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Program Flow
f#

I tried small program based on example written on Prolog. Delete element from list.

In Prolog it looks like

remove(A,[A|B],B).

remove(A,{B|C],[B|D]):-remove(A,C,D).

remove(A,[],[]).

match works next way. If variable has value, the value should match, othervise variable gets value which cannot be changed later. I tried on f# next code

let rec remove a1 list1 =

match list1 with

| a1::t1->t1;//(1)

| h1::t1-> remove a1 t1 ;

| []->list1;;

let rec printInt abc =

match abc with

| a::b-> printfn "%d," a;printInt b;

| []-> printfn "stop";;

let removing a =

let b=remove a [1;3;2;4];

printInt b;;

removing(2);

when i tryed directly remove 2 [1;3;2;4]

it did not work at all. On the code above I expected that in option (1) a1 should match entry value. therefore this match gives fail, and I go to next match. However to my surprise a1 get value 1 and function finished. It is hard to tell if here it accepted a1 as internal value or this is a bug. I spent whole day to find in internet information about program flow and how match works. But I saw only few simple code. However it is very important how program flows.

About video. When you will prepare next video it is important to see your code. but it was very hard.

Thank you. Aleks Kleyn

.

This is a common misunderstanding:

match list1 with

| a1::t1 -> ...

will match all lists with at least one element, and bind the name 'a1' to the head of the list and 't1' to the tail of the list. The previous value of 'a1' gets shadowed.

You want something more like

match list1 with

| h1::t1 when h1 = a1 -> ...

which will test if the head of the list has the same value as a1.

(Don't spend the whole day looking when you can just ask and possibly get a quick response!)

By on 5/29/2009 6:19 PM ()

So the program get form

let rec remove a1 list1 =

match list1 with

| h1::t1 when h1=a1->t1;

| h1::t1-> remove a1 t1 ;

| []->list1;;

let rec printInt abc =

match abc with

| a::b-> printfn "%d," a;printInt b;

| []-> printfn "stop";;

let removing a =

let b=remove a [1;3;2;4];

printInt b;;

removing(2);

As answer I get b=[4]. not [1;3;4]

By on 5/29/2009 7:26 PM ()

On this line of 'remove'

| h1::t1-> remove a1 t1 ;

you are throwing away the head value h1. You want

| h1::t1-> h1 :: (remove a1 t1) ;

That is, for example, the answer to

remove 2 [1;3;2;4]

should be

1 :: (remove 2 [3;2;4])

That is, you want to keep the element at the front, but recurse into the rest of the list to find the element you intend to remove.

By on 5/29/2009 7:53 PM ()

I will try your sugestion. Whar you have written about match with list is clear for me. However this works when variables in expression a::b do not have value. If imutable variable at match changes value this is a bug. I cannot write

| a::b if a=a

so I need to change variable.

and if there would be information how program flows this question may be not appear.

By on 5/29/2009 7:07 PM ()

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