Strange action with =&

I'm looping through an array and I did this:

$rate =& $mydata[$prefix];

Now, in some cases $mydata[$prefix] wasn't set/defined, so I expected $rate
to not be defined, or at least point to something that wasn't defined.

Instead, PHP 5.1.6 set $mydata[$prefix] to nothing.

If I had:

$mydata[1] = 3;
$mydata[3] = 2;
$mydata[5] = 1;

And did a loop from $i=1; $i++; $i<=5 I'd get:

$mydata[1] = 3;
$mydata[2] = ;
$mydata[3] = 2;
$mydata[4] = ;
$mydata[5] = 1;

Is this expected? A bug? Fixed in 5.2.0? I know I shouldn't set a
reference to a variable that doesn't exist, but the expected result is a
warning/error, not for PHP to populate an array.

Beckman
------------------------------------------------------------ ---------------
Peter Beckman Internet Guy
beckman@purplecow.com http://www.purplecow.com/
------------------------------------------------------------ ---------------

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Re: Strange action with =&

sry i don't get what u mean??
----- Original Message -----
From: "Peter Beckman"
To: "PHP-DB Mailing List"
Sent: Tuesday, February 13, 2007 8:29 AM
Subject: [PHP-DB] Strange action with =&


> I'm looping through an array and I did this:
>
> $rate =& $mydata[$prefix];
>
> Now, in some cases $mydata[$prefix] wasn't set/defined, so I expected
$rate
> to not be defined, or at least point to something that wasn't defined.
>
> Instead, PHP 5.1.6 set $mydata[$prefix] to nothing.
>
> If I had:
>
> $mydata[1] = 3;
> $mydata[3] = 2;
> $mydata[5] = 1;
>
> And did a loop from $i=1; $i++; $i<=5 I'd get:
>
> $mydata[1] = 3;
> $mydata[2] = ;
> $mydata[3] = 2;
> $mydata[4] = ;
> $mydata[5] = 1;

the reason mydata2 empty was because it don't have value in it!!

full source plz
why u don't try this

$txt.="

";
foreach($mydata as $nm=>$val){
$txt.="\n
1. $nm = $val";
   $txt2="
   \$mydata[$nm] = $val";
   }
   $txt.="

Re: Strange action with =&

On Tue, 13 Feb 2007, bedul wrote:

> sry i don't get what u mean??
>
>> I'm looping through an array and I did this:
>>
>> $rate =& $mydata[$prefix];

This is how you assign a variable by reference. $rate should be a
reference to $mydata[$prefix], not a copy. If I change the value of
$rate, the value of $mydata[$prefix] is also changed, and vice versa.

>> Now, in some cases $mydata[$prefix] wasn't set/defined, so I expected
>> $rate to not be defined, or at least point to something that wasn't
>> defined.
>>
>> Instead, PHP 5.1.6 set $mydata[$prefix] to nothing.
>>
>> If I had:
>>
>> $mydata[1] = 3;
>> $mydata[3] = 2;
>> $mydata[5] = 1;
>>
>> And did a loop from $i=1; $i++; $i<=5 I'd get:
>>
>> $mydata[1] = 3;
>> $mydata[2] = ;
>> $mydata[3] = 2;
>> $mydata[4] = ;
>> $mydata[5] = 1;
>
> the reason mydata2 empty was because it don't have value in it!!
>
> full source plz
> why u don't try this
>
> $txt.="

";
> foreach($mydata as $nm=>$val){
> $txt.="\n
1. $nm = $val";
   > $txt2="
   \$mydata[$nm] = $val";
   > }
   > $txt.="

Re: Strange action with =&

Peter Beckman wrote:

> Because I'm trying to point out a problem with PHP, where setting a
> reference when the other side is undefined or not set, PHP creates a
> reference to a previously non-existent array item, just by setting a
> reference. I don't think that should happen.
>
And? what's wrong with that. The reference can be used in the future - I
think thats why it doesnt produce any error message.
ie.
$array=array("1"=>"a","3"=>"c");
$ref=&$array["2"];
$array["2"]="b";
echo($ref);


OKi98

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Re: Strange action with =&

Hello
This (=&) is used in the variable references. in $ref=&$array["2"] line
$ref variable and $array["2"] variable point at the same address. if you
assign any value to this any variables, both of them will change their
values because echo($ref) line display b on the screen.


--
Haydar TUNA
Republic Of Turkey - Ministry of National Education
Education Technology Department Ankara / TURKEY
Web: http://www.haydartuna.net

"OKi98" wrote in message
news:45D184C0.2060208@centrum.cz...
> Peter Beckman wrote:
>
>> Because I'm trying to point out a problem with PHP, where setting a
>> reference when the other side is undefined or not set, PHP creates a
>> reference to a previously non-existent array item, just by setting a
>> reference. I don't think that should happen.
>>
> And? what's wrong with that. The reference can be used in the future - I
> think thats why it doesnt produce any error message.
> ie.
> $array=array("1"=>"a","3"=>"c");
> $ref=&$array["2"];
> $array["2"]="b";
> echo($ref);
>
>
> OKi98

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