For more clarification....try this :-
int i=1,j=1,k=1,l;
l= ++i || j++ && k++;
printf("%d %d %d %d",i,j,k,l);
o/p will be 2 1 1 1
because as vaibhav wrote the equation evaluate as l= ++i || (j++ && k++);
only ++i evaluate not other two increments :)
On Sun, Jul 10, 2011 at 11:31 PM, rShetty <[email protected]> wrote:
> Got it Thanks .....
>
> On Jul 10, 10:40 pm, vaibhav shukla <[email protected]> wrote:
> > associativity comes into play when operators are of same precedence.
> >
> > On Sun, Jul 10, 2011 at 11:08 PM, vaibhav shukla <
> [email protected]>wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > && has higher precedence than ||
> > > the expression is evaluated as
> > > z=j || ( k && i );
> > > hence the output i.e 1 ;)
> >
> > > On Sun, Jul 10, 2011 at 11:06 PM, rShetty <[email protected]>
> wrote:
> >
> > >> #include<stdio.h>
> > >> int main()
> > >> {
> > >> int i=0,j=1,k=1,z=0;
> > >> z = j || k && i ;
> > >> printf("%d",z);
> > >> return 0;
> > >> }
> >
> > >> The output is 1 for the above program .
> >
> > >> But according to associativity of logical operators , the evaluation
> > >> should be from left to right , But is it taking from right to left ?
> > >> What is the exact concept for the program behavior above?
> >
> > >> --
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> >
> > > --
> > > best wishes!!
> > > Vaibhav Shukla
> > > DU-MCA
> >
> > --
> > best wishes!!
> > Vaibhav Shukla
> > DU-MCA
>
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--
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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