调用基类构造函数而不命名其类 [英] call base class constructor without naming its class

问题描述

class MyDerived: public Incredble<Difficult< And<Complicated, Long>>, And<Even, Longer>, BaseClass, Name>
{
public:
  MyDerived();
}


MyDerived::MyDerived
: ???(params)
{}

有什么方法可以调用基本构造函数而不写其全名，也可以不键入typeeffing吗？

Is there any way to call a base constructor without writing its full name and without typedeffing it?

原因显然是避免代码重复，并且如果基类模板参数中的详细信息发生更改，则会引入多个位置进行更改。

The reason is clearly to avoid code duplication and introducing multiple positions to change if a detail in the base class template params changes.

其中的第2级：

template <uint32 C>
class MyDerived: public Incredble<Difficult< And<Complicated, Long>>, And<Even, Longer>, BaseClass, Name>
{
public:
  MyDerived();
}

template <uint32 C>
MyDerived::MyDerived<C> 
: ???(C)
{
}

推荐答案

您可以使用 注入的类名 。 Incredible< ...> :: Incredible 引用自身，并且由于 MyDerived 不是类模板，不合格的查找将在其基类的范围内进行：

You could use injected-class-name. Incredible<...>::Incredible refers to itself, and since MyDerived isn't a class template, unqualified lookup will look in the scope of its base classes:

MyDerived::MyDerived
: Incredble(params)
{}

如果 Incredible 是一个相关名称，那么您需要对其进行限定。实际上，您可以简单地使用派生的类型名称来限定基类的 injected-class-name （h / t Johannes Schaub-litb ）：

If Incredible is a dependent name, then you need to qualify it. You can actually simply use the derived type name to qualify the base class's injected-class-name (h/t Johannes Schaub-litb):

MyDerived::MyDerived
: MyDerived::Incredible(params)
{}

这在所有情况下均有效。

This will work in all cases.

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