std :: exception的C ++异常和继承 [英] C++ Exceptions and Inheritance from std::exception

问题描述

给出此示例代码：

#include <iostream>
#include <stdexcept>

class my_exception_t : std::exception
{
public:
    explicit my_exception_t()
    { }

    virtual const char* what() const throw()
    { return "Hello, world!"; }
};

int main()
{
    try
        { throw my_exception_t(); }
    catch (const std::exception& error)
        { std::cerr << "Exception: " << error.what() << std::endl; }
    catch (...)
        { std::cerr << "Exception: unknown" << std::endl; }

    return 0;
}

我得到以下输出：

Exception: unknown

继承 my_exception_t 从 std :: exception public ，I得到以下输出：

Yet simply making the inheritance of my_exception_t from std::exception public, I get the following output:

Exception: Hello, world!

有人可以向我解释为什么这种类型的继承在这种情况下很重要吗？

Could someone please explain to me why the type of inheritance matters in this case? Bonus points for a reference in the standard.

推荐答案

当您私下继承时，您不能转换为或以其他方式访问该基类的类。由于您要求标准中的某些内容：

When you inherit privately, you cannot convert to or otherwise access that base class outside of the class. Since you asked for something from the standard:

§11.2/ 4：

基类被称为如果基类的发明的公共成员可访问，则可访问。如果一个基类是可访问的，可以隐式地将指向一个派生类的指针转换为指向该基类的指针（4.10,4.11）。

§11.2/4:
A base class is said to be accessible if an invented public member of the base class is accessible. If a base class is accessible, one can implicitly convert a pointer to a derived class to a pointer to that base class (4.10, 4.11).

简单地说，对于类外的任何东西，它就像你从未继承过 std :: exception ，因为它是私有的。 Ergo，它不能被捕获在 std :: exception& 子句中，因为不存在转换。

Simply put, to anything outside the class it's like you never inherited from std::exception, because it's private. Ergo, it will not be able to be caught in the std::exception& clause, since no conversion exists.

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