[mercury-users] Typeclass problem (bug, misunderstanding, oth er?)

[Ralph Becket]

> From: Fergus Henderson [mailto:fjh at cs.mu.OZ.AU]
> 
> In order to typecheck a procedure call, one must rename apart the
> type variables in the called procedure's type declaration before
> unifying them with the calling procedure's type variables.
> I think that in making that deduction, you omitted the renaming
> apart step.

That's a good point.  But I'm not sure it explains the following:

:- typeclass tc(T1, T2) where [
	func e(T1) = T1,
	func f(T1) = T2,
	func g(T1, T2) = T2
].

:- func foo(T1) = T2 <= tc(T1, T2).

foo(A) = g(A, f(A)).

This will work fine.  It's equivalent to

foo(A) = Z :-
	X = A,       % X:T1
	Y = f(A),    % Y:T2
	Z = g(X, Y). % Z:T2

However,

:- func bar(T1) = T2 <= tc(T1, T2).

bar(A) = g(e(A), f(A)).

which is the same as

bar(A) = Z :-
	X = e(A),    % X:T1 surely the same type as A
	             %   with the same relation to T2
	Y = f(A),    % Y:T2
	Z = g(X, Y). % Z:T2

causes an unsatisfied typeclass constraint error.

Sorry to keep harping on about this one, but I am
just totally bemused.  Can you go through the type
inference on this one line by line and point out
where the error is?

Cheers,

Ralph


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end
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