result=((n & ((1<<28)/3))<<1)|((n & ((1<<29)/3))>>1);
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first do logical AND of 10101010 with number for masking odd bit shift it
right 1 times suppose it comes n1
first do logical AND of 01010101 with number for masking even bit shift it
left 1 times suppose it come n2
now do n1 || n2
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Mask all odd bits with 10101010 in binary (which is 0xAA), then shift
them left to put them in the even bits. Then, perform a similar
operation for even bits. This takes a total 5 instructions.
int swapOddEvenBits(int x)
{
return ( ((x & 0x) >> 1) | ((x & 0x) << 1) );
}
On No
For 32-bit integers:
result = ((x & 0x) >> 1) | ((x & 0x) << 1);
Dave
On Nov 12, 2:12 am, sudhir mishra wrote:
> Write a program to swap odd and even bits in an unsigned integer with as few
> instructions as possible (e.g., bit 0 and bit 1 are swapped, bit 2 and bit 3
> are swap
