Find if every even bit is set to 0 using bitwise operators

Question

I have a 32 bit int I can only access it 8 bits at a time. I need to find out if every even bit is set to 0 and return 0 if its true and 1 otherwise.

So far I am going to split my int using shifts into 4, 8 bit variables. int a, b, c, d

Now I am going to not them so now I will test if the bit is set to 1 instead of 0. To test if its set to 1 I will and them by 01010101.

Now I dont know how to tell if every even bit is set to 1. I cannot use if/for/while loops or any conditional statements and need to use bitwise operators. Any ideas????

Answer 1

OK, so you've created a bit mask. (01010101)

 if ((value & bit_mask) == bit_mask)

then you know that each bit that was set in bit_mask is also set in value.

---

UPDATE: (after reading the question properly)

You want to check if every second bit is set to 0. (Not set to 1, as my incorrect answer above checks for)

There are two equally valid approaches: We make the bit mask the opposite (10101010)

Then use the OR operator:

if ((value | bit_mask) == bit_mask)

This checks that each bit that was zero in the bit_mask is zero in value.

The second approach is to make the bit mask the same (01010101) and use the AND operator:

if ((value & bit_mask) == 0)

This checks that each bit that is one in the bit_mask is zero in value.

Answer 2

EDIT: I got confused by the original question and followed OP in the negation thing - so basically this solves the reverse problem. Andrew Sheperd's edited solution starts back from the original problem and solves it in 1 step. Rudy Velthuis also offers an interesting approach.

If your bytevalue AND 01010101 == 01010101 all bits selected by the mask are 1 in the original byte value.

In sortof pseudo C:

unsigned char mask = 0x55;

if ((byteval & mask) == mask) {
    printf ("all set");
}

or a slightly fancier xor based variation

unsigned char mask = 0x55;

if (!((byteval & mask) ^ mask)) {
    printf ("all set");
}

Btw, the if is very easy to get rid of for the final result ...

Answer 3

There is no need to test every individual byte against a mask of 0x55. Just "or" the bytes together and test the result against the mask:

return ((a | b | c | d) & 0x55 != 0);

Any even bit set to 1 will make the result of the "and" not be 0 anymore, so it will return 1. If all even bits are 0, then 0 is returned.