C++ - 覆盖期间调用析构函数时的类成员
我试图理解覆盖对象时构造函数和析构函数调用的顺序。
我的代码是：
class A
{
public:
A(int n): x(n)
{ cout << "A(int " << n << ") called" << endl; }
~A( )
{ cout << "~A( ) with A::x = " << x << endl; }
private:
int x;
};
int main( )
{
cout << "enter main\n";
int x = 14;
A z(11);
z = A(x);
cout << "exit main" << endl;
}
--
输出是：
enter main
A(int 11) called
A(int 14) called
~A( ) with A::xx = 14
exit main
~A( ) with A::xx = 14
--
为什么调用析构函数时 A::xx = 14 ？不应该是11吗？
im trying to understand the order of Constructor and Destructor calls when overwriting an object.
My code is :
class A
{
public:
A(int n): x(n)
{ cout << "A(int " << n << ") called" << endl; }
~A( )
{ cout << "~A( ) with A::x = " << x << endl; }
private:
int x;
};
int main( )
{
cout << "enter main\n";
int x = 14;
A z(11);
z = A(x);
cout << "exit main" << endl;
}
--
The output is :
enter main
A(int 11) called
A(int 14) called
~A( ) with A::xx = 14
exit main
~A( ) with A::xx = 14
--
Why is A::xx = 14 when the destructor is called? Shouldn't it be 11?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
为什么应该是11？您将
z重新分配给A(14)，因此最后是 14。（编辑后：您还会看到临时
A(14)对象的析构函数，该对象在赋值结束时被销毁。）Why should it be 11? You reassign
ztoA(14), so it's 14 at the end.(After your edit: You also see the destructor of the temporary
A(14)object that gets destroyed at the end of the assignment.)