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Re: curried fns
• cheno@m...
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 Date: 1995-11-21 (10:50) From: cheno@m... Subject: Re: curried fns
```
Jocelyn Serot wrote :

>>Hello,
>>
>>Could someone please explain the difference(s) between:
>>
>>        let f x = function y -> y + x;;
>>
>>and
>>
>>        let f x y = y + x;;
>>
>>Both have the same type (int -> int -> int) but they seem to behave
>>distinctly wrt evaluation strategy.
>>
>>For instance, if i use the 1st form and write:
>>
>>        let h x = let z = fact x in fun y -> y + z;;
>>        map (h 30) [1;2;3];; (* note 1 *)
>>
>>fact 30 gets evaluated only once (partial evaluation), while
>>the use of the 2nd form for the h function:
>>
>>        let h x y = let z = fact x in y + z;;
>>        map (h 30) [1;2;3];;
>>
>>causes fact 30 to be evaluated _for each_ element of the list.
>>
>>Is this normal or do i misunderstand sth about curryfied fns ?..
>>
>>Thanks for any help

1. excuse my poor english, thank's

2. I think this can help (I hope so !)

For my example :

#let fact x = print_int x ; print_newline() ; x ;;
fact : int -> int = <fun>

first version

#let h1 x =
let z = fact x
in
function y -> y + z ;;
h1 : int -> int -> int = <fun>

NB : the result of h1 30 is a function, which has been evaluated in a
environment where
z yields for the result of fact 30

So, we can see the result of print_int in *this* call !

#let a1 = h1 30 ;;
30
a1 : int -> int = <fun>

Now, a1 is all defined : no more print ... the function a1 don't print anything

#map a1 [1 ; 2 ; 3] ;;
- : int list = [31; 32; 33]

second version

#let h2 x y =
let z = fact x
in
y + z ;;
h2 : int -> int -> int = <fun>

there is no call for fact : you have specialized the first (and not last)
argument of h2

#let a2 = h2 30 ;;
a2 : int -> int = <fun>

now the evaluation is complete, and there is calls for print_int

#map a2 [1 ; 2 ; 3] ;;
30
30
30
- : int list = [31; 32; 33]
#

Laurent Chéno

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