This is probably blindingly simple but I'm not understanding why @ARGV
is not reduced to () (no args)? in this example
I'm trying to understand what happens with ARGV in s getops loop with
arguments after the getopts Options, or any time there is ARGV, I
guess.
Why doesn't the `for loop' reduce
Harry Putnam wrote:
This is probably blindingly simple but I'm not understanding why @ARGV
is not reduced to () (no args)? in this example
To Perl, @ARGV is just another array. It has some special features but
can be assigned values just like other arrays. You can use it to do
tricks, like
HP == Harry Putnam rea...@newsguy.com writes:
HP print Now lets ditch the rest in a for loop\n;
HP for (@ARGV){
change that for to a while. for will create a list of aliases to the
array elements passed to it. it doesn't check its length as you seem to
think. while will loop until @ARGV is
Harry Putnam wrote:
This is probably blindingly simple but I'm not understanding why @ARGV
is not reduced to () (no args)? in this example
[ SNIP ]
print Now lets ditch the rest in a for loop\n;
for (@ARGV){
my $ditch = shift;
perldoc perlsyn
[ SNIP ]
Foreach Loops
[ SNIP ]
The
Hi,
what's the difference between do a block and eval a block?
sub test {
my $bl = shift;
eval $bl;
# do $bl;
}
test { print hello\n };
from the code, I see both do and eval work the same.
Thanks.
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Hi Xiaolan!
On Sunday 24 Jan 2010 05:37:49 Xiao Lan (小兰) wrote:
Hi,
what's the difference between do a block and eval a block?
sub test {
my $bl = shift;
eval $bl;
# do $bl;
}
test { print hello\n };
Well, if I run:
#!/usr/bin/perl
use strict;
use warnings;
sub test
