Question

I have a pointer to an array and either need to change the code considerably to make this work, or just figure out what the equivilent of the statement below is…

ptr = array;
*ptr++ = value;

So far I have most of it

$ptr = \@array;
$$ptr = $value;

but this doesn’t increment the pointer. What do I do?

Answer 1

There are no pointers in Perl. What you have is:

my $ary_ref = \@array;

$ary_ref is now a reference to @array. You cannot dereference $ary_ref to get a scalar.

You can, however, iterate through the elements of @array in a variety of ways. For example, if you want, you can do:

#!/usr/bin/env perl

use strict; use warnings;

my @array;

my $ptr = sub {
    my ($i, $ref, $val, $size) = (0, @_);
    sub { $i < $size ? $ref->[$i ++ ] = $val : () }
}->(\@array, deadbeaf => 10);

1 while defined $ptr->();

use YAML;
print Dump \@array;

By the way, there is no reason you can’t write

p[i] = value;
i++;

in C. In Perl, it might become:

$array[$_] = $value for 0 .. $#array;

or,

@array = ($value) x @array;

or,

$_ = value for @array;

etc. You should explain what the overall goal is rather than asking about a specific statement in C.