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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