some insights into functional programming
I'm starting to figure out a few things that I didn't "get" about functional programming and monads. I wanted to explain them. I'm not looking for a particular response to this post, other than any elaboration that seems natural. There is an exercise here working with the trivial monad W: http://blog.sigfpe.com/2007/04/trivial-monad.html Write a function g :: W a -> W a -> W a such that g (W x) (W y) = W (x+y) except don't use pattern matching, but >>= instead. The answer is g mx my = mx >>= (\x -> my >>= \y -> W (x+y)) There are a couple things here that threw me off. One is that I didn't expect 'my' to be available inside the first lambda. I somehow thought of lambda as isolated, sealed-off from the rest of the universe. But they aren't. I believe this is the concept of closures, or related to it? Secondly, I didn't expect >>= to be available inside the lambda. This is related to the mistaken conception of >>= as a procedural statement rather than an expression. In Python, where I have previously encountered lambdas, no statements are allowed inside lambdas. Of course, >>= is actually an expression and you can put any expression to the right of a lambda ->. Maybe these are typical beginner misconceptions, or maybe they have more to do with coming from Python and complete beginners actually find it more natural. Mike
Michael,
g mx my = mx >>= (\x -> my >>= \y -> W (x+y))
There are a couple things here that threw me off. One is that I didn't expect 'my' to be available inside the first lambda. I somehow thought of lambda as isolated, sealed-off from the rest of the universe. But they aren't. I believe this is the concept of closures, or related to it?
Yes! You're spot on; 'my' is available precisely because closures encapsulate the part of the surrounding environment that's referenced by bound variables.
Secondly, I didn't expect >>= to be available inside the lambda. This is related to the mistaken conception of >>= as a procedural statement rather than an expression. In Python, where I have previously encountered lambdas, no statements are allowed inside lambdas. Of course, >>= is actually an expression and you can put any expression to the right of a lambda ->.
With all due respect to Python (a language I like but no longer use), it's no place to develop any intuition about lambdas. Python's lambda is an attempt to incorporate some of the syntactic benefit of anonymous functions while refusing them any real meaning or importance.
Maybe these are typical beginner misconceptions, or maybe they have more to do with coming from Python and complete beginners actually find it more natural.
As you can probably guess, I think it's more the latter. But I can't say first-hand because I dabbled in dozens of languages before Haskell, and I got my intuition for lambda from scheme. I wish I'd learned them in Haskell though, because with purity they're more elegant. Cheers, John
The reason my is available in the lambda \x -> my >>= \y -> W (x+y) has to do with scoping rules, Haskell (and almost all programming languages) use static scoping, meaning a variable defined in an outer function is available in the inner function, for instance, (\a -> \b -
(a,b)) 1 2 will evauate to (1,2) since a is bound to 1 in the outer lambda, and the inner one can refer to the outer one, if instead you write (\a -> \a -> (a,a)) 1 2 the result will be (2,2) since the innermost a will be used (no ambiguity here, but if shadowing is done by accident it can be hard to find the error). The book Structure and Interpretation of Computer Programs (freely available online) discusses this subject in detail.
= is available inside the lambda for the same reason, >>= is imported into the modules namespace, and therefore available everewhere, unless a shadowing binding exists.
I wouldn't say that this has anything to do with functional programming specifically. This also exists in Python i might add, you can read variables defined in an outer scope:
a = 1 f() def f(): ... return a ... f() 1
The fact that most languages are scoped like this is nothing obvious, and like I said, Python is the same. I hope this helps! / Adam On Aug 9, 2009, at 9:31 PM, Michael Mossey wrote:
I'm starting to figure out a few things that I didn't "get" about functional programming and monads. I wanted to explain them. I'm not looking for a particular response to this post, other than any elaboration that seems natural.
There is an exercise here working with the trivial monad W:
http://blog.sigfpe.com/2007/04/trivial-monad.html
Write a function g :: W a -> W a -> W a
such that g (W x) (W y) = W (x+y)
except don't use pattern matching, but >>= instead. The answer is
g mx my = mx >>= (\x -> my >>= \y -> W (x+y))
There are a couple things here that threw me off. One is that I didn't expect 'my' to be available inside the first lambda. I somehow thought of lambda as isolated, sealed-off from the rest of the universe. But they aren't. I believe this is the concept of closures, or related to it?
Secondly, I didn't expect >>= to be available inside the lambda. This is related to the mistaken conception of >>= as a procedural statement rather than an expression. In Python, where I have previously encountered lambdas, no statements are allowed inside lambdas. Of course, >>= is actually an expression and you can put any expression to the right of a lambda ->.
Maybe these are typical beginner misconceptions, or maybe they have more to do with coming from Python and complete beginners actually find it more natural.
Mike
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Thanks for the ideas, Adam. I still have a few questions. Adam Bergmark wrote:
The reason my is available in the lambda \x -> my >>= \y -> W (x+y) has to do with scoping rules, Haskell (and almost all programming languages) use static scoping, meaning a variable defined in an outer function is available in the inner function, for instance, (\a -> \b -> (a,b)) 1 2 will evauate to (1,2) since a is bound to 1 in the outer lambda, and the inner one can refer to the outer one, if instead you write (\a -> \a -> (a,a)) 1 2 the result will be (2,2) since the innermost a will be used (no ambiguity here, but if shadowing is done by accident it can be hard to find the error).
Because the lambda is executed by the implementation of >>=, doesn't the concept closure still apply? That value of 'my' has to "get into" the other routine. I am aware that other languages have closures, but in Python they are an advanced, rarely explored concept, so I haven't gotten a good intuition for them. (I don't mean to imply I've only programmed in Python---also Lisp and C++, but only to do boring things, never anything sophisticated from a CS point of view.) It's not that I don't understand the use of 'my'---it's just that it didn't occur to me at first.
The book Structure and Interpretation of Computer Programs (freely available online) discusses this subject in detail.
= is available inside the lambda for the same reason, >>= is imported into the modules namespace, and therefore available everewhere, unless a shadowing binding exists.
The problem is not that I didn't expect >>= to be outside the namespace. The problem is that I am still having to "unlearn" imperative concepts, so it was all too easy to think of >>= as an imperative concept, and in Python procedural statements are not allowed inside lambdas. Also, in the early stages of learning Haskell there are no monads used inside lambdas. So it's not that anyone told me I couldn't do it, just that it didn't occur to me the first time I saw the problem. There is no great revelation here, other than my own satisfaction at seeing a little deeper into the way things are done in Haskell, and finding these problems much easier after revisiting them (I didn't do anything Haskell for a couple months, and just came back to it). -Mike
Michael P Mossey wrote:
Thanks for the ideas, Adam. I still have a few questions.
Adam Bergmark wrote:
The reason my is available in the lambda \x -> my >>= \y -> W (x+y) has to do with scoping rules, Haskell (and almost all programming languages) use static scoping, meaning a variable defined in an outer function is available in the inner function, for instance, (\a -> \b -> (a,b)) 1 2 will evauate to (1,2) since a is bound to 1 in the outer lambda, and the inner one can refer to the outer one, if instead you write (\a -> \a -> (a,a)) 1 2 the result will be (2,2) since the innermost a will be used (no ambiguity here, but if shadowing is done by accident it can be hard to find the error).
Because the lambda is executed by the implementation of >>=, doesn't the concept closure still apply? That value of 'my' has to "get into" the other routine.
I think you are both right. AFAIU Adam comments on *visibility* of the variables, while you look more at the fact that you pass the lambda as an argument to (>>=). The lambda can "see" the variable my due to scoping, and (>>=) can trigger evaluation of the lambda due to closures. I'm looking forward to be corrected by someone who knows more about this than I do. /M -- Magnus Therning (OpenPGP: 0xAB4DFBA4) magnus＠therning．org Jabber: magnus＠therning．org http://therning.org/magnus identi.ca|twitter: magthe
participants (5)
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Adam Bergmark -
John Dorsey -
Magnus Therning -
Michael Mossey -
Michael P Mossey