在 C++11 中使用（阴影）基类数据成员初始化结构

Question

It's possible to add members to a derived struct that shadow same-name members of the base class, whether this is an error or not is addressed by another Q&A . My question is about initializing hidden inherited members. By-value assignment of derived D to base class B is permitted by the is-a relationship, but it's hard to initialize the inherited member that corresponds to the base class. In the following example, initialization is done with a brace-enclosed initializer list for B but this is not possible for D . The only option I found for getting defined D instances was to default-initialize (zero) and assign afterwards:

#include <iostream>
using namespace std;

int main()
{
    struct B { int x; };
    struct D: B { int x; };
    B bas = {1};
    cout << bas.x << endl;
    D der = {};
    der.x = 2;
    bas = der;
    cout << bas.x << endl;
    der.B::x = 3;
    bas = der;
    cout << bas.x << endl;
    static_cast<B&>(der).x = 4;
    bas = der;
    cout << bas.x << endl;
    return 0;
}

output:

1
0
3
4

The following ideas, which I assumed might work, resulted in compiler errors:

    D der = {1, 2};

error: could not convert '{1, 2}' from '<brace-enclosed initializer list>' to 'main()::D'

    D der = {B::x=1, 2};

error: invalid use of non-static data member 'main()::B::x'

error: could not convert '{<expression error>, 2}' from '<brace-enclosed initializer list>' to 'main()::D'

( g++.exe -Weffc++ -pedantic -Wextra -Wall -std=gnu++11 -std=c++11 -fexceptions -g -std=c++11 with gcc 8.1.0)

Is there a way for proper initialization in C++11?

Addendum:
When I asked this question, I was so focused on the difficulties to access shadowed members that I totally missed the limited options for aggregate initialization in C++11. So answers to this question will also apply to derived classes without hidden data members.

Answer 1

D der = {1, 2} would work in C++17, as

D der = {{42}, 2};

Demo

For previous version, you cannot initialize base like that and you need a constructor.