Is $a == $b always equivalent to $b == $a?

I think in JavaScript there's a few weird cases where that's not true, due to casting.

I think ide is correct. I'll ask another question.

Comments

I would say 'reflexive'.

Written by Joe

@Joe: Thanks. Updated.

Written by Mark

@Joe is correct. This is considered reflexivity. Associativity is when (A + B) + C = A + (B + C)

Written by treeface

Don't quote me on this, but I think the weird cases arise for transitivity; i.e., a == b and b == c but a != c.

Written by ide

I have tried a number of variations and cannot find a case where ($a == $b) !== ($b == $a), but I am determined to do so now!

Written by Josh

@ide: Oh... is that where it occurs? Maybe I should have asked that instead then :)

Written by Mark

Accepted Answer

Depends what happens between those two calls. Otherwise yes, those are the same. The order makes no difference. Using 2 equals == A string of 1 and integer of 1, will return true when compared. Type is ignored, only value is compared. So no wierdness.

http://php.net/manual/en/language.operators.comparison.php

<?

$a=(string) 1;
$b=(int) 1;

var_dump($a);
var_dump($b);


echo $a==$b;

Outputs: 1

http://www.ideone.com/JLJWQ

EDIT

To clarify, there is absolutely nothing you can ever put in $a or $b to get a different output on the comparison, just by putting it on the other side of the operator.

$a="1234";
$b="1234";

echo $a==$b;
echo $b==$a;

The output of that, for any $a or $b values, will always without a doubt be true true, or false false.

Written by profitphp
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