Even or odd, count the set bits, multiply by shifting

Four questions about one number, none of them answered with arithmetic. Is 182 odd, what is it times eight, what are its bottom four bits, and how many of its eight bits are ones. The answers land in d, hl, e and c.

The instructions of the Example before this one treat a register as a number. These four treat the same register as eight bits side by side, which is the other way to read one and often the cheaper way.

You need to know: the "8-bit and 16-bit arithmetic, logic and bits" lecture and the "jp, jr and the conditions" lecture. What is new here is the carry flag as a way out of a register, srl drops the bit that falls off the bottom into C and a jump reads it.

bit 0, a is C's n & 1 without building the mask and without changing a, and it sets Z from the bit it found backwards: Z goes to 1 when the bit was 0, which is the opposite of what you expect the first time. So jr z means "the bit was clear", and d comes out at 00 here because 182 is even.

Multiplying by eight is three doublings, since every place a bit moves left doubles what it is worth, and 182 times 8 is 1456, which no byte holds. So the number is widened into hl first, two instructions with ld h, 0 because the byte is unsigned, and then add hl, hl doubles the whole pair in one instruction. hl comes out at 05B0, which is 1456. There is no shift that takes a pair, so add hl, hl is what a program writes when it is hl being doubled.

and 0x0F keeps the four bits the mask has set and clears everything else, so e is 6, the 6 of 0xB6. That is how any field is taken out of a packed value: mask what you want, then shift it down to the bottom if it was not there already.

The loop runs eight times, once per bit, and does C's count += n & 1; n >>= 1; with the & 1 done by the shift itself. srl a moves every bit one place down and the bit that falls off the bottom lands in C, so jr nc skips the inc when it was a zero. c comes out at 05, which is the number of ones in 10110110, and a is empty by the time the loop ends, since eight shifts push every bit out of it.

Try changing N equ 182 to N equ 183, one more. d becomes 01 because the number is now odd, hl becomes 05B8, which is 1464, c becomes 6 and e becomes 7.