Asked  7 Months ago    Answers:  5   Viewed   24 times

I realise the second one avoids the overhead of a function call (update, is actually a language construct), but it would be interesting to know if one is better than the other. I have been using unset() for most of my coding, but I've recently looked through a few respectable classes found off the net that use $var = null instead.

Is there a preferred one, and what is the reasoning?



It was mentioned in the unset manual's page in 2009:

unset() does just what its name says - unset a variable. It does not force immediate memory freeing. PHP's garbage collector will do it when it see fits - by intention as soon, as those CPU cycles aren't needed anyway, or as late as before the script would run out of memory, whatever occurs first.

If you are doing $whatever = null; then you are rewriting variable's data. You might get memory freed / shrunk faster, but it may steal CPU cycles from the code that truly needs them sooner, resulting in a longer overall execution time.

(Since 2013, that unset man page don't include that section anymore)

Note that until php5.3, if you have two objects in circular reference, such as in a parent-child relationship, calling unset() on the parent object will not free the memory used for the parent reference in the child object. (Nor will the memory be freed when the parent object is garbage-collected.) (bug 33595)

The question "difference between unset and = null" details some differences:

unset($a) also removes $a from the symbol table; for example:

$a = str_repeat('hello world ', 100);


Notice: Undefined variable: a in xxx

But when $a = null is used:

$a = str_repeat('hello world ', 100);
$a = null;



It seems that $a = null is a bit faster than its unset() counterpart: updating a symbol table entry appears to be faster than removing it.

  • when you try to use a non-existent (unset) variable, an error will be triggered and the value for the variable expression will be null. (Because, what else should PHP do? Every expression needs to result in some value.)
  • A variable with null assigned to it is still a perfectly normal variable though.
Wednesday, March 31, 2021
answered 7 Months ago

Here ya go ->

$vars = array_keys(get_defined_vars());
for ($i = 0; $i < sizeOf($vars); $i++) {

And to clarify, implode returns "a string representation of all the array elements in the same order".

Unset requires the actual variable as a parameter, not just a string representation. Which is similiar to what get_defined_vars() returns (not the actual variable reference). So the code goes through the array of strings, and returns each as a reference using the extra $ in front - which unset can use.

Wednesday, March 31, 2021
answered 7 Months ago

You should point to your vendor/autoload.php at Settings | PHP | PHPUnit when using PHPUnit via Composer.

This blog post has all the details (with pictures) to successfully configure IDE for such scenario:

Related usability ticket:

P.S. The WI-18388 ticket is already fixed in v8.0

Wednesday, March 31, 2021
answered 7 Months ago

On Mac OS X environment variables available in Terminal and for the normal applications can be different, check the related question for the solution how to make them similar.

Note that this solution will not work on Mountain Lion (10.8).

Saturday, May 29, 2021
answered 5 Months ago

You can unset session variable using:

  1. session_unset - Frees all session variables (It is equal to using: $_SESSION = array(); for older deprecated code)
  2. unset($_SESSION['Products']); - Unset only Products index in session variable. (Remember: You have to use like a function, not as you used)
  3. session_destroy — Destroys all data registered to a session

To know the difference between using session_unset and session_destroy, read this SO answer. That helps.

Sunday, July 18, 2021
answered 4 Months ago
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