CODEWITHSUNDEEP
c++operator-overloading
Karlovsky120
Karlovsky120

Reputation: 6352

Overloading less than operator in c++ for use in std::sort

I have a struct defined like this:

struct IFSFunc {
    int a;

    bool operator<(const IFSFunc& other) {
        return a < other.a;
    }
};

Since IFSfunc is a struct, access modifier for the operator< should be public.

I also have this code:

#include <algorithm>
std::vector<std::pair<double, IFSFunc>> ifsFuncs;

// fill the vector with various data

std::sort(ifsFuncs.begin(), ifsFuncs.end());

I need to sort ifsFuncs based on the first double in the pair. I don't care about IFSFunc structure, if the double is the same.

However, for std::sort to work, which is defined like this:

template <class _Ty1, class _Ty2>
_NODISCARD constexpr bool operator<(const pair<_Ty1, _Ty2>& _Left, const pair<_Ty1, _Ty2>& _Right) {
    return _Left.first < _Right.first || (!(_Right.first < _Left.first) && _Left.second < _Right.second);
}

I have to override the less than operator for the second in this case IFSfunc, which I did. However, trying to compile this code gives me the following error:

Error C2678 binary '<': no operator found which takes a left-hand operand of type 'const _Ty2' (or there is no acceptable conversion)

Why?

Upvotes: 0

Views: 904

Answers (2)

Zan Lynx
Zan Lynx

Reputation: 54325

You need to define that operator as a const member function.

Also, don't just return true for a comparison. That can result in infinite looping.

Upvotes: 3

Karlovsky120
Karlovsky120

Reputation: 6352

I just figure it out. The overloaded function signature was wrong, this is what I need:

struct IFSFunc {
    int a;

    bool operator<(const IFSFunc& other) const {
        return a < other.a;
    }
};

Notice that the operator< is now a const function.

Upvotes: 0

Related Questions