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c ++:通过多级继承从模板子类调用基类的模板方法

[英]c++: Calling a template method of a base class from a template child class though multilevel inheritance

I'm looking to call a method of a base class A from a child class D that inherits it through C::A and B::A. 我想从子类D调用基类A的方法,通过C :: A和B :: A继承它。

template <class PType>
class A
{
public:
    template <class ChildClass>
    void Func( void )
    {
        std::cout << "Called A<PType>::Func<ChildClass>" << std::endl;
    }
};

template <class PType>
class B : public A<PType>
{
};

template <class PType>
class C : public A<PType>
{
};

class D : public B<int>, public C<int>
{
public:
    D()
    {
        static_cast<B<int>*>( this )->A<int>::Func<D>();
        static_cast<C<int>*>( this )->A<int>::Func<D>();
    }
};

This works as expected, D calls both B::A::Func and C::A::Func with a template argument of the child class when initialized. 这按预期工作,D在初始化时使用子类的模板参数调用B :: A :: Func和C :: A :: Func。 This doesn't seem to work when D is a template class, however. 但是,当D是模板类时,这似乎不起作用。

template <class PType>
class D2 : public B<PType>, public C<PType>
{
public:
    D2()
    {
        //expected primary-expression before ‘>’ token
        //expected primary-expression before ‘)’ token
        static_cast<B<PType>*>( this )->A<PType>::Func< D2<PType> >();
        static_cast<C<PType>*>( this )->A<PType>::Func< D2<PType> >();
    }
};

The problem seems to be the template argument D2 to Func, but can't figure it out beyond that. 问题似乎是模板参数D2到Func,但无法解决这个问题。

When you are working with a name whose value/type/ template status is dependent on a template type parameter, you must disambiguate which one it is for the compiler manually. 当您使用其值/类型/ template状态取决于template类型参数的名称时,必须手动消除编译器的名称。

This is to make the parsing far easier, and able to be done long before you pass a type into the template . 这是为了使解析更容易,并且可以在将类型传递到template之前很久就完成。

You may think that is obviously a template function call, but the < and > could be comparisons, and the template functions could be values or types or whatever. 您可能认为这显然是template函数调用,但<>可以进行比较,模板函数可以是值或类型或其他。

By default, such dependent names are assumed to be values. 默认情况下,假定此类从属名称为值。 If you want them to treat one as a type, use typename . 如果您希望将它们视为一种类型,请使用typename If you want to treat one as a template, use template as a keyword. 如果要将其中一个视为模板,请使用template作为关键字。

So something like this: 所以像这样:

   static_cast<B<PType>*>( this )->template A<PType>::template Func< D2<PType> >();

now, when you are interacting with a fully instantiated template this is not required. 现在,当您与完全实例化的template交互时,这不是必需的。 So: 所以:

   static_cast<B<int>*>( this )->A<int>::Func< D2<PType> >();

consists of a fully instantiated template type B<int> , so A 's category is no longer dependent on a (locally undetermined) template argument. 由完全实例化的template类型B<int> ,因此A的类别不再依赖于(本地未确定的) template参数。 Similarly, ->A<int> is a fully instantiated template type, so ::Func is no longer dependent on a (locally undetermined) template argument. 类似地, ->A<int>是完全实例化的template类型,因此::Func不再依赖于(本地未确定的) template参数。

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