how to use std::find with an < operator that is not std::less

Question

say I have

class newVector: public std::vector<T> {
    public:
        bool operator< (const newVector& v) { 
            //..
        }
};

And

a std::set<newVector>;

I can't manage to use a.find(...) properly, I am not sure what to put into the (...) in order to use newVector::operator<. When I just put a.find(element) it uses std::less. Should I change std::less somehow?

The question I have is why are you deriving from `std::vector`? — PaulMcKenzie, Jul 01 '15 at 17:07
Bad idea to derive from `std::vector` or any standard library class — The Vivandiere, Jul 01 '15 at 17:09
I want to add some things to the std::vector. Someone told me composition is better but this should not be forbidden? — myelf, Jul 01 '15 at 17:09
Umm..yes, it is forbidden (in terms of guidelines). Scott Meyers discusses this in Effective C++ — The Vivandiere, Jul 01 '15 at 17:09
@myelf Instead of listening to this "someone", research the reasons why deriving from `std::vector` is not a good idea. — PaulMcKenzie, Jul 01 '15 at 17:10

score 1 · Accepted Answer · answered Jul 01 '15 at 17:10

1

Ignoring for the time being that deriving from std::vector is a bad idea, I can think of the following ways to address the issue:

Define operator< for objects of newVector.

class newVector: public std::vector<T> {
    public:
        bool operator< (const newVector& v) const { 
            //..
        }

and

std::set<newVector> a;
a.find(...);

Define a functor that has appropriate operator() function and use it to create the std::set.

template <typename T>
struct NewVectorLess
{
   bool operator()(newVector<T> const& lhs, newVector<T> const& rhs)
   {
     // ...
   }
};

and

std::set<newVector<int>, NewVectorLess<int>> a;
a.find(...);

answered Jul 01 '15 at 17:10

R Sahu

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to be clear this are 2 options? I don't understand the first one I guess there is a const in there but other than that nothing has changed. – myelf Jul 01 '15 at 17:16
@myelf, Yes. These are two options. In your posted code, you had not only missed the `const` but the argument type is `ImpreciseVector`. – R Sahu Jul 01 '15 at 17:18
Oh, that ImperciseVector was an artifact from bad editing. Thank you. – myelf Jul 01 '15 at 17:22

UmNyobe · Answer 2 · 2015-07-01T17:26:30.973

You don't need to overload the vector, or to change std::less, but to define separately your own std::less compatible function object.

#include <iostream>
#include <vector>
#include <set>
using namespace std;

    struct OppositeVectorComp
    {
        template< class T, class Alloc >
        bool operator()( const std::vector<T,Alloc>& lhs,const std::vector<T,Alloc>& rhs )
        {           
           return   !(lhs < rhs);
        }
    };

int main() {
    std::vector<int> a , b;

    std::set<std::vector<int>> defaultset;
    std::set<std::vector<int>, OppositeVectorComp> myset;

    a.push_back(1);
    b.push_back(2);

    myset.insert(a);
    myset.insert(b);

    defaultset.insert(a);
    defaultset.insert(b);

    std::cout << (*myset.begin())[0] << std::endl; // output 2
    std::cout << (*defaultset.begin())[0] << std::endl; // output 1

    return 0;
}

Here OppositeVectorComp define a new order on vectors where

OppositeVectorComp(a,b) true iff a <b is false

By using the type std::set<std::vector<int>, OppositeVectorComp> we define a set which use the custom std::less.

how to use std::find with an < operator that is not std::less

2 Answers2