Partial class template argument deduction in C++17

Question

In the example below, we use the C++17 feature "Class template argument deduction" to deduce that val is of type Base<int, double, bool>:

template<class T, class U, class V>
struct Base {
    Base(T, U) { };
    Base(T, U, V) { };
    Base(V) { };
};

void func() {
    Base val(1, 4., false);
}

Now, is it possible to partially specify the template arguments, and let the remaining ones be deduced? Effectively something like this:

Base<V = bool> val1(1, 4.);        // U & V deduced --> Base<int, double, bool>
Base<T = bool, T = int> val2(5.);  // V deduced     --> Base<bool, int, double>

I've tried e.g.

template<class T, class U> using Base2 = Base<T, U, double>;

void func() {
    NewBase2 val(1, 2);
}

but it doesn't compile: 'Base2': use of alias template requires template argument list.

Is partial deduction possible somehow? If it is not possible directly, are there any good workarounds?

In the example given, you can specify T and have U and have V deduced, or specify T and U, and have V deduced. See the pattern? It would be nice to have named template parameters in C++, unfortunately we do not. — Sam Varshavchik

Jarod42 Jarod42 · Accepted Answer · 2019-08-19T21:00:34

You might add deduction guide as follow:

template<class T, class U, class V>
Base(T, U) -> Base<T, U, bool>;

template<class V>
Base(V) -> Base<bool, int, V>;

which allows

Base val1(1, 4.); // Base<int, double, bool>
Base val2(5.);    // Base<bool, int, double>

If you want to specify the "default" template, you might use the old way with make_

template <typename V, typename T, typename U>
Base<T, U, V> make_Base(T t, U u)
{
    return Base<T, U, V>{t, u};
}

template <typename T, typename U, typename V>
Base<T, U, V> make_Base(V v)
{
    return Base<T, U, V>{v};
}


auto val1 = make_Base<bool>(1, 4.);   // Base<int, double, bool>
auto val2 = make_Base<bool, int>(5.); // Base<bool, int, double>

Partial class template argument deduction in C++17

2 Answers