Mfrhtyj

I ran into a weird problem when trying to move to C++17. The problem is that something (and I'm not sure what) changed in C++17 that made list-initialization work differently in the case of a default constructor. I tried to search https://en.cppreference.com/w/cpp/language/list_initialization for more info, but I didn't find anything that looks relevant.

Does someone know the reason the code below compiles in C++14 but not in C++17 when calling B{} instead of B()?
(I tried it in both gcc 8.2 and 7.3 and icc 19)

struct A{

protected:

    A() {}

};



struct B : public A {};





B f(){

    return B(); //compilation OK

    //return B{}; //compilation error

}

In C++14, the definition of aggregate was:

An aggregate is an array or a class (Clause [class]) with no user-provided constructors ([class.ctor]), no private or protected non-static data members (Clause [class.access]), no base classes (Clause [class.derived]), and no virtual functions ([class.virtual]).

Hence, B is not an aggregate. As a result B{} is surely not aggregate initialization, and B{} and B() end up meaning the same thing. They both just invoke B's default constructor.

However, in C++17, the definition of aggregate was changed to:

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 ]

The restriction is no longer on any base classes, but just on virtual/private/protected ones. But B has a public base class. It is now an aggregate! And C++17 aggregate initialization does allow for initializing base class subobjects.

In particular, B{} is aggregate initialization where we just don't provide an initializer for any subobject. But the first (and only) subobject is an A, which we're trying to initialize from {} (during aggregate initialization, any subobject without an explicit initializer is copy-initialized from {}), which we can't do because A's constructor is protected and we are not a friend (see also, the quoted note).

Note that, just for fun, in C++20 the definition of aggregate will change again.

From my understanding of https://en.cppreference.com/w/cpp/language/value_initialization

B{} does an aggregate_initialization,

and since C++17:

The effects of aggregate initialization are:

• Each direct public base, (since C++17) [..] is copy-initialized from the corresponding clause of the initializer list.

and in our case:

If the number of initializer clauses is less than the number of members and bases (since C++17) or initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default initializers, if provided in the class definition, and otherwise (since C++14) by empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates). If a member of a reference type is one of these remaining members, the program is ill-formed.

So B{/*constructor of A*/} need to construct base class A, which is protected...

The final draft of C++17 n4659 has a compatibility section which contains the changes with respect to previous versions.

C.4.4 Clause 11: declarators [diff.cpp14.decl]

11.6.1

Change: Definition of an aggregate is extended to apply to user-defined types with base classes.

Rationale: To increase
convenience of aggregate initialization.

Effect on original feature:
Valid C++ 2014 code may fail to compile or produce different results
in this International Standard; initialization from an empty
initializer list will perform aggregate initialization instead of
invoking a default constructor for the affected types:

struct derived;

struct base {

friend struct derived;

private:

base();

};

struct derived : base {};

derived d1{}; // Error. The code was well-formed before.

derived d2; // still OK

I compiled the above example code with -std=c++14 and it compiled but failed to compile with -std=c++17.

I believe that could be the reason why the code in the OP fails with B{} but succeeds with B().