如何解决这个问题才能用GCC编译这段代码
你可能会用 std::conjunction (C ++ 17)保证短路
std::conjunction
template <typename To, typename From> typename std::enable_if< std::conjunction< std::negation<std::is_same<std::string, To>>, std::negation<std::is_same<std::string, From>>, LexicalCastable<From, std::string> LexicalCastable<std::string, To>>::value, To>::type lexicalCast(const From &from) { return lexicalCast<To>(lexicalCast<std::string>(from)); }