Gene A yumagene at gmail.com
Sun Oct 10 02:58:09 EDT 2010

```2010/10/9 André Batista Martins <andre_bm at 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
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