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Re: OT: and even in Dart .........Re: simple session question [message #175746 is a reply to message #175743] Sun, 23 October 2011 03:05 Go to previous messageGo to previous message
Jerry Stuckle is currently offline  Jerry Stuckle
Messages: 2598
Registered: September 2010
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On 10/22/2011 10:40 PM, Norman Peelman wrote:
> 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.
>

It's not being "overwritten" - the operand is being changed twice in the
same expression, which is undefined in C.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex(at)attglobal(dot)net
==================

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