CareerCup

Can we not find MSB of 32 bit integer using
int MSB = num & ox80000000 ?

floor(log2(number)) this answer is posted elsewhere on this site..

bool msb(unsigned int n)
{
         if (n > ~n)
             return true;
         else
             return false;
}

for(int i=sizeof(int);i>=0;i--)
{
if(num & (1<< i))
break;
}
return i;

if (num = (num << 1) >> 1) msb 1 else 0

if (num == (num << 1) >> 1) msb 1 else 0

if((num&(0x80))!=0) then msb=1; else msb=0;

bool msb(unsigned int N)
{
    N= N>>((sizeof(int)*8)-1);
    if(N%2) return true;
    return false;
}

I think the question is not worded correctly. I think we have to find the highest set bit. If so, the fast solution is to set all bits to the right of the hibit and then subtract as below.

int hibit(unsigned int n) {
    n |= (n >>  1);
    n |= (n >>  2);
    n |= (n >>  4);
    n |= (n >>  8);
    n |= (n >> 16);
    return n - (n >> 1);
}

int findMSB(int a){
    return (a & 0x80000000) ? 31 : (findMSB((a << 1) | 1) - 1);
}

int MSB = !!(num & 0x80000000)