Why do I need to use typedef typename in g++ but not VS?

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The typename is required by the standard. Template compilation requires a two step verification. During the first pass the compiler must verify the template syntax without actually supplying the type substitutions. In this step, std::map::iterator is assumed to be a value. If it does denote a type, the typename keyword is required.

Why is this necessary? Before substituing the actual KEY and VALUE types, the compiler cannot guarantee that the template is not specialized and that the specialization is not redefining the iterator keyword as something else.

You can check it with this code:

class X {};
template <typename T>
struct Test
{
   typedef T value;
};
template <>
struct Test<X>
{
   static int value;
};
int Test<X>::value = 0;
template <typename T>
void f( T const & )
{
   Test<T>::value; // during first pass, Test<T>::value is interpreted as a value
}
int main()
{
  f( 5 );  // compilation error
  X x; f( x ); // compiles fine f: Test<T>::value is an integer
}

The last call fails with an error indicating that during the first template compilation step of f() Test::value was interpreted as a value but instantiation of the Test<> template with the type X yields a type.

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