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Home/ Questions/Q 7523447
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T02:48:21+00:00 2026-05-30T02:48:21+00:00

#!/usr/bin/perl A(); B(); sub A { my @array = qw(value_1 value_2); $array_ref = \@array;

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#!/usr/bin/perl

    A();
    B();

    sub A {
        my @array = qw(value_1 value_2);
        $array_ref = \@array;
    }

    sub B {
        foreach my $i ( @{$array_ref} ) {
            print "Array Value: $i \n";
        }
    }

Since the array is declared using the ‘my’ keyword, could the array reference be lost ?
Can any one brief me over this.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T02:48:22+00:00Added an answer on May 30, 2026 at 2:48 am

    No, the scope of the variable expires, but not the memory address. The data will remain.

    Isn’t this something you could have simply tried? =) I just copy/pasted your code and tried it, and it worked fine.

    For proper encapsulation, though, you really should return the array ref instead:

    B(A());
    # Or
    my $aref = A();
    B($aref);
    
    sub A {
            my @array = qw(value_1 value_2);
            return \@array;
    }
    
    sub B {
        my $array_ref = shift;
        foreach my $i ( @$array_ref ) {
            print "Array Value: $i \n";
        }
    }
    
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