Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
(m >>= f) >>= g = m >>= (\x -> f x >>= g)
what’s different from f and \x->f x ??
I think they’re the same type a -> m b. but it seems that the second >>= at right side of equation treats the type of \x->f x as m b.
what’s going wrong?
The expressions
fand\x -> f xdo, for most purposes, mean the same thing. However, the scope of a lambda expression extends as far to the right as possible, i.e.m >>= (\x -> (f x >>= g)).If the types are
m :: m a,f :: a -> m b, andg :: b -> m c, then on the left we have(m >>= f) :: m b, and on the right we have(\x -> f x >>= g) :: a -> m c.So, the difference between the two expressions is just which order the
(>>=)operations are performed, much like the expressions1 + (2 + 3)and(1 + 2) + 3differ only in the order in which the additions are performed.The monad laws require that, like addition, the answer should be the same for both.