Make sure exactly one boolean of a given list is true?

Question

If I have the following booleans

const YESTERDAY = false;
const TODAY = true;
const TOMORROW = false;

What code can I write to make sure exactly one is true?

I've tried this:

$x = self::YESTERDAY ^ self::TODAY ^ self::TOMORROW;

The problem is that with all three constants set to true then $x is true.

Virago

you could put all the combination using OR condition (YESTERDAY=true AND TODAY=false AND TOMORROW=false) or (YESTERDAY=false AND TODAY=true AND TOMORROW=false) or (YESTERDAY=false AND TODAY=false AND TOMORROW=true) – Duky

He need to sure that true is excalty ONE, not AT LEAST ONE. – Sorry

Answer 1

$x = ((int) self::YESTERDAY) + ((int) self::TODAY) + ((int) self::TOMORROW); Then if $x === 1; You've got what you need.

EDITED:

Even without type casts (int), it works well, thanks to @DaveRandom, so:

if (self::YESTERDAY + self::TODAY + self::TOMORROW == 1) {}, as for me.

Sorry

I thought about this but it'd be nice to avoid all that extra processing, is there a bitwise way to do this? – Virago

+1 this - it seems after testing that you don't even need the (int) casts, it happens implicitly. – Combination

This is much tidier than any of the other solutions. @Virago was almost correct in using the XOR (^) operator, but unfortunately true ^ true ^ true will evaluate to true, because PHP interprets it as (true ^ true) ^ true, i.e. false ^ true – Tuberosity

This an amazing solution. Thanks :D – Variable

Answer 2

The neatest way I can think of is array_sum():

if (array_sum(array(self::YESTERDAY, self::TODAY, self::TOMORROW)) == 1) {
  // Do something
}

EDIT Actually, all you need to do it replace the ^ with + in your original attempt, and it achieves the same thing:

$x = self::YESTERDAY + self::TODAY + self::TOMORROW;

This turns $x into the number of TRUE values. So for a boolean output use:

$ok = self::YESTERDAY + self::TODAY + self::TOMORROW === 1;

Combination

Actually a variation on devdRew's answer is the neatest - if you just replace the ^ with + in your original code it does the same thing - you don't need the (int) casts as this happens implicitly. – Combination

Answer 3

Just as an alternative to devdRew's answer

$x = array_count_values(array((int) self::YESTERDAY,(int) self::TODAY,(int) self::TOMORROW));
if (isset($x[1]) && $x[1] == 1) {
    echo 'Only one TRUE';
}

Voss

Answer 4

You can loop over a list of your variables and break when you found a second boolean that is true:

$moreThanOneTrue=false;
$oneTrue;
foreach ($BOOL_VAR_ARRAY as $bool) {
    if ($bool) {
        if($oneTrue) {
            $moreThanOneTrue=true;
            break;
        }
        $oneTrue=true;
    }
}

Like this it's more handy when you have more than three variables.

Real

If there's more than 3 I think the array_sum method would be simpler than this – Virago

Answer 5

x will return true if and only if one is true and others are false.

$x = ($a && !($b || $c)) || ($b && !($a || $c)) || ($c && !($a || $b));

May be a bad code, but works.

Tenorio

Answer 6

a ^ b ^ c ^ (a & b & c) is the expresion you look for.

Gwendolyngweneth