Re: why this function call print 9

For the simple answer, it looks like you're trying to pass an array, @3, to your function. @3 is naturally empty, so @_ picks up the values in the array.

The first value in the array is 9, which $a gets set to in your function. You then change $LOCAL_CODEPAGE to $a at the end. (update: And you return that value, which is still, predictably, 9.)

Though I'm not one to criticize coding usually, I will note that you print "\n" at the end of your function, though you return before that. If for any reason the "return" part of your assignment of $LOCAL_CODEPAGE got taken out, you'd be returning the return value of the print, which I don't think is what you want.

I take it this function is a work in progress. I would rewrite it entirely, but I'm lazy. I'll just show you what I did looking at it:

sub f1{
    
    my ($a) =@_;
    $cp=1;
        $LOCAL_CODEPAGE =10;
        print "This is $a\n";
        print $LOCAL_CODEPAGE;
    if ($a==1){
        $LOCAL_CODEPAGE = $cp;
        break;
    }
    return $LOCAL_CODEPAGE = $a;
    print "\n";
}
@3=(3,10, 13);  ## comment this out and see your value come back as 9
print "\$LOCAL_CODEPAGE=$LOCAL_CODEPAGE: f1(@3,9,3,4,5): ".f1(@3,9,3,4
+,5)."\n";
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Comment on Re: why this function call print 9 Download Code

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Re^2: why this function call print 9 by edwardt_tril (Sexton) on Feb 09, 2006 at 00:57 UTC
Thanks, it is just a test function. What I expect is that it prints nothing. I see other people's example using shift to shit to next element in the array but I have not "shifted" to the next element. Is this an implicit shift?	[reply]
Re^3: why this function call print 9 by GrandFather (Saint) on Feb 09, 2006 at 01:01 UTC
`my ($a) =@_;` picks off the first element of @_. `my ($a) = shift;` would get the same value into $a (bad name for a variable BTW - used by sort) and removes that element for @_. DWIM is Perl's answer to Gödel	[reply] [d/l] [select]
Re^3: why this function call print 9 by SamCG (Hermit) on Feb 09, 2006 at 18:46 UTC
When you say, "I expect it to print nothing", I'm guessing you're thinking your first element is undefined? So you're thinking that the shift was done twice, once for what you think is an undefined element and once for an element with the value. This interpretation is incorrect: you're passing a 4-element array to your function, and it's as if the @3 didn't exist at all. There is no "implicit shift" going on, unless you think of the `my ($a)=@_;` as a "shift". If you merge two arrays, and one has no elements, you don't end up with an undefined element as a placeholder for the empty array (I don't know if I can put it more clearly). `sub f1{ my ($zeroelement, $oneelement, $twoelement, $threeelement) =@_; print scalar(@_); ## prints number of elements -->4 ## each element is in the appropriate value, an +d $zeroelement is 9 $cp=1; $LOCAL_CODEPAGE =10; if ($a==1){ $LOCAL_CODEPAGE = $cp; break; } return $LOCAL_CODEPAGE = $a; print "\n"; } @3=(); print "\$LOCAL_CODEPAGE=$LOCAL_CODEPAGE: f1(@3,9,3,4,5): ".f1(@3,9,3,4 +,5)."\n";` [download]	[reply] [d/l] [select]