如果一个对象在构造函数中抛出异常，基类的析构函数是否会被调用？ [英] Will the destructor of the base class called if an object throws an exception in the constructor?

问题描述

如果一个对象在构造函数中抛出异常，基类的析构函数是否会被调用？

Will the destructor of the base class be called if an object throws an exception in the constructor?

推荐答案

在构造期间抛出，所有先前构造的子对象将被适当地销毁。以下程序证明基础肯定被销毁：

If an exception is thrown during construction, all previously constructed sub-objects will be properly destroyed. The following program proves that the base is definitely destroyed:

struct Base
{
    ~Base()
    {
        std::cout << "destroying base\n";
    }
};

struct Derived : Base
{
    Derived()
    {
        std::cout << "throwing in derived constructor\n";
        throw "ooops...";
    }
};

int main()
{
    try
    {
        Derived x;
    }
    catch (...)
    {
        throw;
    }
}

输出：

throwing in derived constructor
destroying base

$ b b

（注意，本地指针的析构函数不做任何操作，这就是为什么我们更喜欢RAII比原始指针。）

(Note that the destructor of a native pointer does nothing, that's why we prefer RAII over raw pointers.)

这篇关于如果一个对象在构造函数中抛出异常，基类的析构函数是否会被调用？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！