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[英]Overload operator += in template class with std::enable_if metafunction
[英]Template operator friend of template class with enable_if
我将如何与enable_if成为模板类的朋友来创建模板运算符?
这是我的问题的一个例子:
#include <type_traits>
template<typename CHAR_TYPE>
class BasicString;
template<typename CHAR_TYPE, typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(CHAR_TYPE)==sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type = nullptr>
BasicString<CHAR_TYPE> operator+(const BasicString<CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right);
template<typename CHAR_TYPE, typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(CHAR_TYPE)!=sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type = nullptr>
BasicString<CHAR_TYPE> operator+(const BasicString<CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right);
template<typename CHAR_TYPE>
class BasicString
{
private:
CHAR_TYPE* characters;
size_t length;
template<typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(CHAR_TYPE)==sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type>
friend BasicString<CHAR_TYPE> operator+<CHAR_TYPE, OTHER_CHAR_TYPE>(const BasicString<CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right);
template<typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(CHAR_TYPE)!=sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type>
friend BasicString<CHAR_TYPE> operator+<CHAR_TYPE, OTHER_CHAR_TYPE>(const BasicString<CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right);
public:
BasicString()
: length(0),
characters((CHAR_TYPE*)std::malloc(sizeof(CHAR_TYPE)))
{
characters[0] = 0;
}
BasicString(const BasicString& str)
: length(str.length),
characters((CHAR_TYPE*)std::malloc(sizeof(CHAR_TYPE)*(str.length+1)))
{
for(size_t i=0; i<length; i++)
{
characters[i] = str.characters[i];
}
}
~BasicString()
{
delete characters;
}
};
template<typename CHAR_TYPE, typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(CHAR_TYPE)==sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type>
BasicString<CHAR_TYPE> operator+(const BasicString<CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right)
{
BasicString<CHAR_TYPE> newStr;
CHAR_TYPE* characters = newStr.characters;
//do some stuff. irrelevant to question
}
template<typename CHAR_TYPE, typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(CHAR_TYPE)!=sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type>
BasicString<CHAR_TYPE> operator+(const BasicString<CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right)
{
BasicString<CHAR_TYPE> newStr;
CHAR_TYPE* characters = newStr.characters;
//do some stuff. irrelevant to question
}
int main()
{
BasicString<char> str1;
BasicString<wchar_t> str2;
BasicString<char> newStr = str1 + str2;
return 0;
}
我知道我可以使用辅助函数或辅助类,但我宁愿避免使用它。
当我对此进行编译时,我得到一个错误,即运算符无法访问该类的私有成员,这意味着我的友善代码无法正常工作。 我将如何结交这些功能?
因此,发布此问题后不久,我就找到了答案。
声明朋友时,我需要使用另一个模板参数来表示CHAR_TYPE
template<typename _CHAR_TYPE=CHAR_TYPE, typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(_CHAR_TYPE)==sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<_CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type>
friend BasicString<_CHAR_TYPE> operator+<_CHAR_TYPE, OTHER_CHAR_TYPE>(const BasicString<_CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right);
template<typename _CHAR_TYPE=CHAR_TYPE, typename OTHER_CHAR_TYPE,
typename std::enable_if<((sizeof(_CHAR_TYPE)!=sizeof(OTHER_CHAR_TYPE))
&& std::is_integral<_CHAR_TYPE>::value && std::is_integral<OTHER_CHAR_TYPE>::value), std::nullptr_t>::type>
friend BasicString<_CHAR_TYPE> operator+<_CHAR_TYPE, OTHER_CHAR_TYPE>(const BasicString<_CHAR_TYPE>& left, const BasicString<OTHER_CHAR_TYPE>& right);
编辑:在符合标准的编译器(而不是Visual C ++)中进行测试之后,我发现该标准实际上不允许这样做,而且似乎没有任何方法可以实际实现。 我已经承认并参加了一个帮助班,并且对此结了婚。
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