I have to write a c++ function which swaps nth and least significant bit of an int. I found some examples and did this:
v1 = v1 ^ ((((v1&1) ^ (v1>>n)&1) << LSBpos) | (((v1&1) ^ (v1>>n)&1) << n));
cout<<v1;
v1 is an int. v1&1 is the value of LSB. LSBpos should be the position of LSB but I don't know how to get it. There are explanations of how to get the position of LSB that is set or clear, but I just need that position, whether it is set or not.
You don't need to know the position of the LSB. And this is great because due to endianness it could be at multiple places!
Let us find some help: How do you set, clear, and toggle a single bit?:
Checking a bit
You didn't ask for this, but I might as well add it. To check a bit, shift the number n to the right, then bitwise AND it:
bit = (number >> n) & 1U;Changing the nth bit to x
Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:
number ^= (-x ^ number) & (1UL << n);
And go for it!
int swap_nth_and_lsb(int x, int n)
{
// let to the reader: check the validity of n
// read LSB and nth bit
int const lsb_value = x& 1U;
int const nth_value = (x>> n) & 1U;
// swap
x ^= (-lsb_value) & (1UL << n);
x ^= /* let to the reader: set the lsb to nth_value */
return x;
}
In the comment, OP said "I have to write one line of code for getting result". Well, if the one-line condition holds, you can start from the above solution and turn it into a one-liner step by step.
