How to deference a reference to a scalar in an array

Question

I have following piece of code

$var1 = 10;

@arr = (1, \$var1, 3);

print "var1= $$arr[1] \n";

This is not printing value 10, is this syntax $$arr[1] correct? using additional variable i was able to print the value

$r = $var[1];

print "var1 = $$r\n";

Need `print "var1= ${$arr[1]}\n";` I suggest to read over [perlreftut](https://perldoc.perl.org/perlreftut.html) — zdim, Jan 03 '20 at 09:03
`$$arr[1]`, short for `${ $arr }[1]`, and equivalent to `$arr->[1]`, is for when you have a reference to an array, and you want the second element of the referenced array. You have a reference to a scalar. — ikegami, Jan 03 '20 at 09:17

ikegami · Answer 1 · 2020-01-03T09:19:32.667

5

If it's $NAME if you have the name, it's $BLOCK if you have a reference. So,

${ $arr[1] }

or (5.24+)

$arr[1]->$*

or (5.20+)

use experimental qw( postderef );

$arr[1]->$*

References:

edited Jan 03 '20 at 09:19

answered Jan 03 '20 at 09:12

ikegami

score 1 · Answer 2 · answered Jan 03 '20 at 09:03

1

Try this

print "var1 = ${ $var[1] }\n" ;

answered Jan 03 '20 at 09:03

pmqs

2 Answers2