On Fri, Feb 22, 2013 at 07:27:28PM -0500, xiao Ling wrote:
Hi All: How do you define a function of signature h :: M Int -> M Int -> M Int so that h ( M x ) ( M y ) = M ( x + y ), but without unwrapping the value from the monad?
This question is from the article "Trivial Monad" found at http://blog.sigfpe.com/2007/04/trivial-monad.html. The provided answer is
h x y = x >>= (\x -> g x y)
Now I am a little confused, how can you put x in g if it takes an Int as first parameter but x is M Int?
I think your confusion may stem from the fact that there are *two different* things named 'x' in the above code. The x's on the right hand side of the >>= shadow the x on the left hand side. This is confusing and poor style; a better way to write it would be h x y = x >>= (\i -> g i y) If you study the type of >>= you will see that i indeed has type Int, as required for the first argument of g. -Brent