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Re: OT: and even in Dart .........Re: simple session question [message #175752 is a reply to message #175751] Sun, 23 October 2011 08:43 Go to previous messageGo to previous message
Tim Streater is currently offline  Tim Streater
Messages: 328
Registered: September 2010
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Senior Member
In article <j80juf$nct$1(at)news(dot)albasani(dot)net>,
The Natural Philosopher <tnp(at)invalid(dot)invalid> wrote:

> Tim Streater wrote:
>> In article <j7vhmk$420$1(at)news(dot)albasani(dot)net>,
>> Thomas Mlynarczyk <thomas(at)mlynarczyk-webdesign(dot)de> wrote:
>>
>>> Norman Peelman schrieb:
>>>
>>>> $foo = 0;
>>>> $foo = $foo++;
>>>> > It should be equivalent to:
>>>> > $foo = 0; // 0
>>>> $foo = $foo; // 0 = 0
>>>> $foo = $foo + 1; // 0 = (0 + 1)
>>>
>>> No. What you are doing here is assigning first and then incrementing.
>>> The evaluation of $foo++ returns 0 and increments $foo as a side
>>> effect. This evaluation process must be completed before the
>>> assignment can happen:
>>>
>>> [1] Evaluate the expression $foo++ (result: 0, side effect: $foo = 1)
>>> [2] Assign the result to $foo ($foo = 0)
>>>
>>>> $foo++ means that the variable is to be incremented after the
>>> variable > is accessed.
>>>
>>> Yes, it means that the expression $foo++ evaluates to the value which
>>> $foo had before. And it is this value which is assigned to $foo after
>>> the $foo++ step is completed.
>>
>> Yes - as per my JavaScript example. If I say:
>>
>> results[i++];
>>
>> I get the value of results[i]. That value is held in limbo to be
>> assigned to the LHS. Then the post-increment occurs. Then the result
>> from limbo is assigned to LHS.
>>
>> And so with PHP. Which will mean that:
>>
>> $foo = $foo++;
>> $foo = ($foo++);
>>
>> both act the same and set $foo to 0.
>>
> First I heard that []===().....

Come again?

--
Tim

"That excessive bail ought not to be required, nor excessive fines imposed,
nor cruel and unusual punishments inflicted" -- Bill of Rights 1689
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