@Arun: A multiple of 8 has three rightmost bits = 0. I first though
about anding out these bits and adding 8. Code for that would be
answer = (n & (~7)) + 8.
Alternatively, round up. That you can do by adding 8 and then zeroing
out the low-order three bits. Code for that would be
answer = ((n + 8) & (~7)).
Another way to round up is by making the low-order three bits 1s and
then adding 1. The addition will change the three low-order bits to 0s
and add one to the next bit position. That's what I decided to submit
since it is slightly simpler (3 characters) shorter:
answer = (n | 7) + 8.
Dave
On Sep 29, 6:57 am, Arun <yourarunb...@gmail.com> wrote:
> @Dave
>
> your solution is really good.
> It would be helpful if you could just share how you reached at
> solution.
>
> Cheers
> Arun
> On Sep 26, 4:48 pm, Dave <dave_and_da...@juno.com> wrote:
>
>
>
> > @Ashwath: Thanks for the correction.
>
> > Dave
>
> > On Sep 26, 1:20 am, aswath G B <aswat...@gmail.com> wrote:
>
> > > @Dave
>
> > > Slight change u have to do
>
> > > #include<stdio.h>
> > > main()
> > > {
> > > int a = 24;
> > > int b = (a | 7)+1; //This Line U have to change.....& not a || 7.....
> > > printf("%d",b);
> > > return 0;
>
> > > }
>
> > > tats it....
>
> > > btw it is nice one line easy soln...
>
> > > On Sat, Sep 25, 2010 at 10:17 PM, Dave <dave_and_da...@juno.com> wrote:
> > > > answer = (x || 7) + 1;
>
> > > > Dave
>
> > > > On Sep 25, 6:56 am, Krunal Modi <krunalam...@gmail.com> wrote:
> > > > > Q: Write an algorithm to compute the next multiple of 8 for a given
> > > > > positive integer using bitwise operators.
>
> > > > > Example,
> > > > > (a) For the number 12, the next multiple of 8 is 16.
> > > > > (b) For the number 16, the next multiple of 8 is 24.
> > > > > -------------------------
> > > > > I have written a code using bitwise operators...but, I am not happy
> > > > > with the solution....Is there any ONE LINE solution ?
> > > > > Here, is my code.
> > > > > Approach: shift left till LSB is 0 (say n number of times) then, OR
> > > > > with 1, finally shift right (same number of times).
>
> > > > > #include<stdio.h>
>
> > > > > int main(){
> > > > > int count,m,n,ans,t;
>
> > > > > printf("Enter any number :");
> > > > > scanf("%d",&n);
>
> > > > > m=8; //multiple of 8 !!
> > > > > ans=0;
> > > > > while(ans<=n){
> > > > > t = m;
> > > > > while(t){
> > > > > count=0;
> > > > > while(ans&1){
> > > > > count++;
> > > > > ans = ans>>1;
> > > > > }
> > > > > ans = ans|1;
> > > > > ans = ans<<count;
> > > > > t--;
> > > > > }
> > > > > }
>
> > > > > printf("[%d]\n",ans);
>
> > > > > return 0;
>
> > > > > }- Hide quoted text -
>
> > > > > - Show quoted text -
>
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> > > --http://aswath78.blogspot.com-Hidequoted text -
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