CareerCup

int x= ( (x & 0xAAAAAAAA)>>1 | (x & 0x55555555)<<1 );

many of the above solns seem wrong.. this should work just fine..
x = ((x & 0x33333333) << 2) | ((x & 0xcccccccc) >> 2)

The question was for a byte. It would be more appropriate if we do (same as hgk answer)

N = ((N & 0x33) <<2) | ((N & 0xCC) >> 2)

((x<<1)&(10101010)) | ((x>>1)&(01010101))

((x<<1)&(0x10101010)) |(x&0x01010101)

opps, should be

((x<<1)&(0b10101010)) |(x&0b01010101)

logically ok.
karate, by the time the control reaches the second part of the logic x has already been shifted left so right shifting puts it back to its original position with the left most bit missed, possibly. as i know the shifting works in place, so the best option would be to use this way
a = x
left = x<<1 & 0b10101010
right = a>>1 & 0b01010101
x= left | right

@amazonianMonkey
Your logic is not correct, you don't need to shift it back because the send shift is performed on the original x value.

Consider,

int x = 2;
x = x << 1 | x << 1;

if you look at the assembly, assuming %eax has x
add %eax %eax
or %eax %eax

If you print out x, the output is 4 not 12.

#define SWAP_BYTE(N)  \
((((1<<0)&(N))<<1)  \
 | (((1<<1)&(N))>>1)  \
 | (((1<<2)&(N))<<1)  \
 | (((1<<3)&(N))>>1)  \
 | (((1<<4)&(N))<<1)  \
 | (((1<<5)&(N))>>1)  \
 | (((1<<6)&(N))<<1)  \
 | (((1<<7)&(N))>>1)  \
)

is it possible to extend any of these logics to reverse nibble(4 bits) in a number ?

@gus
so you mean to say in case of your solution by the time control reaches the second part (the second part of OR) the value of x is the same as the original, right?
((x<<1)&(0b10101010)) |(x&0b01010101)

i don't think so! How do you explain that...this logic applies to post/pre increment operators in C/C++, but how do you apply the same here, can you write concrete examples? would be nice to make things clear for my understanding. Thank

how about XOR ? answer = input XOR ff;

XOR would not work,XORing with ff will just convert all 1s to 0s and vice versa so 11 will be converted to 00, so not correct

sum = 0;m = 2,l = 1
while(p)
{
if(((p&0x03) != 0x03) && ((p & 0x03) != 0x03))
{
p1 = p ^ 0x03;
sum += ((p1 >> 1) &0x01 )*l + (p1 & 0x01)*m;

}
else
{
sum += ((p >> 1) &0x01 )*l + (p & 0x01)*m;

}
m *=4;
l *=4;
p <<= 2;
}
the result is stored in sum

a = (num&0x01010101)<<1
b = (num&x010101010)>>1
return a|b

What if we XOR with 11110000...?????

v = ((v >> 2) & 0x3333) | ((v & 0x3333) << 2);

p=((p&m)<<1)|((m<<1)&p)>>1;

this is working code

#define ENDSWAP2BITS(x) ((((x)&0x03)<<2)|(((x)&0x0c)>>2)|(((x)&0x30)<<2)|(((x)&0xc0)>>2))

swapped_no = (((x & 0xC0) >> 2) | ((x & 0x30) << 2) | ((x & 0x0C) >> 2) | ((x & 0x3) << 2));

int bitswap(char x)
{
int k,l,m,n;
int mask=1;

for(i=0;i<7;i+=2)
{
k=(x & mask)<<1;
mask<<=1;
l=(x&mask)>>1;
mask<<=1;

m=~(1<<i);
n=m & ~(1<<(i+1));
x=(x&n)|k|l;
}
return x;
}

#include<stdio.h>
main()
{
short no=158;
int x=0,y=0,z=0;
short temp = 0xFFFF, i=0;

while(no>0)
{
x = (no&1);
y = (no>>1)&1;
temp = 1<<i;
y = y<<i;
temp = (temp)&(y);

z=(z)|(temp);

i++;
temp = 1<<i;
x = x<<i;
temp = temp&x;
z=z|temp;
i++;
no=no>>2;
}

printf("z is %d",z);

}

int k = 156; //10 01 11 00
	int t1 = 170;//10 10 10 10
	int t2 = 85; //01 01 01 01

	t1 = t1 & k;
	t1 = t1>>1;

	t2 = t2 & k;
	t2 = t2<<1;

	k = t1 | t2;

This should work for all sets of bytes

for (int i = 0; i < no of bytes / 2; i++)
{
u |= (((x>>1)&01 | (x<<1)&02) << (2*i));
x >>= 2;
}

a = (num<<1 & 0xAAAAAAAA);
b = (num>>1 & 0x55555555);
return a|b;

please tell me how to swap this these numbers.

my num=10110000.
output=10000011.first two bits are moved to 3 and 4 th positions.
please tell me how to do this one.