What does -G do in Perl? - perl

I have searched around the Internet and the Perl doc and can't seem to find the answer.
Help will be appreciated.
EDIT:: He asked me about -G, wrote it down on a piece of paper and when i looked stumped asked me to go read up on the basics.

I agree with JesperE, please show us some code. However, as far as I can tell, this is what's happening:
if(-G) {
Perl sees this, doesn't recognize -G, and so treats it as a string. It becomes:
if('-G') {
Which is equivalent to:
if(1) {
So as far as I can tell, if(-G) does nothing. I've tried using it, and it always seems to return true, which supports my hypothesis. Further support is from the following code (tested on OS X with Perl 5.10.0):
use strict;
use warnings;
my $var = -G;
print "$var\n";
Displays no warnings, compiles and runs, and prints simply "-G".
EDIT: Doing a search I should have done much earlier provides the following from Perldoc's perlop page:
Unary "-" performs arithmetic negation if the operand is numeric. If the operand is an identifier, a string consisting of a minus sign concatenated with the identifier is returned. Otherwise, if the string starts with a plus or minus, a string starting with the opposite sign is returned. One effect 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 attempt to convert the string to a numeric and the arithmetic negation is performed. If the string cannot be cleanly converted to a numeric, Perl will give the warning Argument "the string" isn't numeric in negation (-) at ....
As stated in the comments, B::Deparse appears to show that Perl converts if(-G) to if(-'G'). However, the documentation (and the behavior with print()) are consistent with the documentation, which says that it should convert if(-G) to if('-G'). This doesn't change the result of the program either way.
However, subtle typing differences in the behaviors of unary operators that 99% of people will only ever use on numbers are not what I would call "basic." I don't think anyone should (or would ever need to) use the -bareword to 'bareword' conversion in any practical situation.

There's no switch -G in perl.
perl -G
Unrecognized switch: -G (-h will show valid options).
Edit: OK, there's nothing with -G either - only -g.
-g File has setgid bit set.
http://perldoc.perl.org/perlfunc.html
Otherwise, it's nonsense and the question is misphrased.

I don't know about -G but -g is described here as
-g File has setgid bit set.

This is clearly confusion between [ (test) options and Perl -X file tests. -G is in the former (on my BSD system), but not the latter. -G is a non-posix extension and I guess Perl didn't include all the extensions, just some. So its either, he meant to say -g or he meant [ -G $file ]; (for some superset of POSIX [). It is also in my default shell (pdksh) and bash (the linux default shell, for the most part)
-G in test or as a shell builtin here:
-G file
True if file exists and its group matches the effective group ID
of this process.

One answer says: "I don't think anyone should (or would 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's Question");
$cookie2 = $q->cookie(-name=>'answers', -value=>\%answers);
print $q->header(
-type => 'image/gif',
-expires => '+3d',
-cookie => [$cookie1,$cookie2]
);

Related

Perl interface with Aspell

I am trying to identify misspelled words with Aspell via Perl. I am working on a Linux server without administrator privileges which means I have access to Perl and Aspell but not, for example, Text::Aspell which is a Perl interface for Aspell.
I want to do the very simple task of passing a list of words to Aspell and having it return the words that are misspelled. If the words I want to check are "dad word lkjlkjlkj" I can do this through the command line with the following commands:
aspell list
dad word lkjlkjlkj
Aspell requires CTRL + D at the end to submit the word list. It would then return "lkjlkjlkj", as this isn't in the dictionary.
In order to do the exact same thing, but submitted via Perl (because I need to do this for thousands of documents) I have tried:
my $list = q(dad word lkjlkjlkj):
my #arguments = ("aspell list", $list, "^D");
my $aspell_out=`#arguments`;
print "Aspell output = $aspell_out\n";
The expected output is "Aspell output = lkjlkjlkj" because this is the output that Aspell gives when you submit these commands via the command line. However, the actual output is just "Aspell output = ". That is, Perl does not capture any output from Aspell. No errors are thrown.
I am not an expert programmer, but I thought this would be a fairly simple task. I've tried various iterations of this code and nothing works. I did some digging and I'm concerned that perhaps because Aspell is interactive, I need to use something like Expect, but I cannot figure out how to use it. Nor am I sure that it is actually the solution to my problem. I also think ^D should be an appropriate replacement for CTRL+D at the end of the commands, but all I know is it doesn't throw an error. I also tried \cd instead. Whatever it is, there is obviously an issue in either submitting the command or capturing the output.
The complication with using aspell out of a program is that it is an interactive and command-line driver tool, as you suspect. However, there is a simple way to do what you need.
In order to use aspell's command list one needs to pass it words via STDIN, as its man page says. While I find the GNU Aspell manual a little difficult to get going with, passing input to a program via its STDIN is easy enough and we can rewrite the invocation as
echo dad word lkj | aspell list
We get lkj printed back, as due. Now this can run out of a program just as it stands
my $word_list = q(word lkj good asdf);
my $cmd = qq(echo $word_list | aspell list);
my #aspell_out = qx($cmd);
print for #aspell_out;
This prints lines lkj and asdf.
I assemble the command in a string (as opposed to an array) for specific reasons, explained below. The qx is the operator form of backticks, which I prefer for its far superior readability.
Note that qx can return all output in a string, if in scalar context (assigned to a scalar for example), or in a list when in list context. Here I assign to an array so you get each word as an element (alas, each also comes with a newline, so may want to do chomp #aspell_out;).
Comment on a list vs string form of a command
I think that it's safe to recommend to use a list-form for a command, in general. So we'd say
my #cmd = ('ls', '-l', $dir); # to be run as an external command
instead of
my $cmd = "ls -l $dir"; # to be run as an external command
The list form generally makes it easier to manage the command, and it avoids the shell altogether.
However, this case is a little different
The qx operator doesn't really behave differently -- the array gets concatenated into a string, and that runs. The very fact that we can pass it an array is incidental, and not even documented
We need to pipe input to aspell's STDIN, and shell does that for us simply. We can use a shell with command's LIST form as well, but then we'd need to invoke it explicitly. We can also go for aspell's STDIN by means other than the shell but that's more complex
With a command in a list the command name must be the first word, so that "aspell list" from the question is wrong and it should fail (there is no command named that) ... except that in this case it wouldn't (if the rest were correct), since for qx the array gets collapsed into a string
Finally, apsell nicely exposes its API in a C library and that's been utilized for the module you mention. I'd suggest to install it as a user (no privileges needed) and use that.
You should take a step back and investigate if you can install Text::Aspell without administrator privilige. In most cases that's perfectly possible.
You can install modules into your home directory. If there is no C-compiler available on the server you can install the module on a compatible machine, compile and copy the files.

When I run a file which is begin with "#!/usr/bin/perl -w", I get a error: "syntax error at line 153, near "=~ ?""

When I run a file which is begin with #!/usr/bin/perl -w, I get a error:
syntax error at line 153, near "=~ ?"
I try to add "#!/bin/bash", this error is not append, but I get another
error:
"line 34: syntax error near unexpected token `('"
line 153 in my file:
($output_volume =~ ?^([\S]+).mnc?) && ($base_name = $1) ||
die "sharpen_volume failed: output volume does not appear to be"
." a minc volume.\n";
line34 in my file:
use MNI::Startup qw(nocputimes);
$output_volume =~ ?^([\S]+).mnc?
This used to be valid perl and thus might appear in old code and instructional material.
From perlop:
In the past, the leading m in m?PATTERN? was optional, but omitting it would produce a deprecation warning. As of v5.22.0, omitting it produces a syntax error. If you encounter this construct in older code, you can just add m.
That is Perl code so the first error message is meaningful.
With delimiters other than // in the match operator you must have the explicit m for it, so
$output_volume =~ m?^([\S]+).mnc?
It is only with // delimiters that the m may be omitted; from Regex Quote-Like Operators (perlop)
If "/" is the delimiter then the initial m is optional.
See perlretut for a tutorial introduction to regex and perlre for reference.
Also note that the particular delimiters of ? trigger specific regex behavior in a special case. This is discussed by the end of the documentation section in perlop linked above.
You already have two answers that explain the problem.
? ... ? is no longer valid syntax for a match operator. You need m? ... ? instead.
Until Perl 5.22, your syntax generated a warning. Now it's a fatal error (which is what you are seeing). So I assume you're now running this on a more recent version of Perl.
There are, however, a few other points it is probably worth making.
You say you tried to investigate this by changing the first line of your file from #!/usr/bin/perl -w to #!/bin/bash. I'm not sure how you think this was going to help. This line defines the program that is used to run your code. As you have Perl code, you need to run it with Perl. Trying to run it with bash is very unlikely to be useful.
The m? ... ? (or, formerly, ? ... ?) syntax triggers an obscure and specialised behaviour. It seems to me that this behaviour isn't required in your case, so you can probably change it to the more usual / ... /.
Your regex contains an unescaped dot character. Given that you seem to be extracting the basename from a filename that has an extension, it seems likely that this should be escaped (using \.) so that it matches an actual dot (rather than any character).
If you are using this code to extract a file's basename, then using a regex probably isn't the best approach. Perhaps take a look at File::Basename instead.

Variable not being recognized after "read"

-- Edit : Resolved. See answer.
Background:
I'm writing a shell that will perform some extra actions required on our system when someone resizes a database.
The shell is written in ksh (requirement), the OS is Solaris 5.10 .
The problem is with one of the checks, which verifies there's enough free space on the underlying OS.
Problem:
The check reads the df -k line for root, which is what I check in this step, and prints it to a file. I then "read" the contents into variables which I use in calculations.
Unfortunately, when I try to run an arithmetic operation on one of the variables, I get an error indicating it is null. And a debug output line I've placed after that line verifies that it is null... It lost it's value...
I've tried every method of doing this I could find online, they work when I run it manually, but not inside the shell file.
(* The file does have #!/usr/bin/ksh)
Code:
df -k | grep "rpool/ROOT" > dftest.out
RPOOL_NAME=""; declare -i TOTAL_SIZE=0; USED_SPACE=0; AVAILABLE_SPACE=0; AVAILABLE_PERCENT=0; RSIGN=""
read RPOOL_NAME TOTAL_SIZE USED_SPACE AVAILABLE_SPACE AVAILABLE_PERCENT RSIGN < dftest.out
\rm dftest.out
echo $RPOOL_NAME $TOTAL_SIZE $USED_SPACE $AVAILABLE_SPACE $AVAILABLE_PERCENT $RSIGN
((TOTAL_SIZE=$TOTAL_SIZE/1024))
This is the result:
DBResize.sh[11]: TOTAL_SIZE=/1024: syntax error
I'm pulling hairs at this point, any help would be appreciated.
The code you posted cannot produce the output you posted. Most obviously, the error is signalled at line 11 but you posted fewer than 11 lines of code. The previous lines may matter. Always post complete code when you ask for help.
More concretely, the declare command doesn't exist in ksh, it's a bash thing. You can achieve the same result with typeset (declare is a bash equivalent to typeset, but not all options are the same). Either you're executing this script with bash, or there's another error message about declare, or you've defined some additional commands including declare which may change the behavior of this code.
None of this should have an impact on the particular problem that you're posting about, however. The variables created by read remain assigned until the end of the subshell, i.e. until the code hits a ), the end of a pipe (left-hand side of the pipe only in ksh), etc.
About the use of declare or typeset, note that you're only declaring TOTAL_SIZE as an integer. For the other variables, you're just assigning a value which happens to consist exclusively of digits. It doesn't matter for the code you posted, but it's probably not what you meant.
One thing that may be happening is that grep matches nothing, and therefore read reads an empty line. You should check for errors. Use set -e in scripts to exit at the first error. (There are cases where set -e doesn't catch errors, but it's a good start.)
Another thing that may be happening is that df is splitting its output onto multiple lines because the first column containing the filesystem name is too large. To prevent this splitting, pass the option -P.
Using a temporary file is fragile: the code may be executed in a read-only directory, another process may want to access the same file at the same time... Here a temporary file is useless. Just pipe directly into read. In ksh (unlike most other sh variants including bash), the right-hand side of a pipe runs in the main shell, so assignments to variables in the right-hand side of a pipe remain available in the following commands.
It doesn't matter in this particular script, but you can use a variable without $ in an arithmetic expression. Using $ substitutes a string which can have confusing results, e.g. a='1+2'; $((a*3)) expands to 7. Not using $ uses the numerical value (in ksh, a='1+2'; $((a*3)) expands to 9; in some sh implementations you get an error because a's value is not numeric).
#!/usr/bin/ksh
set -e
typeset -i TOTAL_SIZE=0 USED_SPACE=0 AVAILABLE_SPACE=0 AVAILABLE_PERCENT=0
df -Pk | grep "rpool/ROOT" | read RPOOL_NAME TOTAL_SIZE USED_SPACE AVAILABLE_SPACE AVAILABLE_PERCENT RSIGN
echo $RPOOL_NAME $TOTAL_SIZE $USED_SPACE $AVAILABLE_SPACE $AVAILABLE_PERCENT $RSIGN
((TOTAL_SIZE=TOTAL_SIZE/1024))
Strange...when I get rid of your "declare" line, your original code seems to work perfectly well (at least with ksh on Linux)
The code :
#!/bin/ksh
df -k | grep "/home" > dftest.out
read RPOOL_NAME TOTAL_SIZE USED_SPACE AVAILABLE_SPACE AVAILABLE_PERCENT RSIGN < dftest.out
\rm dftest.out
echo $RPOOL_NAME $TOTAL_SIZE $USED_SPACE $AVAILABLE_SPACE $AVAILABLE_PERCENT $RSIGN
((TOTAL_SIZE=$TOTAL_SIZE/1024))
print $TOTAL_SIZE
The result :
32962416 5732492 25552588 19% /home
5598
Which are the value a simple df -k is returning. The variables seem to last.
For those interested, I have figured out that it is not possible to use "read" the way I was using it.
The variable values assigned by "read" simply "do not last".
To remedy this, I have applied the less than ideal solution of using the standard "while read" format, and inside the loop, echo selected variables into a variable file.
Once said file was created, I just "loaded" it.
(pseudo code:)
LOOP START
echo "VAR_A="$VAR_A"; VAR_B="$VAR_B";" > somefile.out
LOOP END
. somefile.out

Passing a variable to a command in a script

I've been searching all over the place and since I'm taking my first steps in PERL this might be one of he dumbest questions but here it goes.
So I'm creating a script to manage my windows and later bind it to keyboard shortcuts, so I I'm trying to run a command and passing some variables:
my $command = `wmctrl -r :ACTIVE: -e 0,0,0,$monitors->{1}->{'width'}/2,$monitors->{1}->{'height'}`;
But I get an error saying I'm not passing the right parameters to the command, but if I do this, everything works great:
my $test = $monitors->{1}->{'width'}/2;
my $command = `wmctrl -r :ACTIVE: -e 0,0,0,$test,$monitors->{1}->{'height'}`;
So do I really have to do this? assign it first to a variable and then pass it, or there's a more elegant way of doing it?
The backticks operator (or the qx{}) accepts A string which is (possibly) interpolated. So accepts string and not expression like $var/2.
Thats mean than the $variables ($var->{1}->{some} too) are expanded but not the arithmetic expressions.
Therefore your 2 step variant works, but not the first.
If you want evaluate an expression inside the string you can use the next:
my $ans=42;
print "The #{[ $ans/2 ]} is only the half of answer\n";
prints
The 21 is only the half of answer
but it is not very readable, so better and elegant is what you're already doing - calculate the command argument in andvace, and to the qx{} or backticks only pass the calculated $variables.

zsh filename globbling/substitution

I am trying to create my first zsh completion script, in this case for the command netcfg.
Lame as it may sound I have stuck on the first hurdle, disclaimer, I know how to do this crudely, however I seek the "ZSH WAY" to do this.
I need to list the files in /etc/networking but only the files, not the directory component, so I do the following.
echo $(ls /etc/network.d/*(.))
/etc/network.d/ethernet-dhcp /etc/network.d/wireless-wpa-config
What I wanted was:
ethernet-dhcp wireless-wpa-config
So I try (excuse my naivity) :
echo ${(s/*\/)$(ls /etc/network.d/*(.))}
/etc/network.d/ethernet-dhcp /etc/network.d/wireless-wpa-config
It seems that this doesn't work, I'm sure there must be some clever way of doing this by splitting into an array and getting the last part but as I say, I'm complete noob at this.
Any advice gratefully received.
General note: There is no need to use ls to generate the filenames. You might as well use echo some*glob. But if you want to protect the possible embedded newline characters even that is a bad idea. The first example below globs directly into an array to protect embedded newlines. The second one uses printf to generate NUL terminated data to accomplish the same thing without using a variable.
It is easy to do if you are willing to use a variable:
typeset -a entries
entries=(/etc/network.d/*(.)) # generate the list
echo ${entries#/etc/network.d/} # strip the prefix from each one
You can also do it without a variable, but the extra stuff to isolate individual entries is a bit ugly:
# From the inside, to the outside:
# * glob the entries
# * NUL terminate them into a single string
# * split at NUL
# * strip the prefix from each one
echo ${${(0)"$(printf '%s\0' /etc/network.d/*(.))"}#/etc/network.d/}
Or, if you are going to use a subshell anyway (i.e. the command substitution in the previous example), just cd to the directory so it is not part of the glob expansion (plus, you do not have to repeat the directory name):
echo ${(0)"$(cd /etc/network.d && printf '%s\0' *(.))"}
Chris Johnsen's answer is full of useful information about zsh, however it doesn't mention the much simpler solution that works in this particular case:
echo /etc/network.d/*(:t)
This is using the t history modifier as a glob qualifier.
Thanks for your suggestions guys, having done yet more reading of ZSH and coming back to the problem a couple of days later, I think I've got a very terse solution which I would like to share for your benefit.
echo ${$(print /etc/network.d/*(.)):t}
I'm used to seeing basename(1) stripping off directory components; also, you can use echo /etc/network/* to get the file listing without running the external ls program. (Running external programs can slow down completion more than you'd like; I didn't find a zsh-builtin for basename, but that doesn't mean that there isn't one.)
Here's something I hope will help:
haig% for f in /etc/network/* ; do basename $f ; done
if-down.d
if-post-down.d
if-pre-up.d
if-up.d
interfaces