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Why is std::is_aggregate an aggregate?

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7

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?

c++ std c++17 typetraits

share|improve this question

edited 14 hours ago

Lightness Races in Orbit

296k55480820

asked 14 hours ago

x432phx432ph

2145

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  en.cppreference.com/w/cpp/language/aggregate_initialization Non-virtual public base classes are allowed in C++17
  
  – KABoissonneault
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Scroll down on the page you linked; read the whole page ;)
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yeah, didn't realize that the linked text continues over multiple answers ;)
  
  – x432ph
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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
  14 hours ago
  

  
  
  
  

add a comment | 

7

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?

c++ std c++17 typetraits

share|improve this question

edited 14 hours ago

Lightness Races in Orbit

296k55480820

asked 14 hours ago

x432phx432ph

2145

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  en.cppreference.com/w/cpp/language/aggregate_initialization Non-virtual public base classes are allowed in C++17
  
  – KABoissonneault
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Scroll down on the page you linked; read the whole page ;)
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yeah, didn't realize that the linked text continues over multiple answers ;)
  
  – x432ph
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @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
  14 hours ago
  

  
  
  
  

add a comment | 

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?

c++ std c++17 typetraits

share|improve this question

edited 14 hours ago

Lightness Races in Orbit

296k55480820

asked 14 hours ago

x432phx432ph

2145

share|improve this question

edited 14 hours ago

Lightness Races in Orbit

296k55480820

asked 14 hours ago

x432phx432ph

2145

share|improve this question

edited 14 hours ago

Lightness Races in Orbit

296k55480820

asked 14 hours ago

x432phx432ph

2145

edited 14 hours ago

Lightness Races in Orbit

296k55480820

edited 14 hours ago

Lightness Races in Orbit

296k55480820

asked 14 hours ago

x432phx432ph

2145

asked 14 hours ago

x432phx432ph

2145

2 Answers
2

active

oldest

votes

10

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

share|improve this answer

edited 14 hours ago

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  And std::is_aggregate is new in C++17 so OP must be using it 👍
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Pertinent link into the OP's source (they didn't read it all)
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yep, there are answers for all standards published so far - OP over there should re-accept, ideally
  
  – Lightness Races in Orbit
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @LightnessRacesinOrbit I've updated the main answer to point to them to make it more obvious.
  
  – NathanOliver
  14 hours ago
  

  
  
  
  

 | 
show 1 more comment

3

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

share|improve this answer

answered 14 hours ago

SergeyASergeyA

45.3k53990

add a comment | 

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10

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

share|improve this answer

edited 14 hours ago

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  And std::is_aggregate is new in C++17 so OP must be using it 👍
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Pertinent link into the OP's source (they didn't read it all)
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yep, there are answers for all standards published so far - OP over there should re-accept, ideally
  
  – Lightness Races in Orbit
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @LightnessRacesinOrbit I've updated the main answer to point to them to make it more obvious.
  
  – NathanOliver
  14 hours ago
  

  
  
  
  

 | 
show 1 more comment

10

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

share|improve this answer

edited 14 hours ago

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  And std::is_aggregate is new in C++17 so OP must be using it 👍
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Pertinent link into the OP's source (they didn't read it all)
  
  – Lightness Races in Orbit
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yep, there are answers for all standards published so far - OP over there should re-accept, ideally
  
  – Lightness Races in Orbit
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @LightnessRacesinOrbit I've updated the main answer to point to them to make it more obvious.
  
  – NathanOliver
  14 hours ago
  

  
  
  
  

 | 
show 1 more comment

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

share|improve this answer

edited 14 hours ago

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

share|improve this answer

edited 14 hours ago

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

answered 14 hours ago

NathanOliverNathanOliver

99.7k16139220

3

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

share|improve this answer

answered 14 hours ago

SergeyASergeyA

45.3k53990

add a comment | 

3

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

share|improve this answer

answered 14 hours ago

SergeyASergeyA

45.3k53990

add a comment | 

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

share|improve this answer

answered 14 hours ago

SergeyASergeyA

45.3k53990

share|improve this answer

answered 14 hours ago

SergeyASergeyA

45.3k53990

answered 14 hours ago

SergeyASergeyA

45.3k53990

answered 14 hours ago

SergeyASergeyA

45.3k53990