Question

I am reading a code where it is supposed to implement a bit vector using a byte array.
The idea is that the bitvector has the bit set if a number is present at the corresponding position.
E.g. if number 10 is present the bit 10 must be set etc. It is a classic concept and I get it, but I am not sure about the actual implementation.

The part I don’t get is:
bitvector [num / 8] |= 1 << (num % 8);

Where num is the number to set.
If num is 10 then the second byte must be used (num/8 ok so far) but 1 << (num % 8) does not set the second bit of the second byte as it should. Does it?

Answer 1

10 % 8 = 2, therefore 1 << (10 % 8) = bit 2, or the value 4 (100 in binary)

(start counting bits from the right side of the byte, starting at zero). Very simple to verify:

7 -> 7 % 8 = 7, byte[0], Bit 7 (1 << 7).
8 -> 8 % 8 = 0, byte[1], Bit 0 (1 << 0).
9 -> 9 % 8 = 1, byte[1], Bit 1 (1 << 1).
10 -> 10 % 8 = 2, byte[1], Bit 2 (1 << 2).