Why is std::is_aggregate an aggregate?

Question

I always was under the impression that types like std::is_same, std::is_void, or std::is_aggregate are supposed to inherit from std::integral_constant, or more specifically from std::bool_constant.

However, aggregate classes must not have a base class by definition, but when I use these types as T in std::is_aggregate_v<T>, I get true. So apparently, they are not derived from std::bool_constant?

So my question is:

Why is std::is_aggregate_v<std::is_aggregate<void>> true, at least with GCC and Clang? Doesn't the standard specify that std::is_aggregate is derived from std::bool_constant? If not, does this mean it leaves the value of the above line as an implementation detail?

https://en.cppreference.com/w/cpp/language/aggregate_initialization Non-virtual public base classes are allowed in C++17 — KABoissonneault, Apr 23 '19 at 14:05
Scroll down on the page you linked; read the _whole_ page ;) — Lightness Races in Orbit, Apr 23 '19 at 14:10
Yeah, didn't realize that the linked text continues over multiple answers ;) — x432ph, Apr 23 '19 at 14:20
@x432ph It's usually a good idea to read all (or at least most) answers on a page; there's a reason we have a one-to-many Q&A model! — Lightness Races in Orbit, Apr 23 '19 at 14:21

score 12 · Accepted Answer · edited Jun 20 '20 at 09:12

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However, aggregate classes must not have a base class by definition

This is no longer true. [dcl.init.aggr]/1 defines an aggregate as

An aggregate is an array or a class with

no user-provided, explicit, or inherited constructors ([class.ctor]),

no private or protected non-static data members (Clause [class.access]),

no virtual functions, and

no virtual, private, or protected base classes ([class.mi]).

[ Note: Aggregate initialization does not allow accessing protected and private base class' members or constructors. — end note ]

There is no longer a condition that it does not have a base class like it did in C++14 and earlier. As long as it has a public, non virtual base class that is now allowed. This means that the type traits are now considered aggregates as long as the above holds true for them

edited Jun 20 '20 at 09:12

Community

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answered Apr 23 '19 at 14:07

NathanOliver

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And `std::is_aggregate` is new in C++17 so OP _must_ be using it – Lightness Races in Orbit Apr 23 '19 at 14:10
[Pertinent link into the OP's source](https://stackoverflow.com/a/50786105/560648) (they didn't read it all) – Lightness Races in Orbit Apr 23 '19 at 14:11
@LightnessRacesinOrbit Oh, cool. I was about to write a comment on the top answer to have it edited but if there is a C++17 specific answer I'll just add a link to it so its easier to see that changes have happened. – NathanOliver Apr 23 '19 at 14:12
Yep, there are answers for all standards published so far - OP over there should re-accept, ideally – Lightness Races in Orbit Apr 23 '19 at 14:13
@LightnessRacesinOrbit I've updated the main answer to point to them to make it more obvious. – NathanOliver Apr 23 '19 at 14:18
Delicious! Thanks. – Lightness Races in Orbit Apr 23 '19 at 14:21

score 3 · Answer 2 · answered Apr 23 '19 at 14:08

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Since C++17, classes with non-virtual, not private or protected bases are aggregates: https://en.cppreference.com/w/cpp/language/aggregate_initialization

answered Apr 23 '19 at 14:08

SergeyA

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Why is std::is_aggregate an aggregate?

2 Answers2