Comments on: Zend PHP5 certification Exam study review 7: PHP4 And PHP5

By: Henry Lu

Henry Lu — Fri, 12 Jun 2009 16:03:22 +0000

I agree mcloide for the following reasons:
You should never use parentheses around your return variable when returning by reference, as this will not work. You can only return variables by reference, not the result of a statement. If you use return ($a); then you’re not returning a variable, but the result of the expression ($a) (which is, of course, the value of $a).
If you are attempting to return the result of an expression, and not a variable, by reference. You can only return variables by reference from a function – nothing else. Since PHP 4.4.0 in the PHP4 branch, and PHP 5.1.0 in the PHP5 branch, an E_NOTICE error is issued if the code tries to return a dynamic expression or a result of the new operator.

By: mcloide

mcloide — Tue, 26 Aug 2008 00:01:05 +0000

Hi Heena, about the first question, the correct answer would be 3. PHP4 will throw a notice error for using an undefined variable.

By: admin

admin — Sun, 01 Jun 2008 06:17:34 +0000

The answer is 4>$obj1 and $obj2 must be the same instance in php5.

By: Heena

Heena — Sun, 01 Jun 2008 00:04:30 +0000

yes I tried to post with tag… btw code is here
if($obj1===$obj2){
//do something
}

By: admin

admin — Sat, 31 May 2008 07:10:14 +0000

Do you mean post the code containing the php tags: < ?php and ?>?

By: Heena

Heena — Sat, 31 May 2008 02:36:13 +0000

hey there might be some problem i cant post php code

By: Heena

Heena — Sat, 31 May 2008 02:34:36 +0000

Consider the following php4 code:

what if any potential compatibility problems will this conditional have in php5?
1>this code is undefined in php4
2>none of above
3>$obj1 and $obj2 musr have the same property values in php5
4>$obj1 and $obj2 must be the same instance in php5
5>there are no compatibility issues
please answer
thanks

By: admin

admin — Thu, 29 May 2008 13:53:34 +0000

had the problem. But, to me it seems fine.

As to the question, I think the answer is 4).

This code in PHP5 is definitely cause an error. But, in PHP4, there seems some problem found on “return by reference” in the language itself.

By: admin

admin — Thu, 29 May 2008 13:36:38 +0000

Hi,
First, I am sorry if the for($i=0; $i<10;$i++)…

By: Heena

Heena — Thu, 29 May 2008 10:55:39 +0000

less than 10;$i++){
$retval = myfun();
}
?>

By: Heena

Heena — Thu, 29 May 2008 10:54:59 +0000

By: Heena

Heena — Thu, 29 May 2008 10:54:14 +0000

hey i can not post the code?

By: Heena

Heena — Thu, 29 May 2008 10:53:48 +0000

for($i=0;$i

By: Heena

Heena — Thu, 29 May 2008 10:52:49 +0000

function &myfun(){
$str = “mystr”;
var_dump($str);
return ($undefined);
}
echo “”;
for($i=0;$i

By: Heena

Heena — Thu, 29 May 2008 10:51:00 +0000

hi,

i got lots of information for zend certification. its very helpful… i have one question if can help me.

consider the following code block.
function &myfun(){
$str = “mystr”;
var_dump($str);
return ($undefined);
}
echo “”;
for($i=0;$i none of the above
2> this could would cause an automatic segmentation fault in php4
3> this code would throw a syntax error in php4
4> returning an undefined variable by reference in php4 would cause eventual memory corruption
5> you could not return undefined variable by reference in php4

what could be the answer?

please help
thanks