PHP order of logical operators - php

Question

I'm curious about how PHP handles conditional statements / order of operations with nesting. If I use the following if condition:
if(x == (1 || 2))
{
// do something
}
I would expect it to behave the same as
if(x == 1 || x == 2)
{
// do something
}
...but it doesn't. My first example seems like it would be a handy shorthand that makes pretty good sense, but it doesn't do what I expect. Can anyone shed some light on the issue? What exactly does PHP do with my first statement?

Answer 1

So for this piece of code:
if ($x == ( 1 || 2))
{
// do something
}
In PHP, any non-zero number is considered true. Disclaimer: This fact isn't necessarily true in other languages. So in PHP, 0 is the only number considered false. So you're asking if $x == true in the above piece of code.
Hence, whenever $x is any number other than 0 the statement inside the if will resolve as true. However, when $x = 0 then that is equivalent to saying false == true which of course will resolve as false.
This article might help: PHP: Booleans

Answer 2

Your shorthand is logically invalid. In almost every case you'll have to write out the full logical cases for all possibilities you want to test for.
I say 'almost' because in PHP you can do something ridiculous like:
if( in_array($x, array(1,2)) ) {
// code!
}

Answer 3

x == (1 || 2)
evaluates like this:
(1 (if its false) then testing for 2, if not, the expression returns true)
now it will become:
if(x==true)?
Another example taken from (PHP.NET):
// foo() will never get called as those operators are short-circuit
$b = (true || foo());

Answer 4

See here about the precedence of an operator
http://php.net/manual/en/language.operators.precedence.php

Answer 5

It will behave the same as math (think BEDMAS), with the brackets being executed first. So your example is behaving as:
if (x == ( 1 || 2)) {
//code
}
and because 1 and 2 are both non-zero values (thus both true), you get:
if (x == true) {
//code
}
Unfortunately to get what you want you'll need:
if (x == 1 || x == 2) {
//code
}

Answer 6

how would it make sense? you are asking the computer to execute the following logical expression:
if x == (1 || 2) which is same as x == (the result of 1 || 2)
so your expression would be x == true since 1 || 2 would return true
computers do whatever you tell them to do

Answer 7

if(x == (1 || 2))
{
// do something
}
OR and AND operations come back with TRUE or FALSE
you statement says - if x equals (true) - as 1 or 2 will always be true
- this just doesn't make sense...

Answer 8

In first if evaluate result of 1||2 and check that it equal to x
In second its like this or this or this
however in first you can see that var_dump(1 || 2) returns every time true so
$x = 3;
var_dump($x == 1 || 2);
if($x == 1 || 2){
echo 'inside if';
}
is also true so it will print inside if even $x is 3
so imo second way is way to go