Test if a given number is power of two in one line in C

In binary representation, a number is power of two if it has only one on bit and others bits as off.
For ex: binary representation of 8 is 1000 (contains only one on bit) whereas, that of 5 is 0101(contains two on bits).

If a number is power of two, for example 8, bitwise and of 8 and 7 is 0

Code:

#include <stdio.h>

int main(){
     int n = 8;
     if((!(n & (n-1)))){  // & is bitwise and whereas ! is logical not
          printf("The number is power of two\n");
     }else{
           printf("Not a power of two\n");
     }
     return 0;
}

However, this does not give correct result for n = 0. For n = 0, the above expression would be 1, thus printing, "the number is power of two". To accommodate, the case for n = 0, use the following code.

 #include <stdio.h>

int main(){
     int n = 8;
     if((n && !(n & (n-1)))){     // && is logical and, ! is logical not and & is bitwise and
          printf("The number is power of two\n");
     }else{
           printf("Not a power of two\n");
     }
     return 0;
}