CODEWITHSUNDEEP
php
Eduardo Matos
Eduardo Matos

Reputation: 741

PHP expression evaluating to wrong (?) value

Why does the variable $strange evaluates to true in the following snippet?

$strange = true and false;
var_dump($strange); // true

Upvotes: 1

Views: 74

Answers (5)

user1922137
user1922137

Reputation:

As documentation says:

// foo() will never get called as those operators are short-circuit

$a = (false && foo());
$b = (true  || foo());
$c = (false and foo());
$d = (true  or  foo());

And

// The constant true is assigned to $h and then false is ignored
// Acts like: (($h = true) and false)
$h = true and false;

So the false will be ignored.

Upvotes: 0

AStopher
AStopher

Reputation: 4550

I'm no PHP expert, but this seems to be an issue with simple logic... The expression var_dump($strange); evaluates and outputs the first argument.

Upvotes: 0

Jeff Lambert
Jeff Lambert

Reputation: 24661

See the documentation:

// The constant true is assigned to $h and then false is executed then and its 
// value is and'd with the results of the assignment. 
// Acts like: (($h = true) and false)
$h = true and false;

The above example is taken straight from there, and matches your issue exactly.

Upvotes: 0

ikegami
ikegami

Reputation: 386331

and is a low-precedence version of &&.

$strange = true and false;

is equivalent to

($strange = true) and false;

You want

$strange = (true and false);

or the more appropriate

$strange = true && false;

and and or are best reserved when preceding a flow control statement like break or return.

foo()
   or throw new Exception('foo() returned an error.');

Upvotes: 3

Marc B
Marc B

Reputation: 360762

and and or in PHP have a LOWER operator precedence then && and ||. Your code is being evaluated as

($strange = true) and false;

These two would have worked:

$strange = true && false;
$strange = (true and false);

Upvotes: 0

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