# Why does this work - haskell mysteries?

Michael Wang Michael.Wang at synopsys.com
Fri Oct 3 18:26:37 EDT 2003

I am guessing this is how the data got accumulated:
in you my_store, the function stored is like

\w -> if w == 'a'
then 3
else if w == 'b'
then 5
else if w == 'a'
then 4
else (\w -> 0 ) w

'a' is stored twice here, ( maybe compiler will optimize it out but
basically
you see the lasted 'a' mapping value  )

-----Original Message-----
Sent: Saturday, October 04, 2003 4:48 PM
Subject: Why does this work - haskell mysteries?

Hi;

the proof of the pudding does lies in the eating... but I still wonder why
this code is working (it is taken from the book "The Craft of functional
programming").

The program connects a variable-name to value. The fun initial gives the
initial state, update sets a variable & value reads a value).

I evaluate

value my_store 'b' to 5
and  	        value my_store 'a' to 3

as expected from the text in the book.

But I can't see what is happening here. The book has a parallel example
where
the data is held in a list, and this version is easy to follow, but this
trick with storing a lambda-function inside a newtype beats me.

The problem is that I do not understand where the accumulated data is stored
(not in a list - it seems like something like a chain of functions which can
be pattern-matched, but I am not sure).

And why does not the lambda-function (\w -> if v==w then n else sto w) start
a
endless loop?

(This is not homework - I am a programmer who is curious about Haskell!)

Any clues, anyone?

Cheers,

Petter

-- Var is the type of variables.

type Var = Char

newtype Store = Sto (Var -> Int)
--
initial :: Store

initial = Sto (\v -> 0)

value :: Store -> Var -> Int

value (Sto sto) v = sto v

update  :: Store -> Var -> Int -> Store

update (Sto sto) v n
= Sto (\w -> if v==w then n else sto w)

-- testit --

my_store = update (update (update initial 'a' 4) 'b' 5) 'a' 3)

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