Using the same variable name for a function's argument and its return value*

Question

In

public function bind($query, $input_param, $btypes)
{
    // $input_param = $this->ref_arr($input_param);  // this self assignment gives an error!
    $input_ref = $this->ref_arr($input_param);       // this works 
}

I learned this by trial and error... but I'm trying to figure out why?

I haven't had a chance to form more test cases but if I use $input_param as in input to the function I can not return the result back to $input_param. Once I change the name to something else, in this case $input_ref it works.

It's not very clear what you're asking. Can you try to clean it up a bit? — Roland Mai
– Roland Mai, Commented Sep 12, 2011 at 3:43
What do you mean by "does not work"? Does it give you an error? And what does ref_arr() look like? — NullUserException
– NullUserException, Commented Sep 12, 2011 at 4:33
Based on code shown here, as far as I can tell both of those should work, so the issue is most likely elsewhere. — Jani Hartikainen
– Jani Hartikainen, Commented Sep 12, 2011 at 5:17

dqhendricks · Accepted Answer · 2011-09-12 20:05:31Z

2

The $this keyword references the current object you are in.

So if you are in code that is in a class like this:

class foo {
   public function __construct() {
      $this->bar = 'that'; // works because $this references the foo object
   }
}

Should work. In the case where you are outside of an object however, $this would not work, because there is no object for $this to reference.

class foo {
   public function __construct() {

   }
}
$this->bar = 'that'; // will not work because you are not inside of any object

answered Sep 12, 2011 at 20:05

dqhendricks

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Comments

user656925 · Accepted Answer · 2011-09-19 01:18:08Z

0

I don't have time to reproduce this...simply changing the variable name fixed the problem.

answered Sep 19, 2011 at 1:18

user656925

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