What is the difference between
$obj1 = new Test();
$obj2 = new $obj1;
and
$obj1 = new Test();
$obj2 = new Test();
Both are creating different memory location for objects. Does it have any difference...?
3 Answers
While there's no functional difference, you probably should try to limit your use of the first one as you have no idea what could be inside $obj1.
In both cases, you can then pass arguments and $obj2 is a separate object from $obj1 with the same class type.
Confusion with the new $obj1 format can arise when Test has a __toString method. In this case, the __toString method is not called, even though the usual case for the new $var() syntax is when $var is a string that has been generated or passed in.
Comments
$obj2 = new $obj1;
use like this $obj2 = $obj1;
you have two variables refering to the same instance of Test.
but here
$obj1 = new Test();
$obj2 = new Test();
you have two variables refering to two different instances of Test.
6 Comments
Lekhulal Mathalipara
Thanks Sajad. I just noticed from php.net/manual/en/language.oop5.basic.php. Would you please explain me the below lines of code in detail $obj1 = new Test(); $obj2 = new $obj1; as compared to $obj1 = new Test(); $obj2 = new Test();
Amal
Objects are only equivalent if they both refer to the same instance.
Lekhulal Mathalipara
@simon Are $obj1 and $obj2 the object of same class? And do they have same properties?
Sajad Karuthedath
@LekhulalMathalipara:didn't my answer helped ? why the downvote?
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In the first example, were you trying to set $obj2 as a reference to $obj1? If so, you should not use the new keyword or it will try to create a new object with a type stored in the $obj1 variable.
Creates two references to the same object:
$obj1 = new Test();
$obj2 = $obj1;//Creates a reference to the first object
1 Comment
Ivan Yonkov
No, he is just asking whats the difference
$class = 'Test'; $obj = new $class