If I understand list comprehensions correctly, what you wrote is the same as

do a <- "ab";
      b <- "12";
      [a:[b]]

which is the same as

"ab" >== \a -> do b <- "12"; [a:[b]]

which is the same as

"ab" >>= \a -> "12" >>= \b -> [a:[b]]

which is the same as

concat $ map  (  \a -> "12" >>= \b -> [a:[b]] ) "ab"

.... enough desugaring for now

Point is, yes it's written in stone.

List comprehensions is just syntactic sugar for monad operations.

Good exercise is to take the above expressions and add parenthesis to make it easier to understand order of operations. (Still trips me up often enough).

Thomas.




Maurí­cio <briqueabraque@yahoo.com>
Sent by: haskell-cafe-bounces@haskell.org

10/25/2007 05:59 PM

To
haskell-cafe@haskell.org
cc
Subject
[Haskell-cafe] List comprehension order of evaluation





Hi,

Today, if I write:

[a:[b] | a<-"ab" , b<-"12"]

I get:

["a1","a2","b1","b2"]

Are there any guarantees that I'll never
get ["a1","b1","a2","b2"] instead, i.e.,
that the first list will always be the
last one to be fully transversed? Even
if I use a different compiler or a
future version of Haskell?

Reading how list comprehensions are
translated in the Haskell report it
seems the answer is yes. Is that
written in stone? Can compilers do
it in their own different way?

Thanks,
Maurício

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