CopyBits(x,p,n,y)
copy n LSBs from y to x starting LSB at 'p'th position.
Answers were Sorted based on User's Feedback
Answer / mohammed sardar
Sorry above answer ; I did a mistake
t=0;
for(i=n; i>0; i--)
{
t|=(1<<p);
p++;
}
x=x&~t
t=t&y;
x=x|t;
Is This Answer Correct ? | 1 Yes | 0 No |
Answer / angad
t=0;
for(i=n; i>0; i--)
{
t|=(1<<p);
p++;
}
x=x&~t
t=t&(y<<p);
x=x|t;
}
x=10100110
y=11110111
let,p=3,n=4
after for loop, t=01111000 - mask for the n(=4) bits starting from the p(=3) bit that need to be altered in x
x=x&~t;
x =10100110
~t=10000111
ANDing clears the 4 bits to zero(bits 3-6)
x=1 0000 111
we need to extract the 1st n(=4) bits out of y , and shift them left to align them against the n(=4) bits of x we need to alter, therefore, left shift y by p(=3)
t=t&(y<<p)
y<<p = 1 0111 000
t = 0 1111 000
AND=>t = 0 0111 000
now x = 1 0000 111
t = 0 0111 000
x=x|t =>1 0111 111
Is This Answer Correct ? | 1 Yes | 0 No |
Answer / mohammed sardar
t=0;
for(i=n; i>0; i--)
{
t|=(1<<p);
p++;
}
t=t&y;
x=x&t;
Is This Answer Correct ? | 0 Yes | 0 No |
Answer / rakesh
It should be p--, not p++. for example y = 217 (11011001)
and you want to extract the least 4 SB then n = 4 and p = 3
(remember the number start from 0 to 7 in memory). After
executing this you will get x = 9 i.e 1001.
t=0;
for(i=n; i>0; i--)
{
t |= (1<<p);
p--;
}
x=x&~t;
t=t&y;
x=x|t;
Is This Answer Correct ? | 0 Yes | 0 No |
Hi,
The below code ll giv desired o/p.
#include<stdio.h>
#include<conio.h>
int main()
{
int x, y, n , p, i, j, temp;
printf("ENTER X, Y, NO OF BITS AND BIT POSITION: \n");
scanf("%d %d %d %d",&x, &y, &n, &p);
for(i = p, j = 0; i < n+p; i++, j++)
{
if(x & (0x01 << i))
x = x^(0x01<<i);
temp = y & (0x01<<j) ? 1 : 0;
x = x | (temp << i-1);
}
printf("VALUE OF X:%d \n",x);
getch();
}
Is This Answer Correct ? | 0 Yes | 0 No |
Hi,
Small bug is there in the above code, which i have posted.
But the same has been resolved here.
#include<stdio.h>
#include<conio.h>
int main()
{
int x, y, n , p, i, j, temp;
printf("ENTER X, Y, NO OF BITS AND BIT POSITION: \n");
scanf("%d %d %d %d",&x, &y, &n, &p);
for(i = p, j = 0; i < n+p; i++, j++)
{
if(x & (0x01 << i-1))
x = x^(0x01 << i-1);
temp = y & (0x01 << j) ? 1 : 0;
x = x | (temp << i-1);
}
printf("VALUE OF X:%d \n",x);
getch();
}
Is This Answer Correct ? | 0 Yes | 0 No |
Answer / intfail
all the above answers are wrong...
never use loops. immediate rejection
CopyBits(x, p, n, y)
First get n bits from pos p from Y
bitsFromy = y >> (p-n+1) & (~(~0<<n))
Now, get a mask such that we can 0 out bits in x at pos p and n bits to the right
startpos = p -n +1
create a mask from (startpos, p)
mask = (~0 << p - startpos +1)<<startpos | ~(~0 << startpos)
Now, 0 out the the bits in the locations (starpos, p) in x
and apply the bits extracted from y
x = (x & mask) | (bitsFromy << startpos)
that is all it takes.
Is This Answer Correct ? | 0 Yes | 0 No |
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