重载操作符在使用std :: map时会产生错误 [英] Overloaded operator generates errors when using std::map

问题描述

class A
{
private:
    int a;
public:
    A( int set )
    {
        a = set;
    };
    ~A();
    bool operator <(const A& ref )
    {
        return this->a < ref.a;
    };
    bool operator ==(const A& ref )
    {
        return this->a == ref.a;
    };
};

int _tmain(int argc, _TCHAR* argv[])
{

    map<A,int>m;
    A a( 1 );
    m.insert( make_pair( a, 2 ) );
    for( map<A,int>::iterator it = m.begin(); it != m.end(); ++it )
    {

    }
    return 0;
}

生成C2678： http://msdn.microsoft.com/en-us/library/ys0bw32s（v = vs.80）。 aspx

如果我使用 m.find 也会为操作符生成==
如何获取

and if I use m.find will also generate for the operator == How to get around with this?

更具体地说，错误会导致：

To be more specific the error leads to:

template<class _Ty>
    struct less
        : public binary_function<_Ty, _Ty, bool>
    {   // functor for operator<
    bool operator()(const _Ty& _Left, const _Ty& _Right) const
        {   // apply operator< to operands
        return (_Left < _Right);
        }
    };

功能

/ p>

Final case:

struct MASTERPLAYER
{
    int a;
    bool operator==( const MASTERPLAYER& ref ) const 
    {
        return a == ref.a;
    }
};

int _tmain(int argc, _TCHAR* argv[])
{

    MASTERPLAYER m;
    vector<MASTERPLAYER>v;
    v.push_back( m );
    std::find( v.begin(), v.end(), 2 );

}

推荐答案

您要将函数标记为 const ，表示它们不修改调用它们的对象：

You want to mark the functions as const, to signify that they don't modify the object on which they're invoked:

bool operator <(const A& ref ) const
{
    return a < ref.a;
};
bool operator ==(const A& ref ) const
{
    return a == ref.a;
};

这篇关于重载操作符在使用std :: map时会产生错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！