function construction - Named patterns in Except

Bug introduced in 8 or earlier and fixed in 10.4.0

^{Bug isn't present in version 5.2}

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I don't think I quite understand how Except works.

I want to define $f(a,b) = (a-b)\ln(a-b)$, with the special case $f(a,a)=0$. Except, the function shouldn't evaluate if either $a$ or $b$ is zero.

I tried:

ClearAll[f]
f[Except[0, a_], Except[0, a_]] := 0;
f[Except[0, a_], Except[0, b_]] := (a - b) Log[a - b];

But for some reason DownValues[f] only reports one definition:

DownValues[f]
(* { HoldPattern[ f[Except[0, a_], Except[0, b_]]] :> (a - b) Log[a - b] } *)

This is bad because if the user inputs f[x,x], I get Infinity::indet errors.

Question What did I do wrong, and how do I make definitions with Except?

Answer

Another way to get two definitions is to only use Except on the first argument for the case where the arguments are identical.

Clear[f]

f[Except[0, a_], a_] := 0

f[Except[0, a_], Except[0, b_]] := (a - b) Log[a - b]

Annoyingly it doesn't matter the order for the definitions it comes out like this:

DownValues@f
(* {HoldPattern[f[Except[0, a_], Except[0, b_]]] :> (a - b) Log[a - b], 
 HoldPattern[f[Except[0, a_], a_]] :> 0} *)

So in order to get the right behaviour you have to do

DownValues[f] = Reverse@DownValues@f
(* {HoldPattern[f[Except[0, a_], a_]] :> 0, 
 HoldPattern[f[Except[0, a_], Except[0, b_]]] :> (a - b) Log[a - b]} *)

Now

f[3, 1]
f[3,0]
f[2,2]
f[0,0]

gives

2 Log[2]
f[3,0]
0

f[0,0]