Hi Brent, Is this how we go about solving these then? Keep adding parameters on the right until we have enough of them that we are able to apply the function definition. In other words, in ex1 = doTwice doTwice the first doTwice needs another parameter, so you added 'y'. ex1 y = (doTwice doTwice) y = doTwice doTwice y Then we can apply the definition. We got stuck again, so you added another parameter 'z'. Then again applied the definition of the left most function. And so on. Is there a similarly systematic approach for the foldl implemented by foldr example? I'm looking for systematic approaches that I can then practice. Thanks! Dimitri Em 20/08/14 14:30, Brent Yorgey escreveu:
On Wed, Aug 20, 2014 at 02:19:16AM -0600, Dimitri DeFigueiredo wrote:
doTwice :: (a -> a) -> a -> a doTwice f x = f (f x)
what does this do?
ex1 :: (a -> a) -> a -> a ex1 = doTwice doTwice
At least, it is clear that there is a parameter to doTwice missing. So, I wanted to do:
ex1 y = (doTwice doTwice) y
but this gets me nowhere as I don't know how to apply the definition of doTwice inside the parenthesis without naming the arguments. Note that function application associates to the left, so
(doTwice doTwice) y = doTwice doTwice y
So, we have
ex1 y = doTwice doTwice y = doTwice (doTwice y) -- definition of doTwice
Now we are stuck again; we can add another arbitrary parameter.
ex1 y z = doTwice (doTwice y) z = (doTwice y) ((doTwice y) z) = doTwice y (doTwice y z) -- remove unnecessary parentheses = y (y (doTwice y z)) = y (y (y (y z)))
Does that help?
What is the systematic way to evaluate these expressions? I actually got really stumped when I considered.
ex2 :: (a -> a) -> a -> a ex2 = doTwice doTwice doTwice doTwice
I assume this is not the same as
ex2 = (doTwice doTwice doTwice) doTwice These ARE exactly the same. It's always the case that
f w x y z ... = (((f w) x) y) z ...
-Brent _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners