where is the problem?

#include <iostream> #include <string> using namespace std; int main() { int numb{2}; int ToPower ; cout << "Enter number : "; cin >> ToPower; for (int i = 1; i<=5; i++) { if (numb^i == ToPower) { cout << "Number" << ToPower<<" is power of 2" << endl; } else { cout << "Number isn*t power of two" << endl; } }
  • It is better to use unsigned, otherwise you still have to handle the minimum negative number, which is not a power of two. - vp_arth

2 answers 2

The problem is that the ^ operation in C ++ (and many other languages) does not mean exponentiation, but bitwise exclusive or .

To build a two to a nonnegative degree k easiest thing to do is to simply

 1 << k

And to check whether the number v a power of two, there is a famous trick :

 bool isPowerOfTwo = v && !(v & (v - 1));

If you are familiar with the binary system, you will be pleased to understand why it works.

  • and if the compiler supports built-in functions ... for a power of two, the Hamming code (popcount, the number of non-zero bits) is 1. Many processors have this as one instruction. - Swift
  • one
    @Swift: In the standard C ++ popcount, if I'm not mistaken, no. Therefore, the code that uses it will be intolerable. If portability is not needed, here is just about this case. - VladD
  • Yes, I know, just added, it may be useful to someone. Moreover, there are cases and methods when portability is provided by the programmer’s code for the required platforms. With rollback to the portable version - Swift

In C ++, there is no exponentiation operator. The ^ operator in C ++ means a bitwise exclusive OR operator.

In addition, the output of messages, whether the number is a power of 2, must be moved out of the cycle.

You can use an approach similar to yours, but when the original number is divided by 2, until you get 1 and the remainder of the division will be equal to 0.

Below is a demonstration program.

 #include <iostream> int main() { while ( true ) { const unsigned int Base = 2; unsigned int n; std::cout << "Enter a non-negative number (0 - exit): "; if ( not ( std::cin >> n ) || ( n == 0 ) ) break; unsigned int i = 0; unsigned int m = n; while ( m != 1 && m % Base == 0 ) { ++i; m /= Base; } if ( m == 1 ) { std::cout << n << " is equal to " << Base << " in power of " << i << std::endl; } else { std::cout << n << " is not power of " << Base << std::endl; } } return 0; }

The output of the program to the console may, for example, look like this.

 Enter a non-negative number (0 - exit): 1 1 is equal to 2 in power of 0 Enter a non-negative number (0 - exit): 2 2 is equal to 2 in power of 1 Enter a non-negative number (0 - exit): 4 4 is equal to 2 in power of 2 Enter a non-negative number (0 - exit): 8 8 is equal to 2 in power of 3 Enter a non-negative number (0 - exit): 16 16 is equal to 2 in power of 4 Enter a non-negative number (0 - exit): 32 32 is equal to 2 in power of 5 Enter a non-negative number (0 - exit): 3 3 is not power of 2 Enter a non-negative number (0 - exit): 4 5 is not power of 2 Enter a non-negative number (0 - exit): 7 7 is not power of 2 Enter a non-negative number (0 - exit): 9 9 is not power of 2 Enter a non-negative number (0 - exit): 0

This approach can be used to check whether a number is a power of some other specified number, that is, not necessarily a two. To do this, it is enough to replace in the program the value of the constant Base with any number at your discretion.

Or you can use the function presented in the noexcept qualifier