[Haskell-cafe] want to post haskell question
miguelimo38 at yandex.ru
Tue Feb 2 05:01:14 EST 2010
> |multSM d m = [[(b*a)| b<-[d], a<-(head m)]]
Well, let's see what do we have here.
We have 's around something. "Something" is [(b*a)| b<-[d], a<-(head m)], which is just a legal Haskell value, so our "mutlSM d m" has to be a one-element list, with the only element being equal to
what we put inside the brackets. It's like  or ["Hello"], just with more complex expression inside.
This unique value is, as we've seen, [(b*a)| b<-[d], a<-(head m)]. Here we also have 's around something, but this new "something" is NOT a legal Haskell value; it's a list comprehension. That means
that we have something like this:
multSM d m = [concatMap (\b -> concatMap (\a -> b*a) (head m)) [d]]
Now, concatMap f [d] = f d, so
multSM d m = [concatMap (\a -> d*a) (head m)]
and if m = [m1:ms], then
multSM d m = [concatMap (\a -> d*a) m1]
So, you take first row ("head" only takes the first element of the list, you know), multiply it by "d", and make a one-element list of it.
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