c++ - 委托构造函数抛出时是否调用析构函数？

Question

众所周知，如果构造函数抛出，那么所有完全构造的子对象都会以相反的顺序被销毁，包括成员数据和各种基类。但是，非委托构造函数不会调用析构函数。对于委托构造函数，当进入构造函数主体时，对象已经被构造，但构造仍在继续。因此，如果委托构造函数从其体内抛出异常，是否会调用类的析构函数？

class X
{
public:
    X();
    X(int) : X() { throw std::exception(); } // is ~X() implicitely called?
    ~X();
};

Answer 1

规则是为所有完全构造的对象调用析构函数。一旦任何构造函数完成，包括委托的构造函数（即使程序在另一个构造函数中继续），对象就被认为是完全构造的。

Answer 2

The lifetime of an object begins when any constructor (i.e., in the case of delegation, the ultimate target constructor) is successfully completed. For the purposes of [C++03] §3.8, “the constructor call has completed” means any constructor call. This means that an exception thrown from the body of a delegating constructor will cause the destructor to be invoked automatically.

source.

And here is a nice article about delegating constructors, should anybody want to read it.