this will do the same thing
N = ((N>>(j+1)) << (j+1)) | (M << i) | (~(~0 << i) & N);
On Tue, Oct 11, 2011 at 12:39 PM, Gaurav Kumar <[email protected]> wrote:
>
> This is a question from Laakman.
>
> This code operates by clearing all bits in N between position i and j, and
> then ORing to put M in there
> 1 public static int updateBits(int n, int m, int i, int j) {
> 2 int max = ~0; /* All 1's */
> 3
> 4 // 1's through position j, then 0's
> 5 int left = max - ((1 << j) - 1);
> 6
> 7 // 1's after position i
> 8 int right = ((1 << i) - 1);
> 9
> 10 // 1's, with 0s between i and j
> 11 int mask = left | right;
> 12
> 13 // Clear i through j, then put m in there
> 14 return (n & mask) | (m << i);
> 15 }
>
>
> Gaurav
>
>
> On Oct 1, 2011, at 4:26 AM, rahul sharma <[email protected]> wrote:
>
>
> You are given two 32-bit numbers, N and M, and two bit positions, i and j.
> Write a method to set all bits between i and j in N equal to M (e.g., M
> becomes a substring of N located at i and starting at j).
> EXAMPLE:
> Input: N = 10000000000, M = 10101, i = 2, j = 6
> Output: N = 10001010100
> _
>
>
> #include<stdio.h>
> #include<stdlib.h>
> int main()
> {
> int N,M,i,j;
> printf("Enter value of N \n");
> scanf("%d",&N);
> fflush(stdin);
> printf("Enter value of M \n");
> scanf("%d",&M);
> fflush(stdin);
> printf("Enter value of i \n");
> scanf("%d",&i);
> fflush(stdin);
> printf("Enter value of j \n");
> scanf("%d",&j);
> fflush(stdin);
> int a=0,k;
> for( k=0;k<j;k++)
> {
> a= a<<1;
> a=a|1;
> }
> for(k =0;k<i;k++)
> {
> a=a<<1;
> }
>
> N = N &(~a);
> printf("value of N is %d",N);
> for(k=0;k<i;k++)
> M=M<<1;
> N=N|M;
> printf("value of N is %d",N);
> getchar();
> }
>
>
> isnt it give us wrong mask????
>
> say i=2;
>
> j=6;
>
> it gives mask as(i.e ~a)
>
> 1111111100000011
>
> but i think from 2 to 6 5 0's are needed????plz tell the above prog is
> ok???or not???check by giving any input whose 7thy bit is set...thnx in
> advance
>
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Regards,
Rahul Patil
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