && op.IsStr()) // string subtraction
        return GetStr();

    return Value();
}

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