CODEWITHSUNDEEP
phpbooleanboolean-logic
Felipe
Felipe

Reputation: 11887

Weird boolean evaluation behaviour in PHP

Let me share some PHP code with you:

$var1 = '';

$var2 = 0;

echo '<pre>';
var_dump($var1 == $var2); //prints bool(true)
echo '</pre>';

echo '<pre>';
var_dump($var1 != $var2); //prints bool(false)
echo '</pre>';

echo '<pre>';
var_dump(!$var1 == $var2); //prints bool(false)
echo '</pre>';

echo '---<br />';

echo '<pre>';
var_dump($var1 === $var2); //prints bool(false)
echo '</pre>';

echo '<pre>';
var_dump($var1 !== $var2); //prints bool(true)
echo '</pre>';

echo '<pre>';
var_dump(!$var1 === $var2); //prints bool(false) .. WTFF????
echo '</pre>';

Question is... why does that last statement (! $var1 === $var2) NOT yield the same result as ($var !== $var2) ??? I mean, it's what we would expect, no?

I used to use both ways interchangeably but now I only use !== although I still don't know why the other form does not work...

Upvotes: 0

Views: 163

Answers (3)

JvdBerg
JvdBerg

Reputation: 21856

This is a matter of precedence:

The not operator is first applied to $var1 and results in a boolean true. And a boolean true is not exact equal to a int 0, so it evaluates to false.

Upvotes: 0

user149341
user149341

Reputation:

The last statement is evaluated as:

(!$var1) === $var2

Since $var1 is a falsey value (empty string), inverting it gives you a truthy value (1), which is not equal (and certainly not identical!) to 0. Thus, the comparison is false.

(Note that I'm deliberately using the terms "truthy" and "falsey" here, as '' and 0 are not quite true and false.)

Upvotes: 1

Tchoupi
Tchoupi

Reputation: 14681

!$var1 is TRUE, $var2is 0.

They are not equal, so the result is false.

Maybe you were confused with !($var1 === $var2)

Upvotes: 4

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