如何正确公开底层模板类型以用作类型

Question

我有一些字符串类IMyString，它可以是 wstring 或普通字符串。它有两个派生MyString 和MWString。如果模板化代码想要同时接受这两者，它当前需要字符串类型参数和字符类型参数。我想改变它。

这是我的代码：

// This string doesn't work, it's just an example
template <typename CharType, typename Derived>
class IMyString
{
public:
  // Define value type so that it can ba accessed by other templates (does not work)
  using ValueType = typename CharType;

  const CharType* c_str() const { return rawData; }
  // Does not actually do the adition for brevity, just imagine it instead
  Derived&& operator+(const CharType* moreData) const { return Derived(/*should add my own data here and also copy everything etc*/moreData); }

protected:
  IMyString(const CharType* rawData) : rawData(rawData) {}

  const CharType* rawData = nullptr;
};

// Normal string
class MyString : public IMyString<char, MyString>
{
public:
  MyString(const char* data) : IMyString(data) {}

};

// Wide char string
class MyWString : public IMyString<wchar_t, MyString>
{
public:
  MyWString(const wchar_t* data) : IMyString(data) {}

};

// Method that doesn't really do anything actually
template <typename StringType>
// I want to be able to use the value type directly as type
void DoSomethingWithAnyString(StringType string, const StringType::ValueType* dataToAdd)
{
  StringType string2 = string + dataToAdd;
  // even if the addition worked, it has no effect because the value is not returned
}

int main()
{
  MyString someString("Hello");
  DoSomethingWithAnyString(someString, " world!");
}

我现在得到的 GCC 错误：

UnderlyingTemplateType.cpp:9:30: error: expected nested-name-specifier before 'CharType'
   using ValueType = typename CharType;

但是当我删除 typename 时，这也是一个错误：

UnderlyingTemplateType.cpp:40:56: error: need 'typename' before 'StringType::ValueType' because 'StringType' is a dependent scope
 void DoSomethingWithAnyString(StringType string, const StringType::ValueType* dataToAdd)

如何正确地做到这一点？我想使用IMyString::ValueType，就像它是一个嵌套类/结构/枚举一样。

我可以看到这是可行的，但我宁愿不要在我使用ValueType 的任何地方写typename：

// Method that doesn't really do anything actually
template <typename StringType>
// I want to be able to use the value type directly as type
void DoSomethingWithAnyString(StringType string, const typename StringType::ValueType* dataToAdd)
{
  StringType string2 = string + dataToAdd;
  // even if the addition worked, it has no effect because the value is not returned
}

是否有一个简洁的定义允许ValueType 充当真实类型？

Answer 1

你能做的最好的事情就是这样 - 编写包装被鄙视的类型名的助手别名。 Live example

template<class T>
using char_t = typename T::ValueType;

// Method that doesn't really do anything actually
template <typename StringType>
// I want to be able to use the value type directly as type
void DoSomethingWithAnyString(StringType string, const char_t<StringType>* dataToAdd)
{
  StringType string2 = string + dataToAdd;
  // even if the addition worked, it has no effect because the value is not returned
}