CODEWITHSUNDEEP
c++initialization
Serve Laurijssen
Serve Laurijssen

Reputation: 9753

Aggregate initialization of a derived class

The following code does not compile with either Visual Studio2017 or online GDB. I expected it to compile as iterator is just a class with types and it is inherited from publicly. Is this not allowed or doesnt this work in VS2017?

template<typename T>
struct Gen : public std::iterator<std::input_iterator_tag, T>
{
    T value;
};

int main()
{   
    Gen<int> g = Gen<int>{ 10 }; // this doesnt
    Gen<int> g2 = Gen<int>{ {}, 10 }; // neither does this
}

The error is

Error C2440 'initializing': cannot convert from 'initializer list' to 'Gen'

Upvotes: 10

Views: 3168

Answers (1)

YSC
YSC

Reputation: 40110

What

Gen<int> g = Gen<int>{ 10 };

tries to do is calling the non-existent Gen<int>(int) constructor. What you want to do is aggregate initialization whose syntax is:

Gen<int> g = { {}, 10 };

And only works since C++17 for derived types:

If the initializer clause is a nested braced-init-list (which is not an expression), the corresponding array element/class member /public base (since C++17) is list-initialized from that clause: aggregate initialization is recursive.

For more information, aggregate initialization is defined in the following standard sections.

[dcl.init.list]/3

  1. List-initialization of an object or reference of type T is defined as follows:
    3.1 If the braced-init-list contains a designated-initializer-list, T shall be an aggregate class.

and

[dcl.init.aggr]/1

An aggregate is an array or a class (Clause 12) with
1.1 no user-provided, explicit, or inherited constructors (15.1),
1.2 no private or protected non-static data members (Clause 14),
1.3 no virtual functions (13.3), and
1.4 no virtual, private, or protected base classes (13.1).

Because in inherits from std::iterator<std::input_iterator_tag, T>.

Upvotes: 14

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