Boolean (Bitwise Shift Operators)

[makan007]

Need a boolean function using bitwise shift operators to test if an integer is even or odd.

int m = n >> 1
m = m << 1

if (n == m) then even

else

odd

---
How to write in c++?

[Bazzy]

You almost have it:

int m = n >> 1; //simply added semicolon
m = m << 1;

return n == m; //returns true if n == m, false otherwise 

You only need to add the function prototype now

[hajen]

I have a suggestion:

bool odd(int n) {
/* Returns true is n is odd */

return (n & 1)
}

The logic of it:

An odd number can be written as 2n + 1, where n is an integer (even or odd). Since all powers of two are divisible by two (i.e. can be written as 2n), except 2^0 == 1 which is odd, we can prove that an odd binary number must have its least significant bit set (end with a binary 1).


0 1 0 1 0 0 1 1
0 0 0 0 0 0 0 1 &
----------------
0 0 0 0 0 0 0 1 // The result is the integer 1, which is freely convertable to bool (true) under C++.



This might save you some cycles, although that was probably not an issue to beging with. You could also do this:

return (n % 2);

Which is basically the same thing but perhaps more clear.

[Duthomhas]

Nice, but it doesn't conform to the OP's requirements.

[hajen]

Duoas, you are probably right. But I try to elaborate on these things so that I learn something myself.