这张地图有什么问题？

Question

I am a bit puzzled, why the following code is not working as expected...

#include <iostream>
#include <map>

struct Foo {
    int x,y,z;
    Foo(int x,int y,int z) : x(x),y(y),z(z) {}
    bool operator<(const Foo& other) const {
       if (x > other.x) return false;
       if (y > other.y) return false;
       if (z > other.z) return false;
       return true;
    }
    bool operator==(const Foo& other) const {
        if (other.x != x) return false;
        if (other.y != y) return false;
        if (other.z != z) return false;
        return true;
    }
};

int main() {
    Foo f(1,2,3);
    std::map<Foo,double> map;
    map[f] = 1.0;
    std::cout << map[f] << "\n";
}

It prints 0 and not 1 . What did I do wrong?

code also here: http://ideone.com/fork/HMwPQ7

Answer 1

It is because your operator< has been implemented incorrectly. Here is the correct version:

bool operator<(const Foo& other) const {
   if (x != other.x) return x < other.x;
   else if (y != other.y) return y < other.y;
   return z < other.z;
}

That basically says if x are equal, then compare y , if that too equals, then compare z .

Read about Strict Weak Ordering .

A shorter implementation could be this:

bool operator<(const Foo& other) const {
   return std::tie(x, y,z) < std::tie(other.x, other.y, other.z);
}

Hope that helps.

Answer 2

Foo::operator < does not define the strict weak ordering required by std::map .
In particular, given two identical Foo s a and b , both a < b and b < a are true, when they should both be false.
Since you broke std::map 's contract, the behaviour is undefined.