From: musiphil@bawi.org   
      
   On 2013-02-25 12:44, Walter Mascarenhas wrote:   
   > The code below does not compile in g++ 4.7.2: it says   
   > that the template parameter S is not used in Foo's   
   > specialization below. Is g++ correct on this? If so,   
   > how would I specialize the struct Foo for   
   > T = std::vector::iterator in a general way?   
   >   
   >   
   > #include    
   > template    
   > struct Foo   
   > {   
   > };   
   >   
   > template    
   > struct Foo< typename std::vector::iterator >   
   > {   
   > };   
      
   It may not always be possible to determine S, given std::vector::iterator.   
      
   Suppose you have:   
      
   template  struct X { typedef void type; };   
      
   template  struct Foo { typedef T* ptr; };   
   template  struct Foo::type> { typedef S* ptr; };   
      
   template <> struct X { typedef int type; };   
   template <> struct X { typedef int type; };   
      
   Now, suppose you instantiate Foo. Since X::type is int for   
   both S=short and S=long, what should S be in the specialization?   
      
   --   
   Seungbeom Kim   
      
      
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