c++ - 非POD的零初始化

Question

为什么在下面的非 POD 类中 x 被初始化为零？

class test {
public:
    void print() {
        cout << x << endl;
    }
private:
    int x;
};

int main(int argc, char** argv)
{
    test * tst = new test();
    tst->print();
    cout << is_pod<test>::value << endl;
}

tst->print() 和 is_pod() 都返回 0

Answer 1

这是 value-initialization 的结果没有用户提供的构造函数的类。

在本例中，T<b>()</b>和new T<b>()</b>首先执行零初始化:

if T is a class type with a default constructor that is neither user-provided nor deleted (that is, it may be a class with an implicitly-defined or defaulted default constructor), the object is zero-initialized and then it is default-initialized if it has a non-trivial default constructor;

zero-initialization的影响是:

if T is an non-union class type, all base classes and non-static data members are zero-initialized, and all padding is initialized to zero bits. The constructors, if any, are ignored.

和

if T is a scalar type, the object's initial value is the integral constant zero explicitly converted to T.