I have the following error:
ls -lt | awk 'BEGIN NR > 1 { print $2, $9 }'
Syntax Error The source line is 1.
The error context is
BEGIN >>> NR <<< > 1 { print $2, $9 }
awk: 0602-500 Quitting The source line is 1.
What I want to do is ls a directory, skip the first... (3 Replies)
Hello,
I have the following command that does 2 searches.
awk '{if ($0 ~ /STRING1/) {c++} }{if ( c == 2 ) {sub(/STRING1/,"NEWSTRING") } } { print }' FILE
How do I search up after the first search?
thanks (4 Replies)
i have a little awk script that I use looks this:
awk '{if (FNR==1){print FILENAME; print $0}else print $0}' file1...file2....fi... > bundled.
i have completely forgotten how to unbundle this. I have tried several different approaches and still can not remember how to unbundle the file bundled.... (2 Replies)
I am trying to read through a file, gather the states in that file and change it from an abbreviation to the ful text.
Can anyone provide some assistance.
Thanks!! (4 Replies)
How I can rid of the following presentation du -sk /u*/oradata/TEST/*.dbf |awk '{print total+=$1} 1.28003e+06
4.35109e+06
4.36134e+06
4.4535e+06
5.47752e+06
5.48777e+06
7.52554e+06
7.73036e+06
9.06158e+06
:confused: thank you (3 Replies)
Can anyone help with this this one liner:
nawk -v RS='' '$1=$1' InputFile
What I have in the file:
0.0013985457223116
-0.0002338180925628
0.0
0.0003709430584958
-0.0005763523138347
0.0
And the output I want:
0.0013985457223116 -0.0002338180925628 0.0
0.0003709430584958... (1 Reply)
I have a script problem that I am not able to solve due my very limited understanding of unix/awk.
This is the contents of test.sh
awk '{print $1}'
From the prompt if I enter:
./test.sh Hello World
I would expect to see "Hello" but all I get is a blank line. Only then if I enter "Hello... (2 Replies)
Use and complete the template provided. The entire template must be completed. If you don't, your post may be deleted!
1. The problem statement, all variables and given/known data:
im using ls -l | xargs | awk '{what ever files here}'
im trying to get something that looks like this... (7 Replies)
Hi Experts,
I am trying to get system output to capture inside awk , but not working:
Please advise if this is possible :
I am trying something like this but not working, the output is coming wrong:
echo "" | awk '{d=system ("date") ; print "Current date is:" , d }'
Thanks, (5 Replies)
Discussion started by: rveri
5 Replies
LEARN ABOUT DEBIAN
devel::refcount
Devel::Refcount(3pm) User Contributed Perl Documentation Devel::Refcount(3pm)NAME
"Devel::Refcount" - obtain the REFCNT value of a referent
SYNOPSIS
use Devel::Refcount qw( refcount );
my $anon = [];
print "Anon ARRAY $anon has " . refcount($anon) . " reference
";
my $otherref = $anon;
print "Anon ARRAY $anon now has " . refcount($anon) . " references
";
DESCRIPTION
This module provides a single function which obtains the reference count of the object being pointed to by the passed reference value.
FUNCTIONS
$count = refcount($ref)
Returns the reference count of the object being pointed to by $ref.
COMPARISON WITH SvREFCNT
This function differs from "Devel::Peek::SvREFCNT" in that SvREFCNT() gives the reference count of the SV object itself that it is passed,
whereas refcount() gives the count of the object being pointed to. This allows it to give the count of any referent (i.e. ARRAY, HASH,
CODE, GLOB and Regexp types) as well.
Consider the following example program:
use Devel::Peek qw( SvREFCNT );
use Devel::Refcount qw( refcount );
sub printcount
{
my $name = shift;
printf "%30s has SvREFCNT=%d, refcount=%d
",
$name, SvREFCNT($_[0]), refcount($_[0]);
}
my $var = [];
printcount 'Initially, $var', $var;
my $othervar = $var;
printcount 'Before CODE ref, $var', $var;
printcount '$othervar', $othervar;
my $code = sub { undef $var };
printcount 'After CODE ref, $var', $var;
printcount '$othervar', $othervar;
This produces the output
Initially, $var has SvREFCNT=1, refcount=1
Before CODE ref, $var has SvREFCNT=1, refcount=2
$othervar has SvREFCNT=1, refcount=2
After CODE ref, $var has SvREFCNT=2, refcount=2
$othervar has SvREFCNT=1, refcount=2
Here, we see that SvREFCNT() counts the number of references to the SV object passed in as the scalar value - the $var or $othervar
respectively, whereas refcount() counts the number of reference values that point to the referent object - the anonymous ARRAY in this
case.
Before the CODE reference is constructed, both $var and $othervar have SvREFCNT() of 1, as they exist only in the current lexical pad. The
anonymous ARRAY has a refcount() of 2, because both $var and $othervar store a reference to it.
After the CODE reference is constructed, the $var variable now has an SvREFCNT() of 2, because it also appears in the lexical pad for the
new anonymous CODE block.
PURE-PERL FALLBACK
An XS implementation of this function is provided, and is used by default. If the XS library cannot be loaded, a fallback implementation in
pure perl using the "B" module is used instead. This will behave identically, but is much slower.
Rate pp xs
pp 225985/s -- -66%
xs 669570/s 196% --
SEE ALSO
o Test::Refcount - assert reference counts on objects
AUTHOR
Paul Evans <leonerd@leonerd.org.uk>
perl v5.14.2 2011-11-15 Devel::Refcount(3pm)