CareerCup

m^ pow(2, n-1)

I can't agree it more. Perfect answer.

you should also check to make sure n > 0. Otherwise this doesn't hold.

Good answer. Here are versions to set, clear and flip n-th bit of m:

int set_nth_bit(const int m, const int n)
{
    return m | (1 << n);
}

int clear_nth_bit(const int m, const int n)
{
    return m & ~(1 << n);
}

int flip_nth_bit(const int m, const int n)
{
    return m ^ (1 << n);
}

this is assuming it's a big endian machine.

int func1(int n, int m)
{
if ( (n>>(m-1)) % 2==0)
return (n | (1<<(m-1)));
else
return (n ^ (1<<(m-1)));
}

Sorry do you mind explaining your solution a bit. I dont really know how to use these shifting bits left right and stuff

shifting n right by m-1 bit will bring m-th bit at lease significant position.

If the resultant number is even then m-th bit is 0, OR with one will alter the bit. that in in the IF part.

else the resultant number is odd , XOR with 1 will alter the bit.

I would be lying if i said i really did understand that.. What resources can i use to learn bit manipulation in full? I mean books or online tuts

Ke solution is more perfect, I should have realized in both the cases XOR will work. no need to check even and odd..

One lil hint I can give about shifting left and right.. rest you can search on google.. I learned like that only..

in numbers Right shift by 3 ( n >> 3) mean dividing n by 2*2*2
in binary if n = 00100011 n>>3= 00000100

in numbers left shift by 3 ( n << 3) mean multiplying n by 2*2*2.
in binary if n = 00011000 n<<3= 11000000

If its 1 change it to 0. Thats what the interviewer meant.

If its 1 change it to 0. Thats what the interviewer meant.

Funtion flip(m, n){
Return m^=1<<n
}

should be m ^=1<<(n-1)