'xor' operator is not a sibling to 'or' and 'and'?

rsFalse has asked for the wisdom of the Perl Monks concerning the following question:

Hello,

I have used 'xor' operator as lower precedence comma, because I understood a logic that it cannot short-circuit as do 'or' and 'and', and the last expression must be evaluated at the end. E.g. I used 'xor' in short sentences like this, to avoid unnecessary parentheses:

#!/usr/bin/perl -l

use strict;
use warnings;

push @_, $_ xor print $_ for 'a' .. 'b';
print @_;
[download]

Output:

Useless use of logical xor in void context at ./perlmonks_xor_vs_or_an
+d.pl line 6.
a
b
ab
[download]

Today I found (for myself) that 'xor' doesn't work in the same way as 'or' and 'and' do, and it is strange for me, why. It seems that, 'xor' not only evaluates one or both expressions surrounding it, but also evaluates logical XOR of these expressions, when 'or' and 'and' operators don't do this, i.e.:

#!/usr/bin/perl -l

use strict;
use warnings;

print "[$_]" for 
    ( 2 and 3 ),
    ( 2 or 3 ),
    ( 2 xor 3 ),
    ;
[download]

Output:

[3]
[2]
[]
[download]

At the last line of output I expected to get '3'!
I could get '3' by using a comma instead (and forcing scalar context): 'scalar( 2, 3 ),'.
I'm not sure this difference of behaviour of operators is written somewhere in documentation, but I haven't find it in 'perlop'.

Comment on 'xor' operator is not a sibling to 'or' and 'and'? Select or Download Code

Replies are listed 'Best First'.
Re: 'xor' operator is not a sibling to 'or' and 'and'? by GrandFather (Saint) on Dec 18, 2019 at 21:29 UTC
xor can only return true (1) or false ('' - empty string) and must evaluate both the lhs and the rhs expressions to determine the result. `true xor true` is `false` so you got back an empty string. `and` and `or` short circuit evaluate and return the last expression evaluated. `use strict; use warnings; print "$_->[0]: [$_->[1]]\n" for map {[$_, eval]} '0 and 1', '2 and 0', '3 and 4', '0 or 5', '6 or 0', '7 or 8', '0 xor 9', '10 xor 0', '11 xor 12', '0 xor 0', ;` [download] Optimising for fewest key strokes only makes sense transmitting to Pluto or beyond	[reply] [d/l] [select]
Re: 'xor' operator is not a sibling to 'or' and 'and'? by ikegami (Patriarch) on Dec 18, 2019 at 20:33 UTC
It seems that, 'xor' not only evaluates one or both expressions surrounding it, but also evaluates logical XOR of these expressions, when 'or' and 'and' operators don't do this huh? That's exactly what `or` and `and` do (aside from evaluating logical OR and AND respectively, of course). At the last line of output I expected to get '3'! TRUE XOR TRUE = FALSE, so `2 xor 3` can't possibly return `3`, a true value.	[reply] [d/l] [select]
Re^2: 'xor' operator is not a sibling to 'or' and 'and'? by rsFalse (Chaplain) on Dec 18, 2019 at 20:43 UTC
>> TRUE XOR TRUE = FALSE, so 2 xor 3 can't possibly return 3. But in the same way 'and' operator does not output '1', despite TRUE AND TRUE = TRUE.	[reply]
Re^3: 'xor' operator is not a sibling to 'or' and 'and'? by Fletch (Bishop) on Dec 18, 2019 at 21:04 UTC
`and` and `or` return the last expression evaluated; `xor` always evaluates both sides and then returns the exclusive or of those two expressions (not either of the surrounding expressions) so it's always going to be a boolean value either 1 or `''/0`. See perlop Update: On further reflection I think I see what your expectations for the behavior might be based on `and/or`'s behavior: if `A xor B` is true return whichever of A or B was true, otherwise return false. The cake is a lie. The cake is a lie. The cake is a lie.	[reply] [d/l] [select]
Re^4: 'xor' operator is not a sibling to 'or' and 'and'? by rsFalse (Chaplain) on Dec 18, 2019 at 21:11 UTC
Re^5: 'xor' operator is not a sibling to 'or' and 'and'? by Fletch (Bishop) on Dec 18, 2019 at 21:23 UTC
Some notes below your chosen depth have not been shown here
Re^3: 'xor' operator is not a sibling to 'or' and 'and'? by LanX (Saint) on Dec 19, 2019 at 05:24 UTC
> But in the same way `and` operator does not output '1', despite TRUE AND TRUE = TRUE. That's the feature of short circuit `and` and `or` which can't be replicated with `xor` Grandfather and ikegami already explained it perfectly. Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery FootballPerl is like chess, only without the dice}	[reply] [d/l]
Re^4: 'xor' operator is not a sibling to 'or' and 'and'? by rsFalse (Chaplain) on Dec 19, 2019 at 07:56 UTC
Re^5: 'xor' operator is not a sibling to 'or' and 'and'? by soonix (Canon) on Dec 19, 2019 at 09:36 UTC
Re^5: 'xor' operator is not a sibling to 'or' and 'and'? by LanX (Saint) on Dec 19, 2019 at 14:41 UTC
Some notes below your chosen depth have not been shown here
Re^3: 'xor' operator is not a sibling to 'or' and 'and'? by ikegami (Patriarch) on Dec 19, 2019 at 20:17 UTC
No, that's completely wrong. The reason `$TRUE and $TRUE` doesn't necessarily return `1` has nothing to do with why `$TRUE xor $TRUE` doesn't return `1`. `$TRUE and $TRUE` doesn't necessarily return `1` because it's more useful to return its RHS. `$TRUE xor $TRUE` doesn't return `1` because that would be wrong.	[reply] [d/l] [select]
Re: 'xor' operator is not a sibling to 'or' and 'and'? (short-circuit and other languages.) by LanX (Saint) on Dec 19, 2019 at 17:53 UTC
This WP article: Short-circuit evaluation covers most if not all aspects of this thread and lists languages with similar behavior. (See table entries marked "last value", like Perl, Ruby, JS, Python,...) Quotes: In loosely typed languages that have more than the two truth-values True and False, short-circuit operators may return the last evaluated subexpression. The expression `x and y` is equivalent to `y if x else x;` the expression `x or y` is equivalent to `x if x else y` (without evaluating x twice). This is called "last value" in the table below. For some Boolean operations, like exclusive or (XOR), it is not possible to short-circuit, because both operands are always required to determine the result. see also Re^5: 'xor' operator is not a sibling to 'or' and 'and'? Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery FootballPerl is like chess, only without the dice}	[reply] [d/l] [select]
Re: 'xor' operator is not a sibling to 'or' and 'and'? by BillKSmith (Monsignor) on Dec 18, 2019 at 22:55 UTC
These operators are called 'Logical' for a reason. They are meant to return a logical value (true or false). Code that depends on the string or numeric value of the result, which arguments are evaluated, or even in the order in which they are evaluated could fail under a future version of perl. Note that the idiom `open ... or die "...";` does depend on the open being done first, but is so common I am sure it will always be supported. Bill	[reply] [d/l]
Re^2: 'xor' operator is not a sibling to 'or' and 'and'? by GrandFather (Saint) on Dec 18, 2019 at 23:29 UTC
I can't imagine the expression result behavior of Perl's logical operators ever going away. There is a large amount of Perl idiom that bakes in that behavior so huge amounts of code would break if that changed. I'm sure the behavior isn't an accidental result of a weird coding decision, but was fully thought out with a view to allowing nice fall back processing using the logical or operators. Optimising for fewest key strokes only makes sense transmitting to Pluto or beyond	[reply]
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