How can this if-statement evaluate to true?

Question

if ($opts['width'] == 'fs' || $opts['height'] == 'fs' || $opts['ratio'] == 'fs') {
    var_dump($opts); // result of this see bellow
}

Result of var_dump($opts) inside (!) if-statement:

array(3) {
   'width' => int(200)
   'height' => int(0)
   'ratio' => int(0)
}

How is this possible? None of the array's values is (stirng) fs?

Try using `===`! http://ideone.com/EpsAH2 – ComFreek Aug 24 '13 at 18:20 — ComFreek, Aug 24 '13 at 18:20

score 3 · Accepted Answer · edited May 23 '17 at 10:24

3

Because 0 == 'fs'. See this conversion table.

PHP has the === operator to compare both value and type.

edited May 23 '17 at 10:24

Community

answered Aug 24 '13 at 18:20

Madara's Ghost

Thank you for the conversion table link. But in my opinion that's in someway weird. 0 should not evaluatue true to (string) 'fs'. Anyway, I've saved the link for the future! – jakub_jo Aug 24 '13 at 18:25
@mcj: In PHP, if `A == B` and `B == C` it doesn't mean that `A == C`. – Madara's Ghost Aug 24 '13 at 18:27

1 Answers1