Robert Dockins wrote:
[snip]
From an earlier post:
Now since f and g compute the same results for the same inputs,
anywhere in a program that you can use f you could just replace f
by g and the observable behaviour of the program would be
completely unaffected. This is what referential transparency means.
My essential claim is that the above statement is in error (but in a
fairly subtle way).
Ok I see now! :-) I was confusing the concept of referential transparency
with a kind of global code equivalence, so the rest of my argument is
irrelevant. Thus I should have said:
" For particular types T1 and T2, if (f (x::T1))::T2 === g x for all x in T1
then f :: T1->T2 and g ::T1->T2 can be freely substituted since the context
T1->T2 cannot tell them apart."
Thanks for pointing out the faulty definition,
Regards, Brian.
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