>> On Tue, 17 Dec 2002 11:59:46, <[EMAIL PROTECTED]> said:
> Also, you might change 'scalar(@url)' to '$#url' for simplicity.
More than just simplicity:
void:chris~ % perl -le '@a=(5,10,15,20); print scalar @a,"\n",$#a'
4
3
$# signifies the index of the last element in an array - and since an
array is zero-indexed in Perl, that's always going to be one less than
the total number of elements in the array, which is what scalar(@array)
gives. Using scalar(@array) as the upper bound in a for loop would give
an extra iteration of the for loop over a non-existant element with the
index $array[$#array+1]. So don't do that. :-)
- Chris.
--
$a="printf.net"; Chris Ball | chris@void.$a | www.$a | finger: chris@$a
| Q. How do you tell an extrovert techie from an introvert techie?
| A. He looks at your feet rather than his own.
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