Re: a perplexing difference
 Pierre Weis
[
Home
]
[ Index:
by date

by threads
]
[ Message by date: previous  next ] [ Message in thread: previous  next ] [ Thread: previous  next ]
[ Message by date: previous  next ] [ Message in thread: previous  next ] [ Thread: previous  next ]
Date:   (:) 
From:  Pierre Weis <weis@p...> 
Subject:  Re: a perplexing difference 
> I'm unable to understand why my fonctions do return these answers The overall problem is that you misunderstood the Caml notion of vector (which is the one commonly used in programming languages such as Pascal, C, Ada, ...): a vector is a mutable data structure, that is, modification are performed in place. In other words, when you change the contents of some element in a vector the vector remains the same. So you cannot get a trace of successive assignments of a vector, unless you require explicit copies of the vector. This is not common in practice, since this is memory consuming. In your case, you may want the computational behaviour of lists and have made a confusion between vectors and lists (replacements in lists instead of assignments in vectors, accesses in list instead of direct accesses in vectors, and so on) ? > (* first a vect swapping function*) > let swap k t= > let prov=t.(k1) in > t.(k1)<t.(k);t.(k)<prov; > t > ;; Warning: swap is a procedure, it has not to return something. In your case returning t is useless, since t is exactly (physically) the same as the argument of the procedure. Thus as far as mathematical computation is concerned swap is the identity function on vectors (more exactely, swap is equivalent to fun k t > t) > (*and as I was doubting of everithing I even tried it!*) > let b=swap 3 [1;2;3;4;5;6];; > let c=swap 1 b;; > b;; In particular here, b and c are identical: what you observe when looking at c is not the new value of b, it's b that has been modified: #c == b;;  : bool = true > (*then an iterator which gives a list of the running results : as with the > Mathematica FoldList*) > let rec fold_list f l a=match l with > []>[a] > p::q>a::fold_list f q (f p a);; This fold_list is a ``memo'' version of the list_it iterator: it returns the list of successive results of the application of list_it. #list_it;;  : ('a > 'b > 'b) > 'a list > 'b > 'b = <fun> fold_list : ('a > 'b > 'b) > 'a list > 'b > 'b list = <fun> > (* first a test with an ordinary function, then another with swap: > the first one does what I expected it would, > the other one yields four times the same list*) > > let h u v=u+10*v;; > fold_list h [5;6;7] 0;; #fold_list h [5;6;7] 0;;  : int list = [0; 5; 56; 567] > fold_list swap [1;3;2] [10;20;30;40;50];; Applied to swap, fold_list returns a list of elements which are successive results of swap applied to the vector [10;20;30;40;50] given as argument: that's a list of vectors physically equal to the input argument. #let orig = [10;20;30;40;50];; orig : int vect = [10; 20; 30; 40; 50] #let [v1; v2; v3; v4] = fold_list swap [1;3;2] orig;; Entrée interactive: >let [v1; v2; v3; v4] = fold_list swap [1;3;2] [10;20;30;40;50];; >^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Attention: ce filtrage n'est pas exhaustif. v1 : int vect = [20; 40; 10; 30; 50] v2 : int vect = [20; 40; 10; 30; 50] v3 : int vect = [20; 40; 10; 30; 50] v4 : int vect = [20; 40; 10; 30; 50] #orig == v1 && v1 == v2 && v2 == v3 && v3 == v4;;  : bool = true > (* so I made another swap_like function, I tried, It works*) > let h x t=[x+t.(1);t.(0)];; As you may guess, here your h function returns a fresh vector at each call, hence the copying behaviour you wanted. > fold_list h [1;3;2] [10;20;30;40;50];; > > I also tried to Nest my swaps on the command line (with several ((...)))): I > got the fold_list I expected (and not the constant one) I don't get this one: if you evaluate (swap 1 (swap 3 (swap 2 orig))) you will perform the side effect and get orig, but won't get the list of sucessive states of vector orig. As if we build it by evaluating an expression equivalent to the call to fold_list, for instance let v1 = swap 2 orig in let v2 = swap 3 orig in let v3 = swap 1 orig in [v3; v2; v1; orig];; will return the same list of 4 vectors identical to orig. > As always, more than an answer, a reference to some place in the > caml books where I might have found the answer would be welcomed. I don't know this place. Well, now you may find it in the caml mailing list archive! :) Pierre Weis INRIA, Projet Cristal, Pierre.Weis@inria.fr, http://pauillac.inria.fr/~weis