Quote:

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(и КАК?)

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В ответ на:

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ierarhy/seq_tree.h:123: warning: `AttGraphAdvanced<VKey, long int, typename
Alphabet::symbol>::Edges' is implicitly a typename
ierarhy/seq_tree.h:123: warning: implicit typename is deprecated, please see
the documentation for details

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Quote:

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SCM::Edges edges=(*scheme).getChildren(*i);

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code:
 
typename SCM::Edges;
SCM::Edges edges=(*scheme).getChildren(*i); 

Quote:

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что ентому компилятору не нравится, что SCM::Edges typename.

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В ответ на:

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const Edges&
getEdgesRev() const
{
return E_rev_;
}

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В ответ на:

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ierarhy/automat_new.h|13| warning: `AttGraphAdvanced<VKey, VVal,
|| EdgeVal>::Edges' is implicitly a typename
ierarhy/automat_new.h|13| warning: implicit typename is deprecated, please see
|| the documentation for details

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code:

	struct K {
	  typedef int mytype_t;
	};
	
	template <class T1> struct A {
	  template <class T2> struct B { 
	      void callme(void);
	    };

	  template <int N> void bar(void) 
	  {
	    // Use 'typename' to tell the parser that T1::mytype_t names
	    //  a type. This is needed because the name is dependent (in
	    //  this case, on template parameter T1).
	    typename T1::mytype_t x;
	    x = 0;
	  }
	};