On Jul 13, Connie Chan said:
>I now have an array like this : @list = ('1234', '4567', '789A', 'BCDE',
>'FGHI', .....);
>
>and now, I want to find where is the location of 'BCDE' in list , so I
>write like this :
>
>my @list = ('1234', '4567', '789A', 'BCDE', 'FGHI');
>my $GetLocation = 0;
>my $value = 'BCDE';
>
>for (my $atLoc = 0; $atLoc <= $#list and ! $GetLocation ; $atLoc++)
>{ $GetLocation = $atLoc if ($value eq $list[$atLoc]) }
>
>ok, so $GetLocation is the result I want...
>but any other smarter method ?
Once you have FOUND the element, you can exit the loop. Also, your loop
is testing !$GetLocation, which returns true if the element is found at
index 0. Your code will end up ASSUMING the element is index 0 if it
isn't found in the array at all. Here's are two solutions:
my @list = ...;
my $wanted = 'foo';
my $pos = -1; # -1 means not found
for (my $i = 0; $i < @list; $i++) {
$pos = $i, last if $list[$i] eq $wanted;
}
Now you'll have $pos being either -1 (not found) or the index in the
array. You'd check via
if ($pos != -1) { ... } # it was found
else { ... } # it wasn't found
Here's the other solution:
my @list = ...;
my $wanted = 'foo';
my $pos = 0; # 0 means not found
for (my $i = "0E0"; $i < @list; $i++) {
$pos = $i, last if $list[$i] eq $wanted;
}
Now you'll have $pos being either 0 (not found), or the index in the
array, BUT! it will be "0E0" if it was the first element. This means you
CAN test it like so:
if ($pos) { ... } # it was found
else { ... } # it wasn't found
because "0E0" is a string that is not false, but is treated as the number
0 when used numerically.
--
Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/
RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/
** Look for "Regular Expressions in Perl" published by Manning, in 2002 **
<stu> what does y/// stand for? <tenderpuss> why, yansliterate of course.
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