Have Geek Brain

What does -G do in Perl?

I've searched the internet and Perl document and can't seem to find an answer.

Help would be appreciated.

EDIT :: He asked me about -G, wrote it down on a piece of paper, and when I looked at me, asked me to read the basics.

-2


a source to share


5 answers


I agree with JesperE, please show us some code. However, as far as I can tell, this is what is happening:

if(-G) {

Perl sees this, does not recognize it, -G

and therefore treats it as a string. This will:

if('-G') {

Which is equivalent:

if(1) {

So, as far as I can tell, if(-G)

does nothing. I tried to use it and it always returns true which confirms my hypothesis. Further support comes from the following code (tested on OS X with Perl 5.10.0):



use strict;
use warnings;

my $var = -G;
print "$var\n";

Displays warnings, compiles and runs, and prints just "-G".

EDIT: Doing a search that I should have done much earlier suggests the following from the Perldoc perlop page:

Unary "-" performs arithmetic negation if the operand is numeric. If the operand is an identifier, a string consisting of the minus sign concatenated with the identifier is returned. Otherwise, if the string starts with a plus or minus sign, a string starting with the opposite sign is returned. One of the consequences of these rules is that -bareword is equivalent to the string "-bareword". If, however, the string begins with a non-alphabetic character (excluding "+" or "-"), Perl will try to convert the string to a numeric value and perform arithmetic negation. If a string cannot be cleanly converted to numeric, Perl will issue a warning . String argument is not numeric in negation (-) at address.

As pointed out in the comments B::Deparse

shows what Perl converts if(-G)

to if(-'G')

. However, the documentation (and behavior from print()

) is consistent with the documentation, which says it should convert if(-G)

to if('-G')

. This does not change the output of the program in any way.

However, the subtle typing differences in the behavior of unary operators, which 99% of people will only use by numbers, are not what I would call "basic." I do not think that someone should (or will ever need) to use a conversion -bareword

to 'bareword'

any practical situation.

+7


a source


There is no -G switch in perl.

perl -G Unrecognized switch: -G (-h will show valid options).

Edit: OK, there's nothing with -G in there - just -g.



-g The setgid bit is set in the file.

http://perldoc.perl.org/perlfunc.html

Otherwise, it is nonsense and the question is wrong.

+2


a source


I don't know about -G, but -g is described here as

-g  File has setgid bit set.
+2


a source


This is clearly a confusion between the [(test) options and the Perl -X file checks. -G is in the former (on my BSD system) but not the latter. -G is a non-positional extension, and I guess Perl doesn't include all extensions, just a few. So he either wanted to say -g, or he meant [-G $ file]; (for some superset of POSIX [). It's also in my default shell (pdksh) and bash (the default linux shell, for the most part)

-G in test or as a wrapper built in here:

-G file True if file exists and its group matches the effective group ID of this process.

+1


a source


One answer says, "I don't think anyone should (or will ever need to) use the -bareword to -'bareword conversion in any practical situation."

This is widely used in one style of named parameters. See the venerable CGI for one:

$cookie1 = $q->cookie(-name=>'riddle_name', -value=>"The Sphynx Question");
$cookie2 = $q->cookie(-name=>'answers', -value=>\%answers);
print $q->header(
    -type    => 'image/gif',
    -expires => '+3d',
    -cookie  => [$cookie1,$cookie2]
);
0


a source






More articles:


All Articles