C++ function objects with explicit template parameters

Question

I have a function object with an explicit(ie non-deduced) template parameter defined like this:

struct foo
{
    template<class T>
    T operator()() const
    {
        return 5;
    }
};

foo bar = {};

When I try to call it like this:

int main()
{
    int i = bar<int>();
    return 0;
}

I get a compile error. Is there no way to call the function object with a template parameter like a regular function? I really need to have it as a function object. Making a free function is not really an option for me(or at least, it is a very messy option).

Answer 1

Unfortunately, you can't call it like that. You need to use the operator() syntax:

int i = bar.operator()<int>();