
13 Jul
2013
13 Jul
'13
11:41 a.m.
cfold’ f z [] = z cfold’ f z (x:xs) = f x z (\y -> cfold’ f y xs) cfold (+) 0 [] So my question comes, is it a bug in the book or something I miss here?
This can't be correct. In expression `f x z (\y ...)`, function 'f' takes three argument but (+) can only take two. May be you are suppose to pass some other function to cfold'. -- Dilawar EE, IITB