Consider the following:

	template <typename T> struct X {};  // #1
	template <typename T> struct X<const T>; //#2
	template struct X<int&>; //#3

Which specialization are we instantiating in #3? The "obvious" answer is #1, because "int&" doesn't have a top level cv-qualification. However, there's also an argument saying that we should actually be instantiating #2. The argument is: int& can be taken as a match for either one (top-level cv-qualifiers are ignored on references, so they're equally good), and given two equally good matches we must choose the more specialized one.

Is this a valid argument? If so, is this behavior intentional?

John Spicer: I don't see the rationale for any choice other than #1. While it is true that if you attempt to apply const to a reference type it just gets dropped, that is very different from saying that a reference type is acceptable where a const-qualified type is required.

If the type matched both templates, the const one would be more specialized, but "int&" does not match "const T".

Nathan Sidwell: thanks for bringing this one to the committee. However this is resolved, I'd like clarification on the followup questions in the gcc bug report regarding deduced and non-deduced contexts and function templates. Here're those questions for y'all,

template <typename T> void Foo (T *); // #1
template <typename T> void Foo (T const *); // #2
void Baz ();
Foo (Baz); // which?

template <typename T> T const *Foo (T *); // #1
void Baz ();
Foo (Baz); // well formed?

template <typename T> void Foo (T *, T const * = 0);
void Baz ();
Foo (Baz); // well formed?

BTW, I didn't go trying to break things, I implemented the cv-qualifier ignoring requirements and fell over this. I could find nothing in the standard saying 'don't do this ignoring during deduction'.