(Difference between revisions)
m (fixed typo)
Latest revision as of 18:06, 29 September 2010
f x = (\y -> x + y)
returns a closure, because the variable
, which is bound outside of the lambda abstraction is used inside its definition. An interesting side note: the context in which
was bound shouldn't even exist anymore, and wouldn't, had the lambda abstraction not closed around x.