Deduce non-type template parameter

Question

Is it possible to deduce a non-type template parameter from a template function parameter?

Consider this simple template:

template <int N> constexpr int factorial()
{
        return N * factorial<N - 1>();
}

template <> constexpr int factorial<0>()
{
        return 1;
}

template <> constexpr int factorial<1>()
{
        return 1;
}

I would like to be able to change factorial so that I can alternatively call it like this:

factorial(5);

and let the compiler figure out the value of N at compile time. Is this possible? Maybe with some fancy C++11 addition?

Answer 1

Your current code would normally be written as follows, I believe:

constexpr factorial (int n)
{
    return n > 0 ? n * factorial( n - 1 ) : 1;
}

If you call it with a constant-expression, such as factorial(5), then all the compiler magic will come into play. But if you do int a = 3; factorial(a), then I think it will fall back on a conventional function - i.e. it won't have built a lookup table of pre-computed answers.

In general, you should mark every function and constructor as constexpr if you can. You lose nothing, the compiler will treat it as a normal function if necessary.