std::set::find vs std::find on std::set with const

Question

I wrote a little (working) test code but I do not understand why in the test1 function I can only pass a int* const as parameter while in the test2 function I can pass a const int*. If I pass a const int* to test1, I get a discard qualifier error.

In my research, I found that both std::find and set::find have a const version so I can't see why they behave differently. I also tried with boost::container::flat_set instead of a std::set and I got the same result. Could someone explain me please?

class myClass
{
public:
    myClass() {};
    ~myClass() {};

    void add(int* ref)
    {
        this->_ref.insert(ref);
    };

    bool test1(int* const ref) const
    {
        return ( this->_ref.find(ref) != this->_ref.end() );
    }

    inline
    bool test2(const int* ref) const
    {
        return ( std::find(this->_ref.begin(), this->_ref.end(), ref) != this->_ref.end() );
    }

    std::set<int*> _ref;
};

int main()
{
    myClass test;
    test.add(new int(18));
    test.add(new int(35));
    test.add(new int(78));
    test.add(new int(156));

    std::cout<<test.test1(0)<<std::endl;
    std::cout<<test.test1(*test._ref.begin())<<std::endl;

    std::cout<<test.test2(0)<<std::endl;
    std::cout<<test.test2(*test._ref.begin())<<std::endl;

    return 0;
}

Answer 1

set::find() gives the answer in O(logN) while std::find() gives the answer in O(N) .

Similarly, map::find() gives the answer in O(logN) while std::find() gives the answer in O(N) .