simple scalars question

mitchreward has asked for the wisdom of the Perl Monks concerning the following question:

hi guys,

Why is the following code print the \n first before the scalars ?

I wanted it like this, print $toto, => new line => print $tutu ...

#!/usr/bin/perl -w

use strict;
use warnings;

my $toto = 'toto';
my $tutu = 'tutu';

my $deux = $toto, print "\n", $tutu;

print $deux
[download]

Comment on simple scalars question Download Code

Replies are listed 'Best First'.
Re: simple scalars question by AppleFritter (Vicar) on Jul 18, 2014 at 10:03 UTC
Hello again, Mitch! To expound on my anonymous brother's answer, what you're doing here: `my $deux = $toto, print "\n", $tutu;` [download] is assigning `$toto` to `$deux`, and then executing `print "\n", $tutu`. In order to embed a newline in `$deux`, you could do either of the following: `$deux = "$toto\n$tutu"; # variable interpolation in double-quot +ed strings $deux = $toto . "\n" . $tutu; # string concatenation` [download] Of course, if you merely care about printing, the following will also work: `print $toto, "\n", $tutu;` [download] This works because `print` is a list operator that can take an arbitrary number of arguments -- but you can't do the same when assigning to a scalar, you have to use either interpolation or string concatenation (the `.` operator).	[reply] [d/l] [select]
Re: simple scalars question by Anonymous Monk on Jul 18, 2014 at 09:35 UTC
Because you wrote `my $deux = $toto, print "\n", $tutu;` [download] and you did not write `print $toto, "\n", $tutu;` [download]	[reply] [d/l] [select]
Re^2: simple scalars question by mitchreward (Acolyte) on Jul 18, 2014 at 09:41 UTC
ok, that's it. thank you	[reply]