Is it possible to pass a pointer to an operator as an argument like a pointer to a function?

Question

I am considering the general case, the following is just a simple example encountered which is easy to handle but has evoked my thoughts.

For example, I am using the sort() function of <algorithm>. Instead of defining a function as

bool cmp (int n1, int n2)
{
    return n1 > n2;
}

and

sort (arr, arr + N, cmp);

in the main function, I am wondering whether I can pass a pointer to the operator >, just as what I do to a pointer to a function, to the sort function. If so, how do I implement it?

Answer 1

You cannot obtain a pointer to a built-in operator. But fortunately, the standard library provides function objects for all standard operators. In your case, that object's name is std::greater:

sort (arr, arr + N, std::greater<int>{});

Since C++14, you can even omit the argument type and it will be deduced from how the object is used:

sort (arr, arr + N, std::greater<>{});

And since C++17, the empty <> can be omitted too:

sort (arr, arr + N, std::greater{});

Answer 2

You cannot do that, but you can use a lambda directly inside the sort, or store the lambda itself in a variable if you need to pass the comparator around

sort (arr, arr + N, [](int a, int b){ return a > b; });

or

auto comp = [](int a, int b){ return a > b; };
sort (arr, arr + N, comp);

or as suggested you can use the std::greater

sort (arr, arr + N, std::greater<>{});