Re: Strange behavior with pointers
From: Bo Persson (bop_at_gmb.dk)
Date: 02/16/05
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Date: Wed, 16 Feb 2005 18:30:06 +0100
"Mythran" <kip_potter@hotmail.comREMOVETRAIL> skrev i meddelandet
news:%23XMjCaEFFHA.2452@TK2MSFTNGP09.phx.gbl...
>
> "John" <John@discussions.microsoft.com> wrote in message
> news:0B171982-A60E-4666-A7A6-E9E11B940F31@microsoft.com...
>>I have created a program that accesses a flat 2 dimensional array and
>>have
>> developed a routine to sum the four neighbors of a specific
>> row/column
>> position.
>>
>> Here is an example, this will sum all the values of row 1, column 1
>> (base 0)
>> which is the number 5
>>
>>
>> int a[9] = {1,2,3,4,5,6,7,8,9};
>> int* pa = &a[3]; // row 1, column 0 (i.e. number 4)
>> int n = *(pa++); // n = 4
>> n += *(++pa); // n = 10 (4 + 6)
>> n += *(--pa - 3); // n = 12 (10 + 2)
>> n += *(pa + 3); // n = 20 (12 + 8)
>>
>> Now here is the strange part if I combine all the code above into one
>> statement
>>
>> n = *(pa++) + *(++pa) + *(--pa - 3) + *(pa + 3);
>>
>> I don't get the same results, instead of 20 I get 18. Am I missing
>> something
>> here is this a compiler or programming error?
>>
>> Any help would be greatly appreciated
>>
>
> Something I see right off the bat is...what happens to pa before and
> after combining this to the statement? If you follow pa through you
> get the following:
>
> int* pa = &a[3]; -- pa == 4
> int n = *(pa++); -- pa == 4
> n += *(++pa); -- pa == 6
> n += *(--pa -3) -- pa == 5
> n += *(pa + 3) -- pa == 5
>
> so, putting that together we get: (4 + 6 + 5 + 5)..
>
> Putting it together on one line like you have it means...
>
> n = *(pa++) + *(++pa) + *(--pa - 3) + *(pa + 3) means the following:
>
> n = pa++ (pa holds a value of 4) + (pa holds a value of 6) + (pa holds
> a value of 4) - 3 + (pa holds a value of 4) + 3
>
> the first pa++ is evaluated then incremented (use in expression then
> add 1 to pa)...value used is 4
> the ++pa is incremented then evaluated (add 1 to pa then use in
> expression)...value used is 6 (previous value used was 4 but previous
> calc for pa was pa++ and therefore it was incremented to 5. This
> ++pa adds 1 to 5 which makes it 6.)
> Then, --pa is decremented then evalueated (subtracts 1 from pa [6 - 1]
> to make 5) ... value used is 5
> adds 3 to 5
> Lastly, pa is 5 and it adds 3 to the 5 which makes 8.
>
> so, we get....
>
> (pa++ [pa evaluated as 4]) + ([pa is 5 before this] ++pa [pa is
> preincremented and now is + 6]) + ([pa is still 6] --pa [now pa is
> 5]) - 3 + ([pa is still 5]5) + 3
>
> 4 + 6 + 5 - 3 + 5 + 3 = 18
>
This is one possible way for the compiler to come to this result.
However, it is not required to chose this particular order. When a
variable is modified more that once in the same expression (formally
"between two sequence points"), any and all results are equally correct
and possible.
For Java, there is probably a defined order of evaluation, but for C and
C++ there is not.
Bo Persson
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