2010/10/9 André Batista Martins <andre_bm@netcabo.pt> Said:

Might have not been clear, but i will try illustrate .

f:: a-> b -> c -> (b,(c,a))
f1 ::  c -> a -> d
-----------------------------
 

I think I would attack this with glue consisting of:

comb f f1 a b c =  arr (\(a,b,c) -> f a b c) >>> arr (\(b,(c,a))) ->f1 c a) $ (a,b,c)

and yes, have to agree that easier to roll your own if only a few functions are like this..
but should be able to parse the type signatures of the functions involved and write a program to automate this process.. using this format as a template..

Actually if you just set it to take all the variables prior to last (->) in sig you can put them
put them together in an uncurried format.. for instance the "a -> b -> c" portion would become always \(a,b,c) -> then the function so arr (\(a,b,c) -> f a b c) then the term (output) would be the last term in this case (b,(c,a)  add that with a "->" between to give that to first part of another lambda construction (\(c,a) -> f1 c a) ... arrowizing the whole thing with arr (first lambda) >>> arr (second lambda) $ and a tuple from all but the last variables in all cases of first function ... so for f it would be (a,b,c).  if for some odd reason it was a single it would just become ((a)) an added parenthesis, which would not hurt a thing for the case where it was a sig like f :: a -> b

So for your case it becomes as shown above:
comb f f1 a b c =  arr (\(a,b,c) -> f a b c) >>> arr (\(b,(c,a))) ->f1 c a) $ (a,b,c)
and say for:

f :: a -> (b,c)
f1:: b -> d

(\(a) -> f a) >>> (\(b,c) -> f1 b) $ (a)   <- it just harmlessly adds the '( ' and ')' around the 'a' even though it doesn't need it as the only parameter prior to the last '->'.  

This is probably clear as mud, on first look, but I think a way forward in automating from
this is possible.. I am sure of it.. but it would be at the source code level and a string parse and output from that ..

cheers,
gene