&& has higher precedence than ||

but why does j and k didn't increase after the statement
l= ++i || j++ && k++;
got executed ?

On Wed, Jul 13, 2011 at 10:07 PM, sagar pareek <[email protected]>wrote:

> 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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