The behaviour of the or operator in PHP

问题:

I'm trying to understand the behavior of or operator. Please see the below examples:

$e = false || true;
var_dump($e);

Output is as expected: bool(true);

$f = false or true;
var_dump($f);

Output is as expected: bool(false). I understood this in a way that the = has a higher precedence than the Or, so that's why the $f is assigned to false.

But the below code works quite opposite of what I thought. I thought that the $foo will be assigned to 5 and then compared to itself. But the $foo is getting assigned only when if the $foo is set that means it is checking if the $foo is assigned to anything before, assign 5 to it.

$foo or $foo = 5; 

Can anyone explain why this is so?

回答1:

The basics:

1. An assignment expression results in the assigned value.

   What does that mean? $foo = 'bar' is an expression, in which the assignment operator = assigns a value. An expression always returns a value itself. Just like the expression 1 + 2 results in the value 3, the expression $foo = 'bar' results in the value 'bar'. That's why this works:

   $foo = $bar = 'baz'; // which is: $foo = ($bar = 'baz');
   

2. Boolean operations are short-circuiting operations. Both sides are not always evaluated if they don't need to be. true || false is always true overall, since the lefthand operand is true, so the whole expression must be true. false is not even being evaluated here.

3. Operator precedence dictates in which order parts of an expression are grouped into sub-expressions. Higher precedence operators are grouped with their operands before lower precedence operators.

Therefore:

$e = false || true;

false || true is being evaluated, which results in the value true, which is assigned to $e. The || operator has a higher precedence than =, therefore false || true is grouped into an expression (as opposed to ($e = false) || true).

$f = false or true;

Here now or has a lower precedence than =, which means the assignment operation is grouped into one expression before or. So first the $f = false expression is evaluated, the result of which is false (see above). So then you have the simple expression false or true which is evaluated next and results in true, but which nobody cares about.

The evaluation works like this:

1. $f = false or true;
2. ($f = false) or true;  // precedence grouping
3. false or true;         // evaluation of left side ($f is now false)
4. true;                  // result

Now:

$foo or $foo = 5; 

Here, again, $foo = 5 has a higher precedence and is treated as one expression. Since it occurs on the right side of the or operator, the expression is only evaluated if necessary. It depends on what $foo is initially. If $foo is true, the right hand side will not be evaluated at all, since true or ($foo = 5) must be true overall. If $foo has a falsey value initially though, the right hand side is evaluated and 5 is assigned to $foo, which results in 5, which is true-ish, which means the overall expression is true, which nobody cares about.

1. $foo or $foo = 5;
2. $foo or ($foo = 5);   // precedence grouping
3. false or ($foo = 5);  // evaluation of left side
4. false or 5;           // evaluation of right side ($foo is now 5)
5. true;                 // result

回答2:

As per the php.net webpage about Logical Operators:

This:

$e = false || true;

Acts like this:

$e = (false || true) // If false is true, then $e = false. Otherwise true

This:

$f = false or true;

Would act like this:

($f = false) or true; // $f = false is true, as the assignment succeeded

This:

$foo or $foo = 5; 

Would act like this:

$foo or ($foo = 5) // foo = undefined or foo = 5, so foo = 5

For the last one, undefined is basically like false, therefore foo equals 5.

Also, here's the link for the operator precedence order: http://www.php.net/manual/en/language.operators.precedence.php

UPDATE:

Ok, now let's get to the main point. Like how we all know when using a fetched query:

while($row = @mysql_fetch_assoc($result))

And we all know while loops only execute on true, therefore $row = @mysql_fetch_assoc($result) returns true.

Same with Daric's question.

$foo or $foo = 5;

Is basically:

$foo or ($foo = 5);

Which is basically:

$foo = undefined or ($foo = 5); // $foo = 5 actually returns true

Which is also

$foo = undefined or true;

And as I have previously mentioned, undefined = false, so therefore $foo = 5 (as that is the true statement).

I hope everyone can understand.

回答3:

$foo or $foo = 5;

Suppose let say $foo=true or $foo=5;

here it will not evaluate after or operator expresion so output will be $foo=1 Now the expression is

$foo=false or $foo=5;

Here it will evaluate after or as = higher precedence so $foo as of which of which it will evaluate $foo=5so output will be 5 But when we evaluate $foo=false or true so here it will consider = higher precedence so the output will be $foo=false but whole expression will evaluate as true because false or true becomes false

回答4:

<?php
$foo = false;
$foo or ($foo = '5');
echo $foo;
?>

Check this you can assign the value "5" for $foo.

Compare than or = has high priority.. Thats the fact.... :)