On Monday 21 Jul 2003 7:21 pm, Alastair Reid wrote:
If I try to run the program (compiled using GHC 6), it calculates all
members of the list and then prints the whole list in the end. Since
Haskell is 'lazy' I was expecting behaviour similar to HUGS where it
prints the numbers as it finds
AJ wrote:
Ok so I added the hSetBuffering stdout NoBuffering line to my main
function and still got the same 'nonlazy' behaviour, or so I thought.
It took me some time to figure out what was going on. I am using emacs
to write my code and I was calling a.out from inside it. Obviously it
On Friday 25 Jul 2003 6:52 pm, you wrote:
AJ wrote:
Ok so I added the hSetBuffering stdout NoBuffering line to my main
function and still got the same 'nonlazy' behaviour, or so I thought.
It took me some time to figure out what was going on. I am using emacs
to write my code and I was
Hi everyone,
I have written the following program to find magic numbers i.e. integers 'n' such
that both (n+1) and (n/2+1) are perfect squares.
-- Program to find magic numbers
Import IO
main :: IO ()
main =
do
print (filter magicP sqs)
sqs :: [Int]
sqs = [x*x | x -
If I try to run the program (compiled using GHC 6), it calculates all
members of the list and then prints the whole list in the end. Since
Haskell is 'lazy' I was expecting behaviour similar to HUGS where it prints
the numbers as it finds them. Does this behaviour have something to do with
