some_function($input) returns the integer 23
$other_int has a value that is not equal to 23
Code snippet 1
if ($int = some_function($input) && $int != $other_int) {...}
Code snippet 2
$int = some_function($input);
if ($int != $other_int) {...}
I coded snippet 1 thinking that the if statement would return true given the above conditions. I was wrong.
The if statement in snippet 2 is true.
Why?
So, assignments in an if clause like that is a bad practice for a reason. That said, you're not assigning what you think you are. This is what you're actually doing:
$int = (some_function($input) && $int != $other_int);
which evaluates to true. To get the result you want, change the code to
if (($int = some_function($input)) && ($int != $other_int)) {...}
But really, don't assign variables in an if() statement.
As others have mentioned, && has higher precedence than assignment, so you're performing the logical operation before you assign to the variable.
Instead of && you can use and. and and or have lower precedence than assignment, and are intended precisely for situations like this.
if ($int = some_function($input) and $int != $other_int)
The && and the != have higher precedence than the assignment operator =.
See http://www.php.net/manual/en/language.operators.precedence.php
So inside the conditional:
if ($int = some_function($input) && $int != $other_int) {...}
It executes the != then the &&, but $int hasn't been initialized yet (assuming you haven't initialized it outside that conditional) so the right side != always returns false.
You can't assign a value like that inside the if statement AND compare that value in the same statement
<?php
function somefunc() {
return 23;
}
if($int = somefunc() && $int == 23) {
echo 'Equal';
} else {
echo 'Not Equal';
}
echo "\n";
var_dump($int);
You'll note the error
Notice: Undefined variable: int in /in/GHVKs on line 7
And at the end $int is false. Assign it outside your if
if statement, the OP is just missing the parenthesis for order of operations, no?. https://3v4l.org/IPDQKInterestingly, a variable cannot be both initialized and used within the condition of an if statement. You may have found a bug/feature of php.
function some_function()
{
return 23;
}
$intA = 0;
if ($intA = some_function()
&& $intA != $other_int
) {
echo 'hello world!'; // works fine
}
if ($intB = some_function()
&& $intB != $other_int
) {
echo 'hello world!'; // PHP Notice: Undefined variable: intB
}
