Question

I am a little bit confused while I’m trying to determine if a number is even / uneven using only logical operators (||, &&).

I’m used to:

if (x%2)
   printf("Number is even");
else
   printf("Number is not even");

Is there any algorithm that determinates the parity of a number ? If there is, can you give me some documentation ?

Best regards,
Duluman Edi

Answer 1

Can you do shifting first? You can accomplish the same “bit” logic with the logical operators if you use just 1 bit, so for example:

int num = 0x14; // 0000 0000 0001 0100
unsigned short shift_num = num << 15; // 0000 0000 0000 0000

if(shift_num && 1)
    printf("it's odd\n");
else
    printf("it's even\n");

So anything non-0 && 1 gives you 1, in this case we shiffted off all but the lowest bit, if it’s even (as above) it leaves you with 0 and 0 && 1 is false so "it's even" is printed. If we use an odd number:

int num = 0x15; // 0000 0000 0001 0101
unsigned short shift_num = num << 15; // 1000 0000 0000 0000

Now there’s a non-0 number there with the 1 so we’ll get "it's odd"