If the inner hashes always just have those 3 keys, there's no need for an inner loop:

my $same = 0;
if (@AoH_curr == @AoH_prev) {
    $same = 1;
    for my $i (0..$#AoH_curr) }
        my $a = $AoH_curr[$i];
        my $b = $AoH_prev[$i];
        if ($a->{node} ne $b->{node}
             || $a->{link} ne $b->{link}
             || $a->{link} ne $b->{link}) {
             $same = 0;
             last;
        }
    }
}

[download]

(Untested).

Perl 6 - the future is here, just unevenly distributed

In reply to Re: Comparing elements in Array of Hashes (AoH) by moritz
in thread Comparing elements in Array of Hashes (AoH) by hmadhi

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