Re: [PHP] strange reference behavior

[Martin Ellingham]

Robert Enyedi wrote:

Hi,

I've been studying the PHP reference mechanism (with PHP 5.2.1) and 
I'm unsure if the following behavior is normal.


This code works as expected:

$a = 2;
$b = $a;
//$c = $a;
$c = $b;
$a = 1;

echo $c.\n; // Prints 2 as expected

but this one does not:

$a = 2;
$b = $a;
$c = $a;
$c = $b; // Should overwrite the previous assignment, so $c
 // should get a copy of $b (and NOT a reference)
$a = 1;

echo $c.\n; // I would expect 2, but prints 1


Could anyone please clarify why this happens?

Regards,
Robert

This is because PHP5 has changed the default behaviour and $a = $b is 
now call by reference as standard.


That's my understanding of it.

Martin

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Re: [PHP] strange reference behavior

[Robert Cummings]

On Sat, 2007-09-01 at 13:06 +0300, Robert Enyedi wrote:
 Hi,
 
 I've been studying the PHP reference mechanism (with PHP 5.2.1) and I'm 
 unsure if the following behavior is normal.
 
 This code works as expected:
 
   $a = 2;
   $b = $a;
   //$c = $a;
   $c = $b;
   $a = 1;
 
   echo $c.\n; // Prints 2 as expected
 
 but this one does not:
 
   $a = 2;
   $b = $a;
   $c = $a;
   $c = $b; // Should overwrite the previous assignment, so $c
// should get a copy of $b (and NOT a reference)
   $a = 1;
   
   echo $c.\n; // I would expect 2, but prints 1
 
 Could anyone please clarify why this happens?

Sure...


1: $a = 2;
2: $b = $a;
3: $c = $a;
4: $c = $b;   // Should overwrite the previous assignment, so $c
5:// should get a copy of $b (and NOT a reference)
6: $a = 1;
7:
8: echo $c.\n; // I would expect 2, but prints 1

By line number...

1: Assign 2 to a variable called $a
2: Assign to $b a reference to $a
3: Assign to $c a reference to $a
4: Assign the value of $b to $c
   (this does NOT break $c's reference to $a)
6: Assign the value 1 to $a
   ($a is currently referenced by $b and $c)
8: Echo $c which should be 1. You will get the same result
   in PHP4

Cheers,
Rob.
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Re: [PHP] strange reference behavior

[Robert Cummings]

On Sat, 2007-09-01 at 11:39 +0100, Martin Ellingham wrote:
 Robert Enyedi wrote:
  Hi,
 
  I've been studying the PHP reference mechanism (with PHP 5.2.1) and 
  I'm unsure if the following behavior is normal.
 
  This code works as expected:
 
  $a = 2;
  $b = $a;
  //$c = $a;
  $c = $b;
  $a = 1;
 
  echo $c.\n; // Prints 2 as expected
 
  but this one does not:
 
  $a = 2;
  $b = $a;
  $c = $a;
  $c = $b; // Should overwrite the previous assignment, so $c
   // should get a copy of $b (and NOT a reference)
  $a = 1;
  
  echo $c.\n; // I would expect 2, but prints 1
 
  Could anyone please clarify why this happens?
 
  Regards,
  Robert
 
 This is because PHP5 has changed the default behaviour and $a = $b is 
 now call by reference as standard.

In the above example no objects have been used. As such, nothing has
changed in the above semantics that do not exist in PHP4.

Cheers,
Rob.
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Re: [PHP] strange reference behavior

[Robert Enyedi]

Thanks for the clarifications.

Regards,
Robert

Robert Cummings wrote:

On Sat, 2007-09-01 at 11:39 +0100, Martin Ellingham wrote:

Robert Enyedi wrote:

Hi,

I've been studying the PHP reference mechanism (with PHP 5.2.1) and 
I'm unsure if the following behavior is normal.


This code works as expected:

$a = 2;
$b = $a;
//$c = $a;
$c = $b;
$a = 1;

echo $c.\n; // Prints 2 as expected

but this one does not:

$a = 2;
$b = $a;
$c = $a;
$c = $b; // Should overwrite the previous assignment, so $c
 // should get a copy of $b (and NOT a reference)
$a = 1;

echo $c.\n; // I would expect 2, but prints 1


Could anyone please clarify why this happens?

Regards,
Robert

This is because PHP5 has changed the default behaviour and $a = $b is 
now call by reference as standard.


In the above example no objects have been used. As such, nothing has
changed in the above semantics that do not exist in PHP4.

Cheers,
Rob.


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