C++14 非聚合的统一初始化

Question

我正在使用 Visual C++ 2013。当 class 是一个聚合时，它是零初始化的。当它是非聚合时，它似乎是默认初始化的并且不确定。这是为什么？

#include <iostream>

using namespace std;

class Test_1
{
public:
    int i;
    void f(){};
};

class Test_2
{
public:
    int i;
    virtual void f(){};
};

int main()
{
    Test_1 t1{};
    Test_2 t2{};

    cout<<t1.i<<endl; //0
    cout<<t2.i<<endl; //-858993460

    getchar();
}

Answer 1

如果你的编译器这样做，它就坏了。

[dcl.init.list]/p3（所有引用均来自 N4140）：

List-initialization of an object or reference of type T is defined as follows:

If T is an aggregate, aggregate initialization is performed (8.5.1).

Otherwise, if the initializer list has no elements and T is a class type with a default constructor, the object is value-initialized.

[...]

[dcl.init]/p8:

To value-initialize an object of type T means:

if T is a (possibly cv-qualified) class type (Clause 9) with either no default constructor (12.1) or a default constructor that is user-provided or deleted, then the object is default-initialized;

if T is a (possibly cv-qualified) class type without a user-provided or deleted default constructor, then the object is zero-initialized and the semantic constraints for default-initialization are checked, and if T has a non-trivial default constructor, the object is default-initialized;

if T is an array type, then each element is value-initialized;

otherwise, the object is zero-initialized.

Test_2 不是一个集合，所以 t2 应该被值初始化。反过来，由于 Test_2 的默认构造函数不是用户提供的，因此 t2 应该首先被零初始化（导致 t2.i 被初始化为 0），然后是默认的构造函数是运行。

C++14 非聚合的统一初始化

C++14 Uniform Initialization on a Non-Aggregate

c++

c++14

uniform-initialization