/prog/ - Haskell — World4ch

Name: Anonymous 2007-10-03 23:44

Hi /prog/, i'm doing a basic programming course at uni and the lecturer has set a challenge to write the id Haskell prelude function recursively:

id :: a -> a
id x = x

I can't think of a way to do this which is meaningful, any ideas (this isn't marked or anything so i'm not in any way cheating).

Name: Anonymous 2007-10-04 1:19

$ shift . ( map ! car unshift $^#%^ ) :: cdr -> sussman

Name: Anonymous 2007-10-04 5:14

Spit in your lecturers face and tell him he 1) Doesn't understand recursion 2) Hasn't read SICP.

Name: Anonymous 2007-10-04 5:59

>>1
id Haskell prelude function recursively
whut?

id is (lambda (x) x)

Name: Anonymous 2007-10-04 6:39

Tell your professor that he can pick between an "id" written recursively, and one that TERMINATES.

Name: Anonymous 2007-10-04 7:33

CAPTAIN ENTERPRISE to the rescue!

enterpriseMissionCriticalIdentityFunction :: a -> a

enterpriseMissionCriticalIdentityFunction x =

  enterpriseNTierIdentityFunction x null null null null null null null 69 -- Scalable!

  where

    enterpriseNTierIdentityFunction x _ _ _ _ _ _ _ n

      | n > 0 = enterpriseNTierIdentityFunction x null null null null null null null (n - 1)

      | otherwise = x



main = do

  print $ enterpriseMissionCriticalIdentityFunction 1

Name: Anonymous 2007-10-04 7:53

>>6
You forgot singleton enterpriseMissionCriticalIdentityFunctionClassFactoryBridge
Wait, what

Name: Anonymous 2007-10-04 9:16

>>6

(OP) Thankyou Captain Enterprise, shall i take this to mean that i was correct in assuming that there is no meaningful way to do this then? I'll post his answer when he gives it so anyone who is interested can take a gander.

Name: Anonymous 2007-10-04 12:35

shall i take this to mean that i was correct in assuming that there is no meaningful way to do this then?

Of course.
____
Sincerely,
Captain. Obvious, on behalf of Captain Enterprise

Name: Anonymous 2007-10-04 14:23



id :: a -> a

id = id -- POINT FREE

Name: Anonymous 2007-10-04 18:30

>>10
Pointless

Name: Anonymous 2007-10-05 10:34

The "correct" solution was

id :: a -> a->
id x = const x id

or

id x = const x (id x)

or

id x = if True then x else id x

so i guess there was no meaningful solution after all.

Name: Anonymous 2010-12-06 9:26

^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^{^Ba}c}k}}t}o}}/}b}/},}}`}`}G}N}A}A}}F}a}g}g}o}t}'}'}

Haskell

1 Name: Anonymous 2007-10-03 23:44

2 Name: Anonymous 2007-10-04 1:19

3 Name: Anonymous 2007-10-04 5:14

4 Name: Anonymous 2007-10-04 5:59

5 Name: Anonymous 2007-10-04 6:39

6 Name: Anonymous 2007-10-04 7:33

7 Name: Anonymous 2007-10-04 7:53

8 Name: Anonymous 2007-10-04 9:16

9 Name: Anonymous 2007-10-04 12:35

10 Name: Anonymous 2007-10-04 14:23

11 Name: Anonymous 2007-10-04 18:30

12 Name: Anonymous 2007-10-05 10:34

14 Name: Anonymous 2010-12-06 9:26