How to function that will test if numbers are power of 2?-CodePudding

I already made program, for one number to test if that number is power of 2. But I have problem, and I don't know how to fix it. But I need to make program that will test more then just one number. I don't know if I explained that good (because I don't know english very well). Do you have any suggestions, what can I change in code ?

#include <stdio.h>

//function prototype for checking power of two
int checkPowerofTwo(int n);

int main()
{
   int num;

   printf("Enter the number you want to test: ");
   scanf("%d", &num);

   if (checkPowerofTwo(num) == 1)
      printf("\n%d is a power of 2\n", num);
   else
      printf("\n%d is not a power of 2\n", num);

   return 0;
}

//function body
int checkPowerofTwo(int x)
{
   //checks whether a number is zero or not
   if (x == 0)
      return 0;

   //true till x is not equal to 1
   while( x != 1)
   {
      //checks whether a number is divisible by 2
      if(x % 2 != 0)
         return 0;
         x /= 2;
   }
   return 1;
}

CodePudding user response：

If I have understood correctly what you need is a loop. For example

int main( void )
{
    while ( 1 )
    { 
       int num;

       printf("Enter the number you want to test (0 - exit ): ");

       if ( scanf("%d", &num) != 1 || num == 0 ) break;

       if ( checkPowerofTwo( num ) )
          printf("\n%d is a power of 2\n", num);
       else
          printf("\n%d is not a power of 2\n", num);
    }

   return 0;
}

Pay attention to that your function will not work correctly if the user will enter a negative number.

So it is better to declare the variable num and the function parameter as having unsigned type unsigned int

CodePudding user response：

How about something like this

for (int i = 0; i<1000; i  ) {
       if (checkPowerofTwo(i) == 1)
            printf("\n%d is a power of 2\n", i);
   }

to test the code?

CodePudding user response：

How to function that will test if numbers are power of 2?

For non-negative integers, there's a trick to do this very quickly:

    if( x & (x - 1) == 0) {
        // Either zero or a power of 2
    }

To also support negative integers you can:

    if(x > 0) {
        if( x & (x - 1) == 0) {
            // It's a power of 2
        }
    } else if(x < 0) {
        if( -x & (-x - 1) == 0) {
            // It's a power of 2
        }
   }

Note that your original code doesn't support negative numbers (and you probably should be using unsigned int everywhere, including using scanf("%u", &num);).

But I need to make program that will test more then just one number.

That's a different problem:

int main() {
    unsigned int num;

    do {
        printf("Enter the positive number you want to test, or 0 to exit: ");
        scanf("%u", &num);
        if(num == 0) {
            return 0;
        }

        if (checkPowerofTwo(num) == 1) {
            printf("\n%u is a power of 2\n", num);
        } else {
            printf("\n%u is not a power of 2\n", num);
        }
    } while(1);   // Loop forever
}

CodePudding user response：

Thank you all. The only thing I did is changing int checkPowerofTwo(int x) to int checkPowerofTwo(int num), changed every intiger x to num and now it works.