/prog/ - /prog/ challenge (S(SKK)(SKK))(S(SKK)(SKK))

Name: Anonymous 2011-04-10 9:36

The challenge: Implement generators in the favourite language of yours.
Are banned: Language native generators (i.e. FIOC's yield), delimited continuations (sorry, Schemers, that would be too easy).

The test program will be an infinite generator that computes facts, and a loop that takes the first five of them.

The deadline is 17/04/2011 00:00 /prog/time.
The most complete, small and easy to use implementation wins.

Name: Anonymous 2012-07-12 14:28

Name: Anonymous 2012-07-12 14:29

bump

Name: Anonymous 2012-07-12 14:30

Name: Anonymous 2012-07-12 14:31

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Name: Anonymous 2012-07-12 15:33

make_counter = (max) ->
current = 0
return () ->
if current <= max
return current++

c = make_counter 5

Name: Anonymous 2012-07-13 19:22

PROC facs generator = PROC INT: (

    HEAP INT g := 1, n := 0;

    INT: g *:= (n +:= 1));

PROC INT fac = facs generator;

TO 5 DO print((fac, new line)) OD

Name: Anonymous 2012-07-14 23:46

bump

Name: Anonymous 2012-07-16 0:16

>>52
holy fucking ENTERPRISE variables nigger

/prog/ challenge (S(SKK)(SKK))(S(SKK)(SKK))

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Name: Anonymous 2011-04-10 9:36

Name: Anonymous 2012-07-12 14:28

Name: Anonymous 2012-07-12 14:29

Name: Anonymous 2012-07-12 14:29

Name: Anonymous 2012-07-12 14:29

Name: Anonymous 2012-07-12 14:30

Name: Anonymous 2012-07-12 14:30

Name: Anonymous 2012-07-12 14:31

Name: Anonymous 2012-07-12 14:31

Name: Anonymous 2012-07-12 15:33

Name: Anonymous 2012-07-13 19:22

Name: Anonymous 2012-07-14 23:46

Name: Anonymous 2012-07-16 0:16