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copied!<p>You won't get around having to specify the return type directly. And in that case, why bother with the <code>basic_string</code> class directly? Just take a <code>typename String</code> and be done:</p> <pre><code>template<char C, class Options, class String, String Options::* Member> struct option_string; </code></pre> <p>And if you still want to make sure that <code>String</code> is indeed a specialization of <code>basic_string</code>, well, take this little helper and <code>static_assert</code> it:</p> <pre><code>#include <type_traits> #include <string> template<class T> struct is_basic_string : std::false_type{}; template<class Ch, class Tr, class Al> struct is_basic_string<std::basic_string<Ch,Tr,Al>> : std::true_type{}; // in 'option_string' static_assert(is_basic_string<String>::value, "Data member must be a 'basic_string' specialization."); </code></pre> <hr> <p>Since there doesn't seem to be a need to pass the member pointer as a template argument, I suggest just passing it in the constructor:</p> <pre><code>template<char C, class Option, class String> struct option_string{ static_assert(is_basic_string<String>::value, "Data member must be a 'basic_string' specialization."); typedef String Option::*member_type; option_string(member_type m, ...) : member(m), ... { ... } private: member_type member; }; template<char C, class Option, class String> option_string<C, Option, String> make_option_string(String Option::*member, ...){ return {member, ...}; } // in code: auto os = make_option_string<'H'>(&some_type::a_string_member); </code></pre>

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