c++ - 使用带有虚函数的类的构造函数进行大括号初始化

Question

我知道聚合不能有虚拟成员函数，但此代码可以编译

struct B {
    B(int yIn) :y(yIn) {}
    int y;
    virtual void f() {}
};

B b = {3}; 

大括号初始化与聚合初始化不同，还是该类因其构造函数而成为有效聚合？

Answer 1

是的B不是aggregate type ，它不能有虚拟成员函数。因此 B b = {3}; 将不会执行 aggregate initialization ，但是copy-list-initialization (since C++11)相反，结果是调用构造函数 B::B(int) 来初始化对象。

• Otherwise, the constructors of T are considered, in two phases:

  • ...

  • If the previous stage does not produce a match, all constructors of T participate in overload resolution against the set of arguments that consists of the elements of the braced-init-list, with the restriction that only non-narrowing conversions are allowed. If this stage produces an explicit constructor as the best match for a copy-list-initialization, compilation fails (note, in simple copy-initialization, explicit constructors are not considered at all).