c++ - g++ doesn't compile certain nested templates -

when break defined, g++ 4.7.2 not compile following, think valid c++. compile break defined if a<u> tmp changed else, a<int> tmp - while makes minimal test case here work, it's no in actual application. there here not legal c++?

template <typename t> class b {  };  template <typename t> class { public:     template <typename u> b<u> *alloc_b( ); };  template <typename t> template <typename u> b<u> *a<t>::alloc_b( ) {     return new b<u>( ); }  #ifdef break template <typename t> class c { public:     template <typename u> void x(b<u> &b) {         a<u> tmp;         b<u> *tmp2;         tmp2 = tmp.alloc_b<u>( );         delete tmp2;     } }; #endif  int main( ) {     a<int> a;     b<float> *bp = a.alloc_b<float>( );     delete bp;  #ifdef break     c<int> c;     b<float> b;      c.x(b); #endif } 

the alloc_b function template dependent name. must call so:

tmp2 = tmp.template alloc_b<u>( ); 

that's problem, , why works when use a<int>, because type no longer depends on template argument u.