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You'll have to extend that a bit so it properly deals with undef vs empty string.

sub ident { my $value = ( values %{ $_[0] } )[0]; for ( values %{ $_[0] } ) { no warnings 'uninitialized'; return 0 unless !( defined $value xor defined $_ ) and ( $valu +e eq $_ ); } return 1; }

It would be better if I could think of a way to skip the $value eq $_ test when both variables are undefined, and still have the entire expression be true, but I can't see a way to do that without an extra test for definedness on one of the values.

Update after bageler's reply: this was wrong:

return 0 unless ( defined $value xor defined $_ ) and ( $value eq $_ ) +;

The condition will only ever be true when comparing an undef with an empty string, but in no other case, because the left expression is only true if the operands are unequal, ie if only one of them is defined. In that case, the right expression can only also be true if the defined value is an empty string.

Makeshifts last the longest.

In reply to Re^3: scanning hash by Aristotle
in thread scanning hash by hotshot

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