What does the following function $f()$ in C return?

int f(unsigned int N)
{
unsigned int counter = 0;

while (N > 0)
{
counter += N & 1;
N = N >> 1;
}

return counter == 1;
}

Given function takes an unsigned integer $N$.

Inside the function:

$counter=0$

Now the loop runs while:

$N>0$

In each iteration:

$counter += N\ \&\ 1$

Here, $N\ \&\ 1$ checks the last bit of $N$.

If the last bit is $1$, then $counter$ increases by $1$.

If the last bit is $0$, then $counter$ does not increase.

After that:

$N=N>>1$

This shifts the bits of $N$ one position to the right.

So, the loop checks all bits of $N$ one by one.

Therefore, $counter$ stores the number of $1$ bits in the binary representation of $N$.

Finally, the function returns:

$counter==1$

This condition is true only when the binary representation of $N$ contains exactly one $1$.

A number which has exactly one $1$ in binary form is a power of $2$.

For example:

$2=(10)_2$

$4=(100)_2$

$8=(1000)_2$

So, the function returns $1$ if $N$ is a power of $2$, otherwise it returns $0$.