I'm seeing some confusing behavior related to reference/assignment in PHP...
private function doSomething($things)
{
foreach ($things as $thing) {
echo $thing->property; // 'foobar'
$copyThing = $thing;
unset($copyThing->property);
echo $thing->property; // undefined
I expect this behavior when passing variables by reference (&$thing) but I'm not trying to do that here and it seems to be happening anyway. What am I missing?
Just explaining my comment:
objects in foreach loops are always passed by reference
When you use a foreach loop for an array of objects the variable that you are using inside the loop is a pointer to that object so it works as a reference, any change on the object inside the loop is a change on the object outside. This is because:
objects are always passed by reference (@user3137702 quote)
Detailed and official explanation here.
When you copy and unset your variable:
$copyThing = $thing; unset($copyThing->property);
you are creating another pointer and unseting it, so the original value is a gone. As a matter of fact, since the foreach loop also uses a pointer the $things array is also affected.
check this ideone (notice the vardump [where the 'a' property is gone], as the output is the same as you got)
I do not know in which version it changed, if ever, as it seems like default object/pointer behavior
As a workaround (some ideas):
$x = clone($obj); (As long as the default copy constructor works for your objects)
unseting object properties I unset keys in an array it would not be by reference?
&$varto overtly pass by reference.cloneyour object, for instance:foreach ($things as $thing) { $copyThing = clone $thing; unset($copyThing->property); }