Question

typedef struct Edge
{
    int v1, v2, w;
    bool operator <(const Edge &rhs) const{
        bool b = v1 < rhs.v1 || v2 < rhs.v2;
        return (b);
    }
} edge;


template <class T>
struct my_less
{
    bool operator()(const T& _Left, const T& _Right) const
    {
        return (_Left < _Right);
    }
};

int main(int argc, char *argv[])
{
    set <edge, my_less<edge> > F;

    edge e3 = { 3, 3, 3};
    edge e4 = { 3, 7, 3};
    edge e5 = { 2, 7, 3};

    F.insert(e3);
    printf("e3 ? %d\n", F.find(e3)!=F.end()); // O
    printf("e4 ? %d\n", F.find(e4)!=F.end()); // O
    printf("e5 ? %d\n", F.find(e5)!=F.end()); // X

    //printf("%d\n", e3<e4);

    return 0;
}

If run this code, I got an error at "F.find(e5)!=F.end()" with following message.
"Debug Assertion Failed!. Expression: invalid operator < "

The condition of two edges of '(x,y), (p,q)' equality is 
 !(x < p || y < q) && !(p < x || q < y)

It can be '(x>=p && y>=q) && (p>=x && q>=y)'

I really don't know why assertion raised.
Is there something wrong?

Answer 1

您的比较不会强制执行严格的排序。例如：

Edge e1;
Edge e2;

e1.v1 = 5;
e1.v2 = 4;

e2.v1 = 4;
e2.v2 = 5;

// e1 < e2 is true
// e2 < e1 is true
// So which one should we really trust? Neither, let's abort the program!

您需要让<运营商真正像<那样工作。如果e1 < e2为真，则e2 < e1必须为假。

我认为这可能是你想要的，但要注意我没有测试过它：

return v1 < rhs.v1 || (v1 == rhs.v1 && v2 < rhs.v2);

理论上应该按v1排序，并使用v2来打破关系。

使用std :: set find和struct data

1 个答案: