c++ aggregates initialization with c-style arrays

Question

In c++14 I have the following type:

std::tuple<int[2], int>;

How can I properly initialize it? This

std::tuple<int[2], int> a {{2,2},3};

gives me this error:

/usr/include/c++/5/tuple:108:25: error: array used as initializer

While this:

std::tuple<std::array<int,2>, int> a {{2,2},3};

works, but I want to be able to work with standard C-style arrays

Why do you want raw arrays? std::array<int,2> solves the problem and is a lot easier to work with. If your worried about speed or efficiency note that std::array is a zero cost abstraction and as long as you compile with optimizations on you get the same assembly code. — NathanOliver
– NathanOliver, Commented Jan 31, 2019 at 16:26
std::tuple is not aggregate, and int[2] is not copyable :-/ ... — Jarod42
– Jarod42, Commented Jan 31, 2019 at 16:26
@NathanOliver, I agree with you. I have to work with existing code which uses them though, so I'm trying to see if there are other ways — Mario Demontis
– Mario Demontis, Commented Feb 1, 2019 at 11:48

YSC · Accepted Answer · 2019-01-31 16:42:34Z

7

std::tuple is not an aggregate and doesn't provide a list-initializer constructor. This makes list initialization with that type impossible to use with C-arrays.

You can however use std::make_tuple:

auto a = std::make_tuple<int[2], int>({2,2},3);

answered Jan 31, 2019 at 16:37

YSC

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3 Comments

or std::make_tuple<int[2]>({2,2},3); :). and make_* is normally to let compiler deduce type...

@Jarod42 That's cleverly bizarre. Won't use ;)

The type turns out to be std::tuple<int*, int>, so we're not there yet unfortunately.