基本模板类数据成员在派生模板类中不可见？ [英] Base template class data members are not visible in derived template class?

问题描述

请考虑以下C ++代码，

Consider the following C++ code,

template <typename Derived>
struct A
{
    bool usable_;
};

template <typename Derived>
struct B : A< B<Derived> >
{
    void foo()
    {
        usable_ = false;
    }
};

struct C : B<C>
{
    void foo()
    {
        usable_ = true;
    }
};

int main()
{
    C c;
}

我遇到编译错误：在成员函数 void B< ; Derived> :: foo（）：

I got compilation error: In member function void B<Derived>::foo():

template_inherit.cpp：12：error：'useful_'was未在此
范围中声明。

template_inherit.cpp:12: error: 'usable_' was not declared in this scope.

为什么？任何好的修复？

Why is that ? Any good fix ?

推荐答案

这是因为可用_ 名称，因此在解析模板时查找，而不是在实例化时查找（当基类已知时）。

That's because usable_ is a non-dependent name, so it is looked up at the time the template is parsed, instead of being looked up at instantiation (when the base class is known).

未限定名称查找将不会查找，而从属名称也不会在依赖基类中查找。您可以使名称可用_ 依赖如下，它也将摆脱无限制名称查找

Unqualified name lookup will not lookup and non-dependent names are never looked up in dependent base classes. You can make the name usable_ dependent as follows, which will also get rid of unqualified name lookup

this->usable_ = false;

// equivalent to: A<B>::usable_ = false;
A< B<Derived> >::usable_ = false;

B::usable_ = false;

所有这些都可以工作。或者您可以使用声明在派生类中声明名称

All of these will work. Or you can declare the name in the derived class with a using-declaration

template <typename Derived>
struct B : A< B<Derived> >
{
    using A< B<Derived> >::usable_;

    void foo()
    {
        usable_ = false;
    }
};

请注意，在 C 问题 - 它只影响 B 。

Note that in C there will be no problem - it only affects B.

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