Is it possible to expand non-variadic arguments in a variadic template function?

Question

It is probably easier to explain what I mean by an example. Imagine a following template:

template <class... Args> std::tuple<Args...> foo();

It can be invoked, for example, like this:

auto ret = foo<int, bool>();

But what if I want to pass additional arguments to the function, based on the number of variadic template arguments? For example, let's say I want to pass a character string literal for every Args:

auto ret = foo<int, bool>("a", "b");

The problem with this, is that it does not seem possible to expand non-variadic arguments, so the following obviously doesn't compile:

template <class... Args> std::tuple<Args...> foo(const char*... names);

Is there any sensible way to implement this?

There's no need for tuple in this case, because all types are of the same type. I could use initializer_list, but the question is specifically about an arguments. — ovk
– ovk, Commented May 21, 2018 at 14:08
You can still use variadic template, and than have a static assert to make sure those arguments are of the allowed types. — SergeyA
– SergeyA, Commented May 21, 2018 at 14:10

Dan M. · Accepted Answer · 2018-05-21 14:05:35Z

You can do this with something like

template <class... Args> std::tuple<Args...> foo(proxy<Args, const char*>... names);

where proxy is

template<class T, class E> using proxy = E;

You can see this in action here: https://godbolt.org/g/SHBYzy

Very elegant solution. No checks needed at all to make sure the list have the same amount of elements. +1

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