Milos "baze" Bazelides, baze_at_baze_au_com
last updated 29.03.2006
I decided to create this collection of Z80 routines for one simple reason - I like magic. And of course, I was fed up with bad code one can find in many embedded devices, web pages, tutorials and such. The routines presented here is what I believe to be the best of its kind or at least very close to it.
However, if you find a bug or optimisation please let me know. My objective is to know and share the best stuff out there. Also, if you feel you've got something that should be posted here drop me a mail. Keep in mind though that any code you submit should be machine independent and of general use. Of course, you'll be guaranteed a honourable mention in the Credits :)
Please don't complain about lack of comments. This is not a coding tutorial but rather a collection of tricks for (more or less) experienced coders. I'm sure it's not that hard to figure out what's going on.
Note: This document is by far not finished yet. I'll continue to add new code in near future. I also think of providing binary images of look-up tables in cases where table generator is not trivial. Also, I'd be glad if some native English speaker would help me to correct numerous grammar and spelling mistakes :)
Input: H = Multiplier, E = Multiplicand, L = 0, D = 0
Output: HL = Product
sla h ; optimised 1st iteration jr nc,$+3 ld l,e add hl,hl ; unroll 7 times jr nc,$+3 ; ... add hl,de ; ...
Input: A = Multiplier, DE = Multiplicand, HL = 0, C = 0
Output: A:HL = Product
add a,a ; optimised 1st iteration jr nc,$+4 ld h,d ld l,e add hl,hl ; unroll 7 times rla ; ... jr nc,$+4 ; ... add hl,de ; ... adc a,c ; ...
Input: DE = Multiplier, BC = Multiplicand, HL = 0
Output: DE:HL = Product
sla e ; optimised 1st iteration rl d jr nc,$+4 ld h,b ld l,c add hl,hl ; unroll 15 times rl e ; ... rl d ; ... jr nc,$+6 ; ... add hl,bc ; ... jr nc,$+3 ; ... inc de ; ...
Input: B = Multiplier, C = Multiplicand (both in range -128..127)
Output: HL = Product
Note: Routine uses one of these two formulas: 2ab = (a + b)^2 - a^2 - b^2 or 2ab = a^2 + b^2 - (a - b)^2, depends if (a + b) overflows or not. Powering by 2 is done by table lookup. 512 bytes long table is aligned to 256 byte boundary and contains entries of form SqrTab[x] = x^2. If we treat one of the operands as fractional number -1..1 premultiplied by 128, 2ab performs native shift of the result into register H. That's especially useful e.g. for x * sin(y). Otherwise we have to shift HL right (divide it by 2). We could divide table entries by 2 instead but that causes loss of precision.
Mul8x8 ld h,SqrTab/256 ld l,b ld a,b ld e,(hl) inc h ld d,(hl) ; DE = a^2 ld l,c ld b,(hl) dec h ld c,(hl) ; BC = b^2 add a,l ; let's try (a + b) jp pe,Plus ; jump if no overflow sub l sub l ld l,a ld a,(hl) inc h ld h,(hl) ld l,a ; HL = (a - b)^2 ex de,hl add hl,bc sbc hl,de ; HL = a^2 + b^2 - (a - b)^2 ; sra h ; uncomment to get real product ; rr l ret Plus ld l,a ld a,(hl) inc h ld h,(hl) ld l,a ; HL = (a + b)^2 or a sbc hl,bc or a sbc hl,de ; HL = (a + b)^2 - a^2 - b^2 ; sra h ; uncomment to get real product ; rr l ret
Square table generator is based on observation that differences between consecutive squares (0, 1, 4, 9, 16, 25, ...) form a sequence of odd numbers. (1, 3, 5, 7, 9, ...). Thus, by adding successive odd numbers iteratively we generate integer squares.
ld hl,SqrTab ; must be a multiple of 256 ld b,l ld c,l ; BC holds odd numbers ld d,l ld e,l ; DE holds squares SqrGen ld (hl),e inc h ld (hl),d ; store x^2 ld a,l neg ld l,a ld (hl),d dec h ld (hl),e ; store -x^2 ex de,hl inc c add hl,bc ; add next odd number inc c ex de,hl cpl ; one byte replacement for NEG, DEC A ld l,a rla jr c,SqrGen
The first thing that should be pointed out here is that the topic is not particularly correct. Actually, the routine is able to multiply any pair of numbers x, y as long as (x + y) <= 127 and (x - y) >= -128. But if x, y are signed 6-bit values these rules are never violated, no overflows occur and no specific checking is needed.
The routine is based on a formula 4xy = (x + y)^2 - (x - y)^2 and uses the same lookup table (see previous chapter) except all table entries are pre-divided by 4 to avoid division (shifting) at the end. An explanation why it works can be found here. In case we leave the table as is, routine nicely handles fixed point multiplications. That means, if we treat one of the operands as fractional number in range (-1, 1) pre-multiplied by 64, integer part of the result gets shifted handily into register H.
Note: SqrTab must be aligned to 256 byte boundary.
Input: B = Multiplier, C = Multiplicand
Output: BC = Product
Mul6x6 ld h,SqrTab/256 ld d,h ld a,b add a,c ; A = x + y ld l,a ld a,b sub c ; A = x - y ld e,a ld a,(de) ; subtract lower byte sub (hl) ; lower byte of (x + y)^2 - (x - y)^2 ld c,a inc h inc d ld a,(de) sbc a,(hl) ; higher byte of (x + y)^2 - (x - y)^2 ld b,a
This is the fastest version I could come up with but there's also slightly slower one which preserves one register pair:
Input: B = Multiplier, C = Multiplicand
Output: HL = Product
Mul6x6 ld h,SqrTab/256 ld a,b sub c ; A = x - y ld l,a ld a,b add a,c ; A = x + y ld c,(hl) inc h ld b,(hl) ; BC = (x - y)^2 ld l,a ld a,(hl) dec h ld l,(hl) ld h,a ; HL = (x + y)^2 or a sbc hl,bc ; HL = (x + y)^2 - (x - y)^2
It's also possible to speed up this routine by having two consecutive look-up tables where first table is negated. Question is, however, if 4 cycles are worth wasting another 512 bytes.
Mul6x6 ld h,SqrTab/256 ld a,b sub c ; A = x - y ld l,a ld a,b add a,c ; A = x + y ld c,(hl) inc h ld b,(hl) ; BC = -(x - y)^2, that's the trick inc h ld l,a ld a,(hl) inc h ld h,(hl) ld l,a ; HL = (x + y)^2 add hl,bc ; HL = (x + y)^2 - (x - y)^2
Division is an awkward arithmetic operation even if it's directly supported by hardware. Thus, we can't expect blazing speed even from well written code. Before you attempt to use any of these routines, please consider these hints:
The routines presented here are mostly based on so-called restoring division algorithm. Although more sophisticated methods exist, implemetation of this algorithm (particularly its left-rotating version) on Z80 is very efficient and straightforward.
Some routines use "undocumented" instruction SLIA (Shift Left Inverted Arithmetic), sometimes also denoted as SLL (Shift Left Logical). SLIA shifts register left and sets the least significant bit to 1 (operation code is CBh 30h..37h). It is used in routines which subtract 16-bit divisor from the 16-bit remainder. As there is no simple way to test whether such subtraction will be successful without actually performing it, it's better to assume it will be successful (hence 1 in the LSB) and possibly make a correction to 0. Doing it the opposite way introduces the overhead of Carry complement (see bellow).
Note: Most division routines leave register B untouched. It enables you to create loops using DJNZ in case you prefer compact code and couple of additional cycles is not an issue.
Input: D = Dividend, E = Divisor, A = 0
Output: D = Quotient, A = Remainder
sla d ; unroll 8 times rla ; ... cp e ; ... jr c,$+4 ; ... sub e ; ... inc d ; ...
The most awkward "feature" of this algorithm is the relation between Carry and the newly determined bit of the result. Successful subtraction (Carry = 0) means that we should set the bit to 1 and vice versa. One possible workaround is to leave Carry as is and complement whole result at the end. This introduces a little overhead but also saves one instruction in the main loop.
Input: D = Dividend, E = Divisor, A = 0, Carry = 0
Output: A = Quotient, E = Remainder
rl d ; unroll 8 times rla ; ... sub e ; ... jr nc,$+3 ; ... add a,e ; ... ... ; ld e,a ; save remainder (if needed) ld a,d ; complement all bits of the result cpl
In case you are really looking to save each cycle you can do so by using another version of the algorithm which never restores remainder nor does it need a Carry complement. The downside is that routine splits into two almost identical tracks of code (I wrote them to columns for better readability).
Input: D = Dividend, E = Divisor, A = 0
Output: D = Quotient, A = Remainder
sla d rla cp e jr c,C1 NC0 sub e slia d C1 sla d rla rla cp e cp e jr c,C2 jr nc,NC1 NC1 sub e slia d C2 sla d rla rla cp e cp e jr c,C3 jr nc,NC2 NC2 sub e slia d C3 sla d rla rla cp e cp e jr c,C4 jr nc,NC3 NC3 sub e slia d C4 sla d rla rla cp e cp e jr c,C5 jr nc,NC4 NC4 sub e slia d C5 sla d rla rla cp e cp e jr c,C6 jr nc,NC5 NC5 sub e slia d C6 sla d rla rla cp e cp e jr c,C7 jr nc,NC6 NC6 sub e slia d C7 sla d
Input: HL = Dividend, C = Divisor, A = 0
Output: HL = Quotient, A = Remainder
add hl,hl ; unroll 16 times rla ; ... cp c ; ... jr c,$+4 ; ... sub c ; ... inc l ; ...
Input: A:C = Dividend, DE = Divisor, HL = 0
Output: A:C = Quotient, HL = Remainder
slia c ; unroll 16 times rla ; ... adc hl,hl ; ... sbc hl,de ; ... jr nc,$+4 ; ... add hl,de ; ... dec c ; ...
We can use the forementioned trick with complementing the result also here but it's not that obvious how removed DEC C balances against additional overhead. In any case, this routine doesn't contain any "undocumented" instructions and might be preferable (?) for that reason.
Input: A:C = Dividend, DE = Divisor, HL = 0
Output: BC = Quotient, HL = Remainder
rl c ; unroll 16 times rla ; ... adc hl,hl ; ... sbc hl,de ; ... jr nc,$+3 ; ... add hl,de ; ... ... rl c rla cpl ld b,a ld a,c cpl ld c,a
Input: E:HL = Dividend, D = Divisor, A = 0
Output: E:HL = Quotient, A = Remainder
add hl,hl ; unroll 24 times rl e ; ... rla ; ... cp d ; ... jr c,$+4 ; ... sub d ; ... inc l ; ...
Input: A:BC = Dividend, DE = Divisor, HL = 0
Output: A:BC = Quotient, HL = Remainder
slia c ; unroll 24 times rl b ; ... rla ; ... adc hl,hl ; ... sbc hl,de ; ... jr nc,$+4 ; ... add hl,de ; ... dec c ; ...
Input: DE:HL = Dividend, C = Divisor, A = 0
Output: DE:HL = Quotient, A = Remainder
add hl,hl ; unroll 32 times rl e ; ... rl d ; ... rla ; ... cp c ; ... jr c,$+4 ; ... sub c ; ... inc l ; ...
Although square roots don't appear in the code as often as multiplication or division, they represent a challenging problem when there's actual need for such calculation. From a coder's point of view it's interesting to explore the set of different approaches one can take and evaluate their pros and cons. Talking about integer square root, let's be specific what it means: given radicant N, the square root of N is the largest integer X that satisfies the condition X * X <= N.
As mentioned earlier in the chapter Integer Multiplication, integer squares can be generated iteratively by adding consecutive odd numbers. The simplest way to calculate square root is then just a reverse of this process; take a number and subtract successive odd integers from it. Number of iterations it takes until the result is negative is the radix.
Input: A = Radicand
Output: B = Radix
ld b,-1 Sqrt8 inc b inc b sub b jr nc,Sqrt8 ; use JP to save 2 T srl b ; translate (2X + 1) to X
This routine is based exactly on the same method, except in 16-bits it's more effective to add negative number rather than to subtract it. The simplicity and virtually no memory requirements are the main advantages of this algorithm. The obvious shortcoming is running time; the time used is proportional to the magnitude of radix.
Input: HL = Radicand
Output: A = Radix
ld a,-1 ld d,a ld e,a Sqrt16 add hl,de inc a dec e dec de jr c,Sqrt16 ; use JP to save 2 T
This elegant algorithm is similar to restoring division. One root digit is generated per iteration such that the partial root converges to the correct answer. Unlike in division, the "divisor" changes as solution progresses and two new digits are shifted into the partial remainder at a time.
Let R be the current remainder and X be the square root approximation found so far. The area that remains is then (R - X^2) and our new remainder becomes R' = 4 * (R - X^2) + AB, where AB are the next two radicand digits. We want to find a better root approximation by adding the largest new digit D such that (2X + D)^2 <= R' + 4X^2 (note that 2X represents shift one digit left in binary and 4X^2 is the area we subtracted in the previous step). This is identical to 4X^2 + 4XD + D^2 <= R' + 4X^2, or even better (4X + D) * D <= R'. In binary, the largest D we can try is 1 so our formula reduces to (4X + 1) <= R' and the trial step becomes R' - (4X + 1). If the result is non-negative, a "1" is generated. Otherwise a "0" is generated and the remainder is restored. Another simplification introduced by binary is that (4X + 1) is nothing else but "01" appended to the partial root.
I recommend searching the web for "integer square root algorithm" if you want to gain more thorough understanding of the math behind it. I created a figure that might help you to understand the basic idea from a geometric point of view.
Input: L:A = Radicand, Carry = 0
Output: D = Radix
ld de,0040h ; 40h appends "01" to D ld h,d sbc hl,de ; unroll 7 times jr nc,$+3 ; ... add hl,de ; ... ccf ; ... rl d ; ... rla ; ... adc hl,hl ; ... rla ; ... adc hl,hl ; ... ... sbc hl,de ; optimised last iteration ccf rl d
Although this method works best on processors with native multiply instruction, it's applicable to Z80 too. The algorithm is a trial and error process: starting with the most significant bit of the intermediate result X, we set the bit to "1" and check whether X * X <= radicand; if the test fails, we correct the bit to "0". Then we progress to the next lower bit (if any) and repeat the procedure. On a Z80, squaring is done best with 512 byte long lookup table (the algorithm in fact performs bisection search on it). This method should be marginally faster than the restoring algorithm but requires more memory.
Note: We can achieve better code optimisation if the table entries are negative (ADD HL,DE is faster than SBC HL,DE). This requires only a minor modification to the square table generator presented in the chapter Integer Multiplication. Table must be aligned to 256 byte boundary.
ld hl,NegSqrTab ; must be a multiple of 256 ld b,l ld c,l ; BC holds odd numbers ld d,l ld e,l ; DE holds squares SqrGen ld (hl),e inc h ld (hl),d ; store x^2 dec h ex de,hl dec bc add hl,bc ; add next odd number dec c ex de,hl inc l jr nz,SqrGen
Now for the actual square root code. First, we will take a look at compact (but slower) version:
Input: DE = Radicand
Output: A = Radix
xor a ld b,128 BiSqrt xor b ; set the trial bit ld h,NegSqrTab/256 ld l,a ld c,(hl) inc h ld h,(hl) ld l,c ; HL = A * A add hl,de ; HL < DE ? jr c,BitOK xor b ; reset bit if the test fails BitOK srl b jr nc,BiSqrt ; use JP to save 2 T
And here's the full-blown unrolled version. The immediate operands in each iteration are 128, 64, 32, 16, 8, 4, 2 and 1.
ld b,NegSqrTab/256 ld a,128 ; optimised 1st iteration ld h,b ld l,a ld c,(hl) inc h ld h,(hl) ld l,c add hl,de jr c,$+3 xor a xor 64 ; unroll 7x (but change immediate operands!) ld h,b ; ... ld l,a ; ... ld c,(hl) ; ... inc h ; ... ld h,(hl) ; ... ld l,c ; ... add hl,de ; ... jr c,$+4 ; ... xor 64 ; ...
This is a very simple linear congruential generator. The formula is x[i + 1] = (5 * x[i] + 1) mod 256. Its only advantage is small size and simplicity. Due to nature of such generators only a couple of higher bits should be considered random.
Input: none
Output: A = pseudo-random number, period 256
Rand8 ld a,Seed ; Seed is usually 0 ld b,a add a,a add a,a add a,b inc a ; another possibility is ADD A,7 ld (Rand8+1),a ret
This generator is based on similar method but gives much better results. It was taken from an old ZX Spectrum game and slightly optimised.
Input: none
Output: HL = pseudo-random number, period 65536
Rand16 ld de,Seed ; Seed is usually 0 ld a,d ld h,e ld l,253 or a sbc hl,de sbc a,0 sbc hl,de ld d,0 sbc a,d ld e,a sbc hl,de jr nc,Rand inc hl Rand ld (Rand16+1),hl ret
Input: HL = number to convert, DE = location of ASCII string
Output: ASCII string at (DE)
Num2Dec ld bc,-10000 call Num1 ld bc,-1000 call Num1 ld bc,-100 call Num1 ld c,-10 call Num1 ld c,b Num1 ld a,'0'-1 Num2 inc a add hl,bc jr c,Num2 sbc hl,bc ld (de),a inc de ret
Hexadecimal conversion operates directly on nibbles and takes advantage of nifty DAA trick.
Input: HL = number to convert, DE = location of ASCII string
Output: ASCII string at (DE)
Num2Hex ld a,h call Num1 ld a,h call Num2 ld a,l call Num1 ld a,l jr Num2 Num1 rra rra rra rra Num2 or F0h daa add a,A0h adc a,40h ld (de),a inc de ret
As this is one of the most typical tasks, why not to do it tricky way? The code snippet here takes a byte from (HL) and prints it. Note that it uses another (shorter) DAA trick as we know that Half Carry is cleared before DAA.
Input: HL = address of data
Output: memory dump
Note: You'll have to replace the PRINT_CHAR macro by actual platform-specific code. Don't forget to preserve the contents of HL!
xor a rld call Nibble Nibble push af daa add a,F0h adc a,40h PRINT_CHAR ; prints ASCII character in A pop af rld ret
The following routine calculates standard CRC-CCITT bit-by-bit using polynomial 1021h. Another common scheme CRC-16 uses polynomial A001h and starts with value 0 (so it's likely that you misinterpret bunch of zeros as valid data). It might be useful to extend the code to use 16-bit byte counter.
Input: DE = address of input data, C = number of bytes to process
Output: HL = CRC
Crc16 ld hl,FFFFh Read ld a,(de) inc de xor h ld h,a ld b,8 CrcByte add hl,hl jr nc,Next ld a,h xor 10h ld h,a ld a,l xor 21h ld l,a Next djnz CrcByte dec c jr nz,Read ret
Note: I haven't tested the results yet so there might be a bug somewhere (most likely wrong polynomial producing bad results).
This is a much faster equivalent of the previous routine. It processes one byte at a time using 512 byte long table. There is a change in algorithm though. Intermediate results are shifted right and polynomial is reversed. It means that even results are reversed (the most significant bit is actually the least significant one and vice versa). Depending on actual use this might be a problem or not (for example, it's suitable if you interoperate with hardware as UARTs send least significant bit first). Even if you decide to adjust result back to correct value, you should still gain more than you loss.
Input: HL = address of input data, BC = number of bytes to process
Output: DE = CRC
Note: CrcTab must be aligned to 256 byte boundary. Table generator uses reverse of 1021h, that is 8408h.
Crc16 ld d,FFh ld a,e Read xor (hl) ex de,hl ld l,a ld a,h ld h,CrcTab/256 xor (hl) inc h ld h,(hl) ex de,hl cpi jp pe,Read ld e,a ret
CRC table generator:
ld hl,CrcTab CrcGen ld d,0 ld e,l ld b,8 CrcByte srl d rr e jr nc,Next ld a,d xor 84h ld d,a ld a,e xor 08h ld e,a Next djnz CrcByte ld (hl),e inc h ld (hl),d dec h inc l jr nz,CrcGen ret
My thanks goes to the following people who contributed to this document:
Slavomir "Busy" Labsky