RE: strtok

[CWilt@xxxxxxxxxxxx]

Scott,

Thanks for the explanation.  I (thought I) knew what (*) did, but what I
didn't realize the effect of having a procedure on the right-hand of the
assignment.

In other words, I was thinking gettok() would be called once and every
element of tokens would be assigned the same value.  But what you're saying
is that gettok() gets called multiple times, once for every element. 

Cool.....

Charles



> -----Original Message-----
> From: Scott Klement [mailto:rpg400-l@xxxxxxxxxxxxxxxx]
> Sent: Thursday, August 19, 2004 11:57 AM
> To: RPG programming on the AS400 / iSeries
> Subject: RE: strtok
> 
> 
> 
> Hello Charles,
> 
> > Usually, I follow what you are doing.  However, the following has me
> > scratching my head:
> [SNIP]
> > >           x = 0;
> > >           tokens(*) = gettok( blah : '|' : x );
> 
> I guess the answer lies in the question  "What does (*) do?"  When you
> assign all values of an array in this manner, array(*) = 
> 'something'; what
> you're really doing is looping through the entire array, assinging
> 'something' to every element.
> 
> So, the following two code snippets are equivalent:
> 
>           x = 0;
>           tokens(*) = gettok( blah : '|' : x );
> 
> and:
>           x = 0;
>           for i = 1 to %elem(tokens);
>               tokens(i) = gettok( blah : '|' : x );
>           endfor;
> 
> Normally, I'd use the latter.  I think it's easier to tell what's
> happening -- but having been told earlier this week that the 
> SCAN op-code
> is easier than the %scan() BIF because it can return an array 
> in one line
> of code, I figured I'd do it this way. :)
> 
>