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 <sivavikne...@gmail.com> wrote:
> i compiled .. the answer is 17 !! ... its 5+5+7 ..evaluate from left
> to right
>
> On Nov 6, 8:42 pm, bipul21 <bipuljai...@gmail.com> 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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