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Re: OT: and even in Dart .........Re: simple session question [message #175730 is a reply to message #175724] Sat, 22 October 2011 21:52 Go to previous messageGo to previous message
The Natural Philosoph is currently offline  The Natural Philosoph
Messages: 993
Registered: September 2010
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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.
>

No: increment is after assignment and the NEW value of foo is not being
incremented.

Without the brackets $foo=$foo++ should mean an unchanged foo..thats
perfectly legal.

Its the fact that the brackets don't force taking the value after
increment that is worrisome


I.e.

$ php -r ' $foo=0; $bar=(($foo++)); print $bar."\n";'
0
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