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 $var2;'04.56bash-2.01$ I am expecting 456 in the ouput instead of 4.56 Am I missing anything...? Thanks,Arijit
Do you Yahoo!?
With
I think the –p is the one causing
the problem
Tried
echo 4.56 | perl -e 'my $var1 =
* 100; print $var1;'
and the result is 456
bash-2.01$ echo 4.56 | perl -p -e 'my $var1 = ; $var2 =
$var1 * 100; print $var2;'
04.56
I only remove th
Arijit Das wrote:
> I am just wondering why is this giving a strange result. Any clues...?
>
> bash-2.01$ 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...?
Try it
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 (<>) {
.
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...?
>
Remov
