answer is 17............i got in gcc comp

On Sun, Nov 7, 2010 at 7:23 PM, Piyush <[email protected]> wrote:

> C standard says that between two sequence points object's stored value can
> be modified only once ( by evaluation of expression),
> A sequence point occurs in following conditions:
> 1. at the end of full expression. ( the case for your answer)
> 2. at the && , || , ?: operators
> 3. at a function call (after the evaluation of all arguments i.e just
> before the actual call)
>
> Since the value of your variable is getting modified more than once between
> the two sequence points the result is undefined.
>
>
> On Sun, Nov 7, 2010 at 5:33 PM, siva viknesh <[email protected]>wrote:
>
>> i compiled .. the answer is 17 !! ... its 5+5+7 ..evaluate from left
>> to right
>>
>> On Nov 6, 8:42 pm, bipul21 <[email protected]> wrote:
>> > #include<stdio.h>
>> > int main()
>> > {
>> >    int a=5;
>> >         a=a+(a++)+(++a);
>> >         printf("%d",a);
>> >     return 0;}
>> >
>> > answer is 19 but how??
>>
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