Here's the subroutine I came up with. Thanks for everyone who responded. I could never get the flattening the array method to work. It is not very elegant and probably slow, but this is the only solution I could find that prints the right answer:
equal non equal equal
#!/usr/bin/perl $HASH_SIZE = 100; %hash1; %hash2; for ($i=0; $i < $HASH_SIZE; $i++) { $hash1{"i=$i"} = "j=$i"; } %hash2 = %hash1; if (&hasheq(\%hash1, \%hash2)) { print "\nequal\n"; } else { print "\nnon equal\n"; } $hash2{"anotherone"} = "anotherONE"; if (&hasheq(\%hash1, \%hash2)) { print "\nequal\n"; } else { print "\nnon equal\n"; } $hash1{"anotherone"} = "anotherONE"; if (&hasheq(\%hash1, \%hash2)) { print "\nequal\n"; } else { print "\nnon equal\n"; } sub hasheq { my ($ha1, $ha2) = @_; my %h1 = %$ha1; my %h2 = %$ha2; my @k1 = keys(%h1); my @k2 = keys(%h2); # do they have the same number of elements? if (@k1 != @k2) { return 0; } # are the keys the same? if ((join '/' , sort @k1 ) ne (join '/' , sort @k2)) { return 0; } # are the values the same? if ( scalar grep { $h1{$_} ne $h2{$_} } @k1 ) { return 0; } return 1; }

In reply to Re: Re: How to test equality of hashes? by acser
in thread How to test equality of hashes? by acser

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