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Re: OT: and even in Dart .........Re: simple session question [message #175743 is a reply to message #175725] Sun, 23 October 2011 02:40 Go to previous messageGo to previous message
Norman Peelman is currently offline  Norman Peelman
Messages: 126
Registered: September 2010
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Senior Member
On 10/22/2011 04:28 PM, Jerry Stuckle wrote:
> On 10/22/2011 4:08 PM, Norman Peelman wrote:
>> On 10/22/2011 01:34 PM, Thomas Mlynarczyk wrote:
>>> Luuk schrieb:
>>>> i did send a bug-report:
>>>> https://bugs.php.net/bug.php?id=60114
>>>
>>> I do not see any bug here. I was confused because it's a crappy way to
>>> code, but it's clear why it works the way it does:
>>>
>>> $foo = 0;
>>> $foo = $foo++;
>>>
>>> is definitely identical to
>>>
>>> $foo = 0;
>>> $foo = ($foo++);
>>>
>>> since ++ has higher precedence than =. In fact, ($foo = $foo)++ does
>>> rightfully throw a parse error, since you can only increment a variable,
>>> not an expression.
>>>
>>> As mentioned before in this thread, the above code is equivalent to
>>>
>>> $foo = 0;
>>> $tmp = $foo; // $foo++ yields the previous value, which is 0
>>> $foo = $foo + 1; // then $foo is incremented...
>>> $foo = $tmp; // ...and then re-assigned the old value
>>>
>>> It's the exact same procedure as it would be with $foo = $bar++.
>>>
>>> Greetings,
>>> Thomas
>>>
>>
>> No they are not the same.
>>
>> $foo = 0;
>> $foo = $foo++;
>>
>> It should be equivalent to:
>>
>> $foo = 0; // 0
>> $foo = $foo; // 0 = 0
>> $foo = $foo + 1; // 0 = (0 + 1)
>>
>> $foo++ means that the variable is to be incremented after the variable
>> is accessed. Something is clobbering that increment.
>>
>> ++$foo means that the variable is to be incremented piror to the
>> variable being accessed. Works as expected.
>>
>>
>> Haven't seen the source code but there must be some temporary variable
>> that's getting clobbered for the $foo++ example.
>>
>>
>
> Actually, I take my previous statement about the behavior being defined
> back. Officially in C/C++, the results of this operation is undefined.
>
> Precedence and associativity define the order in which operators are
> processed. But the order of operand processing is not defined.
>
> For
>
> $foo = $foo++;
>
> we have the same operand ($foo) being set twice. The $foo++ will return
> 0, and this value will be assigned into $foo.
>
> But what is NOT defined is whether the assignment will occur before or
> after the value of $foo is actually incremented. Either is possible.
>
> Now in the case of
>
> $foo = ($foo++);
>
> The results are different, but they are still undefined.
>
> In the first case the increment is obviously performed before the
> assignment (although the assignment correctly uses the pre-increment
> value).
>
> In the second case it looks like the assignment is still using the
> pre-increment value, but the increment is done after the assignment.
>
> Since the results are undefined, both are correct (or incorrect, as you
> may look at it), and may change with changes to the interpreter (or
> potentially the same version on different OS's).
>

The problem seems to be that the post increment is being overwritten
as if:

$foo = 0;
$tmp = $foo;
$foo = $foo + 1;
$foo = $tmp;

Which I think someone else may have said.

--
Norman
Registered Linux user #461062
-Have you been to www.php.net yet?-

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