PHP check | (OR) operator in variable assignment

Question

I want to check two variables a and b and assign both the variable to new variable "c" and want to get the OR result from both the variables. for example if a=1 and b=0, c must be 1, while if a=0 and b=1, c must be 1, if a=0 and b=0 then c=0, for this purpose i am using the following | operator, which returns the required result, but i am not sure if i am doing it correct or not

<?php 
$a = 0; 
$b = 1;     
$c = $a | $b;
echo("Value in $c = ".$c); 
?>

EDIT: i have gone through the PHP.NET website and find that:

 $a | $b    Or (inclusive or)   Bits that are set in either $a or $b are set.

Reference: http://php.net/manual/en/language.operators.bitwise.php

| is bitwise or. you might be looking for || (which is a logical or) — Dekel
– Dekel, Commented Dec 29, 2016 at 16:35
but i have just checked the || operator which is not return the required result? — Abdul Rahman
– Abdul Rahman, Commented Dec 29, 2016 at 16:37
Make sure you understand the difference between bitwise (|, &, ...) and logical (&&, ||,...) operators. — syck
– syck, Commented Dec 29, 2016 at 16:39
you are doing it right if you mean bitwise OR. except you don't need the last ?> — bansi
– bansi, Commented Dec 29, 2016 at 16:40
In your scenario you really need the bitwise or | and you are doing it right. — bansi
– bansi, Commented Dec 29, 2016 at 16:47

Machavity · Accepted Answer · 2016-12-29 17:03:23Z

2

Assuming you only have the states you have in your question, you can use a ternary to do this. It might help others understand what you're doing in the future

$c = ($a || $b) ? 1 : 0;

There's nothing wrong with the way you did it in your question, tho.

edited Dec 29, 2016 at 17:03

answered Dec 29, 2016 at 16:43

Machavity♦

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4 Comments

Abdul Rahman Over a year ago

what if $a=1 and $b = 1, then $c will be "0" right?

Abdul Rahman Over a year ago

while i need to get $c = 1 if either is "1" in both the variable "a" and "b"

Abdul Rahman Over a year ago

@Machavity the code you wrote returns the same result like my code? isn't?

Machavity Over a year ago

@AbdulRahman There's nothing wrong with your way. Mine is just a different way that might be easier to understand. Same result

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