Question

鉴于这一样本代码:

#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;
}

我获得以下产出:

Exception: unknown

Yet simply making the inheritance of my_exception_t from std::exception public, 我获得以下产出:

Exception: Hello, world!

请允许我向我解释为什么在这种情况下继承事项类型? 在标准中注明参考点。

Answer 1

当你私下继承时,你不能转换成或以其他方式进入该类。由于你要求从标准中找到一些东西:

§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).

简言之,对于从<条码>中继承下来的那种类别之外的任何东西:除外,因为它是私人的。 Ergo公司将无法在std:Exgion&上登记。条款,因为不存在转换。

Answer 2

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

继承类型确实如此。这只是你可以轻易转换成一种捕获类型的事项。诚然,由于这不是公共继承,因此没有公众可以进入的转变。

<>Explanation:

你可以在此看到同样的行为:

class B
{
};

class C1 : B
{
};

class C2 : public B
{
};

int main(int argc, char** argv)
{
    B& b1 = C1();//Compiling error due to conversion exists but is inaccessible
    B& b2 = C2();//OK
    return 0;
}

如果:

The catch block has a matching type, or
The catch block is for a type that has an accessible conversion
The catch block is a catch(...)

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