Wrapper hash function to return hash of wrapped type

Question

I have the following code:

#include <bits/stdc++.h>
using namespace std;

// wrapper class on type E
template <typename E>
class g {
public:
    E val;
    g(E x) : val(x) {};
};

// hash for g<E> should be hash for E
template<typename E> struct std::hash<g<E>> { 
    std::size_t operator()(const ::g<E> &t) const { 
        return std::hash<E>()(t); 
    } 
};

int main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    unordered_set<g<int>>{g<int>(3)};
}

Basically, the idea is that I have a wrapper around a templated type, and I want to be able to use this wrapper class in an unordered_set/map. But I get the following error: no match for call to '(std::hash<int>) (const g<int>&)'. This is strange - doesn't c++ implement this hash? What am I doing wrong?

Answer 1

Your code is attempting to invoke std::hash<int>() on t, which has type g<int>. You need to "unwrap" the wrapper and call std::hash<int>() on the enclosed int:

return std::hash<E>()(t.val);