circular boolean constants

More humorous than truly obfuscated. Credit to Andrew Pimlott on fwp@perl.org.

Is is obvious to you how this works?

#!/usr/bin/perl -wl
use constant FALSE, !TRUE;
use constant TRUE, !FALSE;
use strict;

print "false is \"", FALSE, "\" in string context";
print "false is ", 0+FALSE, " in numeric context";
print "true is ", TRUE;
__END__
output:
false is "" in string context
false is 0 in numeric context
true is 1
[download]

Update: the print statements are only intended to demonstrate that the constants created act like perl's built in boolean return values (where false is dual-valued as 0 and ""). There's nothing complex about them.

Comment on circular boolean constants Download Code

Replies are listed 'Best First'.
Re: circular boolean constants by Coruscate (Sexton) on Jan 22, 2004 at 05:27 UTC
Your putting the strict after the constants gives away a lot :). Put simply, the line `use constant FALSE, !TRUE` (where TRUE is a bareword, because it is not yet defined) gives FALSE the negation of string 'TRUE'. Then, we go and declare constant TRUE, and since FALSE is now defined, FALSE is not a bareword in line 3. So we assign TRUE the negated value of FALSE. FALSE is 0/"", so TRUE becomes 1. Regarding the print statements then. In the first one, you are doing the equivalent of replacing `FALSE` with `!'TRUE'`, which in scalar context is an empty string. Print statement number two, you are simply taking !'TRUE' and making it numerical, which is 0. The last one is simply because TRUE == 1 because of above statement of how TRUE is defined. (Guesses verified with a quick B::Deparse output check. I did have one thing wrong that I updated. I figured that FALSE would be declared as 0, but is it actually hardcoded as !'TRUE' (unless Deparse just doesn't go far enough into the works and FALSE is actually empty string.)	[reply] [d/l] [select]

Replies are listed 'Best First'.

Re: circular boolean constants
by Coruscate (Sexton) on Jan 22, 2004 at 05:27 UTC

Your putting the strict after the constants gives away a lot :). Put simply, the line use constant FALSE, !TRUE (where TRUE is a bareword, because it is not yet defined) gives FALSE the negation of string 'TRUE'. Then, we go and declare constant TRUE, and since FALSE is now defined, FALSE is not a bareword in line 3. So we assign TRUE the negated value of FALSE. FALSE is 0/"", so TRUE becomes 1.

Regarding the print statements then. In the first one, you are doing the equivalent of replacing FALSE with !'TRUE', which in scalar context is an empty string. Print statement number two, you are simply taking !'TRUE' and making it numerical, which is 0. The last one is simply because TRUE == 1 because of above statement of how TRUE is defined.

(Guesses verified with a quick B::Deparse output check. I did have one thing wrong that I updated. I figured that FALSE would be declared as 0, but is it actually hardcoded as !'TRUE' (unless Deparse just doesn't go far enough into the works and FALSE is actually empty string.)

[reply]
[d/l]
[select]