为什么派生类找不到基类的类型别名？ [英] Why can't my derived class find my base class's type alias?

问题描述

让我们从代码示例开始，因为应该很容易看到正在发生的事情：

Let's start with the code example, because it should be easy to see what's going on:

template <typename T>
struct Base
{
    using Type = int;
};

template<typename T>
struct Derived : public Base<T>
{
    // error: unknown type name 'Type'
    using NewType = Type;
};

int main()
{}

预期派生找到基本类型别名。但是，我尝试过的所有编译器（MSVC，Clang和GCC）似乎都不喜欢此代码。

I would have expected Derived to find Base's Type alias. However, none of the compilers I've tried (MSVC, Clang, GCC) seem to like this code.

更令人惊讶的是，将Derived的继承更改为：

More surprisingly, changing Derived's inheritance to:

struct Derived : public Base<int>

解决了这个问题。

是否存在我可以更改某些内容以允许派生找到Base的别名吗？

Is there something I can change to allow Derived to find Base's alias?

推荐答案

因为 Type 是从属名称（取决于模板 Base< T> ）。您需要限定它并使用 typename 。您还可以使用 Derived :: （如 op 自己找出来）：

because Type is a dependent name (it depends on the template Base<T>). You need to qualify it and use typename. You can also qualify it with Derived:: (as the op himself figure it out):

template<typename T>
struct Derived : public Base<T>
{
    using NewType = typename Base<T>::Type;
    // or
    using NewType2 = typename Derived::Type;
};

您可以在此处详细了解相关名称：

You can read more about dependent names here:

https://en.cppreference.com/w/cpp/language/dependent_name

您如何理解C ++中的从属名称

我必须在何处以及为什么要将模板放在哪里？和类型名称

Why do I have to access template base class members through the this pointer?

未在此范围内声明模板和继承错误

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