C++ STL fails to find comparator for nested class -

i expected code work, fails compile gcc. compile if lift inner class out.

#include <algorithm>  template <typename t> struct outer {   struct inner   {     int x;   };   inner vec[3]; };  template <typename t> bool operator <(const typename outer<t>::inner& lhs, const typename outer<t>::inner& rhs) {   return lhs.x < rhs.x; }  int main() {   outer<int> out;   outer<int>::inner in;   std::lower_bound(out.vec, out.vec + 3, in); } 

gcc 4.4 has say:

... bits/stl_algo.h:2442: error: no match ‘operator<’ in ‘* __middle < __val’ 

gcc 4.7 prints lot more stuff, including above, ending this:

... bits/stl_algobase.h:957:4: note: couldn't deduce template parameter ‘t’ 

i'm willing believe it's not well-formed c++, why not?

the problem mentioned compiler couldn't deduce template parameter t. because typename outer::inner nondeduced context context t.

when template parameter used in nondeduced context, it's not taken consideration template argument deduction. details @ section 14.8.2.4 of c++ standard (2003).

why? if specialize outer as:

template <> struct outer<int> {    struct inner    {        double y;    };    inner vec[3]; };  

there no way compiler deduce if outer::inner should reference definition or previous one.

now, onto solutions. there couple of possible resolutions:

1. move operator< inside inner , make friend function.
2. define operator< inside inner m m. suggested.
3. suggested johannes schaub - litb: derive inner crtp base parameterized inner. make parameter type crtp class , cast parameter reference derived inner class, type of given template argument deduce.

i tried out crtp approach since sounds cool!:

template <typename inner> struct base { };  template <typename t> struct outer {   struct inner : base<inner>   {     t x;   };   inner vec[3]; };  template <typename t> bool operator< (const base<t>& lhs, const base<t>& rhs) {   return static_cast<const t&>(lhs).x < static_cast<const t&>(rhs).x; } 

related answers: 1, 2, 3