[Haskell-beginners] some insights into functional programming

Michael P Mossey mpm at alumni.caltech.edu
Mon Aug 10 18:36:13 EDT 2009

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 

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).


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