Multiply and divide, with the remainder

Two unit conversions, one in each direction. The first turns 365 days into 8760 hours and the second turns 1000 seconds into 16 minutes and 40 seconds, which answers both questions at once because a division produces the remainder on its way to the quotient.

The M68K does each of these in one instruction, mulu and divu, and so do MIPS and RISC-V. The Z80 has neither. There is no multiply instruction and no divide instruction anywhere in the set, so both of them are loops you write, and this is the rung of the ladder where the machine costs you the most.

You need to know: the "8-bit and 16-bit arithmetic, logic and bits" lecture and the "Loops and djnz" lecture. What is new here is that both algorithms are the long multiplication and long division you were taught at school, done in base 2, where a digit is either 0 or 1 and multiplying by the base is a shift.

The multiplication looks at the multiplier one bit at a time from the bottom up. srl a drops the lowest bit of 24 into C, jr nc skips the addition when that bit was a 0, and sla e with rl d under it doubles the multiplicand so that the next bit up is worth twice as much. Eight passes, whatever the numbers, and the product is kept in hl because 365 times 24 needs a great deal more than the eight bits a has. de comes out at 2238, which is 8760.

sla e and rl d are the 16 bit shift from the arithmetic lecture: sla e puts the top bit of e into C and rl d brings it in at the bottom of d, so the two of them are one shift of the pair. The low half goes first, because the carry has to be produced before the instruction that consumes it.

The division is the same idea run backwards. hl holds what is left of the dividend and a holds the remainder being built, and every pass shifts the whole 24 bit thing one place left: add hl, hl moves the top bit of hl into C and rla brings it into the bottom of a. Then cp c asks whether the divisor fits in what the remainder has become, and when it does, sub c takes it away and inc l writes a 1 into the quotient. The quotient is written into the bottom of hl as the dividend leaves the top of it, which is why one register does both jobs and why inc l is enough: the bit it sets was shifted in as a 0 a moment earlier.

hl comes out at 0010, which is 16 minutes, and a at 28, which is the 40 seconds left over. On the M68K those two answers arrive packed into one register and are pulled apart with a swap; here they are simply in the two registers the loop used, and the remainder is free.

The whole program is 134 instructions for two sums the other machines do in two. A multiplication by a constant is much cheaper, because you know the bits in advance: 24 is 16 plus 8, so four add hl, hl, a copy kept when the number has been doubled three times, and one add hl, de at the end would do it with no loop at all.

Try changing ld b, 8 to ld b, 4. The loop looks at only the bottom four bits of 24, which are 1000, so de comes out at 0B68, which is 2920: the multiplier became 8 and the bits above the fourth were never looked at.