That does the first line in function:
my($what, @array) = @_;
That puts to $what the first parameter passed to function and to @array
the rest of parametres..in this case @names. So you have a copy of
@names in @array. But I am begginer too, so maybe i undersand it
wrong;-) And for begginings, I reccomend O'Reilly's Llama Book...
Good luck with perl
Jirka
Joseph Mwesigwa Bbaale wrote:
Hello,
I am a complete beginner - no programming background.
I want to begin with Perl and I decided to by Randal's book "*Learning Perl*".
I seem to have been progressing smoothly till when I arrived at the code
below on page 65.
my @names = qw/ tom fred dan betty roy /;
my $result = &which_element_is("dan" @names);
sub which_element_is {
my($what, @array) = @_;
foreach (0..$#array) {
if ($what eq $array[$_]) {
return $_;
}
}
-1;
}
The author states or seems to imply that, "... *...@array* is a copy of
*...@names
*"! But I don't understand how and why this is so.
I am stuck.
Please, can someone help and explain to me how and why *...@array* is a copy of
*...@names?
*
Thanking you in advance.
Kind Regards,
Joe.
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