simple scalars question

[mitchreward]

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

[AppleFritter]

Hello again, Mitch! To expound on my anonymous brother's answer, what you're doing here:

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

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

Of course, if you merely care about printing, the following will also work:

print $toto, "\n", $tutu;

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).

[Anonymous Monk]

Because you wrote

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

and you did not write

print $toto, "\n", $tutu;

[mitchreward]

ok, that's it. thank you