Strange behaviour in Perl

RE: Strange behaviour in Perl

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The -p switch is causing Perl to iterate over the inputs as if they were
filename arguments.



Try this instead...



echo 4.56 | perl -e '$var1 =; $var2 = $var1 * 100; print "$var2\n"'



cn



_____

From: activeperl-bounces@listserv.ActiveState.com
[mailto:activeperl-bounces@listserv.ActiveState.com] On Behalf Of Arijit Das
Sent: Wednesday, January 25, 2006 5:47 AM
To: activeperl@listserv.activestate.com;
perl-unix-users@listserv.activestate.com
Subject: Strange behaviour in Perl



I am just wondering why is this giving a strange result. Any clues...?



bash-2.01$ echo 4.56 | perl -p -e 'my $var1 = ; $var2 = $var1 * 100;
print $var2;'
04.56
bash-2.01$



I am expecting 456 in the ouput instead of 4.56



Am I missing anything...?



Thanks,
Arijit



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Re: Strange behaviour in Perl

Arijit Das wrote:
> I am just wondering why is this giving a strange result. Any clues...?
>
> bash-2.01$ echo 4.56 | perl -p -e 'my $var1 = ; $var2 =
$var1 * 100; print $var2;'
> 04.56
> bash-2.01$
>
> I am expecting 456 in the ouput instead of 4.56
>
> Am I missing anything...?
>

Remove -p.

dZ.
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Re: Strange behaviour in Perl

Arijit Das wrote:

>I am just wondering why is this giving a strange result. Any clues...?
>
> bash-2.01$ echo 4.56 | perl -p -e 'my $var1 = ; $var2 = $var1 * 100; print $var2;'
>04.56
>bash-2.01$
>
> I am expecting 456 in the ouput instead of 4.56
>
> Am I missing anything...?
>
> Thanks,
>Arijit
>
>
>
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> With a free 1 GB, there's more in store with Yahoo! Mail.
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Works fine without the -p.

Tony

--

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Re: [Perl-unix-users] Strange behaviour in Perl

Arijit Das wrote:
> I am just wondering why is this giving a strange result. Any clues...?
>

This is from the Perl In A Nutshell book:
------
-p: Causes Perl to assume the following loop around your script, which
makes it iterate over filename arguments:

LINE:
while (<>) {
... # your script goes here
} continue {
print;
}

The lines are printed automatically. To suppress printing, use the -n
switch. If both are specified, the -p switch overrides -n. BEGIN and END
blocks may be used to capture control before or after the implicit loop.
------

So your one-liner is actually:

LINE:
while (<>) {
my $var1 = ;
$var2 = $var1 * 100;
print $var2;
} continue {
print;
}

It's already looping through the arguments with while(<>), so reading
will yield nothing. Also, -p implies printing each argument
read, so that's the output that you are getting, not $var2 as you may think.

You have 2 choices, either remove -p to read STDIN as you are doing, or
change -p to -n to supress automatic printing of the input strings and
remove the assignment of STDIN and read $_, which would be the next
argument read from the command line:

echo 4.56 | perl -e 'my $var1=; $var2=$var1*100;print $var2;'

or

echo 4.56 | perl -n -e '$var2=$_*100;print $var2;'

If you keep -p, it'll print the input string right after printing $var2,
like this:

4564.56

Which you do not want.

dZ.
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RE: Strange behaviour in Perl

From: activeperl-bounces@listserv.ActiveState.com
[mailto:activeperl-bounces@listserv.ActiveState.com] On Behalf Of Arijit
Das
Sent: 25 January 2006 11:47
To: activeperl@listserv.activestate.com;
perl-unix-users@listserv.activestate.com
Subject: Strange behaviour in Perl


> I am just wondering why is this giving a strange result. Any clues...?
>
> bash-2.01$ echo 4.56 | perl -p -e 'my $var1 = ; $var2 = $var1 *
100; print $var2;'
> 04.56
> bash-2.01$
>
> I am expecting 456 in the ouput instead of 4.56
>
> Am I missing anything...?

Possibly. Either remove the -p switch, or change your command to echo
4.56 | perl -p -e '$_ *= 100'.

Have you checked 'perldoc perlrun' for what -p does? That may give you a
clue as to what is happening.

HTH

--
Brian Raven



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RE: Strange behaviour in Perl

Arijit Das wrote:
> I am just wondering why is this giving a strange result. Any clues...?
>
> $ echo 4.56 | perl -p -e 'my $var1 = ; $var2 = $var1 * 100; print
$var2;'
> 04.56
> $
>
> I am expecting 456 in the ouput instead of 4.56
>
> Am I missing anything...?

The '-p' option is causing the problem. It is adding a loop into the
program. It looks something like this:

while (<>) {

my $var1 = ;
$var2 = $var1 * 100;
print $var2;

print $_;
}

So the while loop picks up your input into $_ leaving nothing for you to
read into $var1. Since $var1 is empty, $var2 becomes 0. So you
effectively wind up with this:

print 0;
print 4.56;

which outputs:

04.56

Remove the '-p' and your program will work as expected.

Or use the '-p' and do it like this:

echo 4.56 | perl -p -e '$_ *= 100';

--
Bowie
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Re: [Perl-unix-users] Strange behaviour in Perl

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Thanks a lot folks.

My mistake...I should have looked at "-p"'s significance before posting it.

Thanks again,
Arijit

DZ-Jay wrote:
Arijit Das wrote:
> I am just wondering why is this giving a strange result. Any clues...?
>

This is from the Perl In A Nutshell book:
------
-p: Causes Perl to assume the following loop around your script, which
makes it iterate over filename arguments:

LINE:
while (<>) {
.... # your script goes here
} continue {
print;
}

The lines are printed automatically. To suppress printing, use the -n
switch. If both are specified, the -p switch overrides -n. BEGIN and END
blocks may be used to capture control before or after the implicit loop.
------

So your one-liner is actually:

LINE:
while (<>) {
my $var1 = ;
$var2 = $var1 * 100;
print $var2;
} continue {
print;
}

It's already looping through the arguments with while(<>), so reading
will yield nothing. Also, -p implies printing each argument
read, so that's the output that you are getting, not $var2 as you may think.

You have 2 choices, either remove -p to read STDIN as you are doing, or
change -p to -n to supress automatic printing of the input strings and
remove the assignment of STDIN and read $_, which would be the next
argument read from the command line:

echo 4.56 | perl -e 'my $var1=; $var2=$var1*100;print $var2;'

or

echo 4.56 | perl -n -e '$var2=$_*100;print $var2;'

If you keep -p, it'll print the input string right after printing $var2,
like this:

4564.56

Which you do not want.

dZ.




---------------------------------

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Arijit Das wrote:
> I am just wondering why is this giving a strange result. Any clues...?
>

This is from the Perl In A Nutshell book:
------
-p: Causes Perl to assume the following loop around your script, which
makes it iterate over filename arguments:

LINE:
while (<>) {
... # your script goes here
} continue {
print;
}

The lines are printed automatically. To suppress printing, use the -n
switch. If both are specified, the -p switch overrides -n. BEGIN and END >blocks may be used to capture control before or after the implicit loop.
------
R>So your
one-liner is actually:

LINE:
while (<>) {
my $var1 = ;
$var2 = $var1 * 100;
print $var2;
} continue {
print;
}

It's already looping through the arguments with while(<>), so reading
will yield nothing. Also, -p implies printing each argument
read, so that's the output that you are getting, not $var2 as you may think.

You have 2 choices, either remove -p to read STDIN as you are doing, or
change -p to -n to supress automatic printing of the input strings and
remove the assignment of STDIN and read $_, which would be the next
argument read from the command line:

echo 4.56 | perl -e 'my $var1=; $var2=$var1*100;print $var2;'

or

echo 4.56 | perl -n -e '$var2=$_*100;print $var2;'

If you keep -p, it'll print the input string right after printing $var2,
like this:
>4564.56

Which you do not want.

dZ.