c++ - Get type of outer class in template - 


struct outer {     struct inner {     }; };  template <typename t> void test() {     ??outer?? foo; // how type of t's outer class ? }  test<outer::inner>();

inside test() have template argument of type t , want declare variable of whatever type outer class of t.

i think should trivial compiler info, far couldn't find how it.

it can't done. there indeed information available compiler cannot used in metaprogramming context. ongoing work on adding compile-time reflection language might improve situation, don't hold breath.

in meantime, you'll have old-fashioned way.

struct outer {     struct inner {         typedef outer outer_class;           }; }

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