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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
我將如何與enable_if成為模板類的朋友來創建模板運算符?
這是我的問題的一個例子:
#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;
}
我知道我可以使用輔助函數或輔助類,但我寧願避免使用它。
當我對此進行編譯時,我得到一個錯誤,即運算符無法訪問該類的私有成員,這意味着我的友善代碼無法正常工作。 我將如何結交這些功能?
因此,發布此問題后不久,我就找到了答案。
聲明朋友時,我需要使用另一個模板參數來表示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);
編輯:在符合標准的編譯器(而不是Visual C ++)中進行測試之后,我發現該標准實際上不允許這樣做,而且似乎沒有任何方法可以實際實現。 我已經承認並參加了一個幫助班,並且對此結了婚。
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