c++ - how can I use std::enable_if in a conversion operator? -


basically want range type implicitly convertible range<const char> range<const unsigned char>. std::enable_if seems impossible because function takes no arguments , has no return. whats work around?

here tried:

template<typename t> class range{     t* begin_;     t* end_; public:     range(t* begin,t* end):begin_{begin},end_{end}{}     template<int n>     range(t (&a)[n]):begin_{static_cast<t*>(&a[0])},end_{static_cast<t*>(&a[n-1])}{}     t* begin(){return begin_;}     t* end(){return end_;}     operator typename std::enable_if<std::is_same<t,const char>::value,range<const unsigned char>&>::type (){         return *reinterpret_cast<range<const unsigned char>*>(this);     } };

make template dummy parameter defaults t - postpone type deduction point function gets instantiated, otherwise sfinae doesn't work. check want in default value of parameter.

template<     typename u = t,     typename = typename std::enable_if< std::is_same<u,const char>::value >::type > operator range<const unsigned char>() {     return *reinterpret_cast<range<const unsigned char>*>(this); }

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