LLX > Neil Parker > Apple II > Multiplying and Dividing on the 6502

One of the more common questions asked by new 6502 programmers seems to be, "Hey...there's no multiply instruction! How do I multiply on the 6502?" This article presents some answers to this question, and as a bonus, answers to the companion question, "How do I divide on the 6502?"

Of course the methods shown below are not the only ways to multiply and divide, nor even the fastest. They are, however, probably the easiest ways to understand.

Most of the routines below are for unsigned integers, since these are the easiest kinds of numbers to work with. Extending them to signed integers is discussed briefly near the end. (The next logical step, numbers with fractional parts, would require a whole new article.)

The easiest way to multiply on the 6502 is not to do it at all.

Actually, that statement isn't nearly as useless as it seems. Sometimes the nature of the problem is such that with a little rewriting, the multiplication can be eliminated entirely. This happens often in loops, if the loop index has to be multiplied by something. As an example, consider the following BASIC-like code fragment:

FOR I = 5 TO 100 STEP 5 J = I*23 Do something with J NEXT I

This can be rewritten like this:

J = 115: REM 115 = 5*23 FOR I = 5 TO 100 STEP 5 Do something with J J = J+115 NEXT I

This kind of rewriting is called *strength reduction*.
Optimizing compilers will usually notice when this is possible, and
automatically rewrite the code to take advantage of it.

The problem with this, of course, is that many multiplications
*can't* be eliminated by rewriting. For these cases, the rule
is, "Don't work any harder than you have to." If all you need is to
multiply
by a constant, the code for that will usually be much more efficient
than a general routine to multiply any two arbitrary numbers.

Skipping over the trivial cases of multiplying by zero or one, the
easiest constant to multiply by is two, because all you need are the
`ASL`

and `ROL`

instructions. For a single-byte
result:

ASL NUM

Or for a two-byte result:

ASL NUM ROL NUM+1

Extending this to wider numbers should be obvious.

Multiplying by 4, 8, 16, or higher powers of two is simply a matter of multiplying by two enough times to get the result. For example, to multiply a one-bye number by four:

ASL NUM ASL NUM

Or for two-byte numbers:

ASL NUM ROL NUM+1 ASL NUM ROL NUM+1

Multiplying by a constant other than a power of two is more difficult, but it can be done by adding up the results of several power-of-two multiplications. This will generally require the use of temporary memory locations to hold intermediate results.

For example, consider multiplication by 3. To do this, we note that
for any number `x`

, `3x = 2x + x`

- that is, we
can multiply by three by multiplying by two, and adding the original
number to the result. Of course this means the original number has to
be kept around somewhere so it will be available for the addition. For
two-byte numbers:

LDA NUM ;Start with RESULT = 2*NUM ASL A STA RESULT LDA NUM+1 ROL A STA RESULT+1 CLC LDA NUM ADC RESULT STA RESULT LDA NUM+1 ADC RESULT+1 STA RESULT+1 ;RESULT = 3*NUM

Or we can multiply by 10 using the fact that ```
10x = 8x +
2x
```

(or, factoring out a 2, `10x = 2(4x + x)`

):

LDA NUM ;Start with RESULT = NUM STA RESULT LDA NUM+1 STA RESULT+1 ASL RESULT ROL RESULT+1 ;RESULT = 2*NUM ASL RESULT ROL RESULT+1 ;RESULT = 4*NUM CLC LDA NUM ADC RESULT STA RESULT LDA NUM+1 ADC RESULT+1 STA RESULT+1 ;RESULT = 5*NUM ASL RESULT ROL RESULT+1 ;RESULT = 10*NUM

So how do we figure out what powers of two we need to add to get the answer? Easy: Just write the binary equivalent of the constant, and look for the 1 bits. The position of each 1 bit shows which powers of two need to be added to get the answer. For example,

3 (decimal) = 11 (binary) |+-- 1 +--- +2 -- 3, i.e. 1x + 2x = 3x 10 (decimal) = 1010 (binary) | +-- 2 +---- +8 -- 10, i.e. 2x + 8x = 10x 25 (decimal) = 11001 (binary) || +-- 1 |+----- 8 +------ +16 --- 25, i.e. x + 8x + 16x = 25x.

But what if you need to multiply two numbers, and you don't know what either one is in advance? In this case, you need a general multiply-anything-by-anything routine.

A reasonable way to multiply arbitrary numbers is to do it the way you learned in school, with pencil and paper. Consider an example:

654 x 321 ----- 654 1308 1962 ------ 209934

If the steps are written out as an algorithm, it goes something like this:

- Set the answer to 0.
- Repeat as many times as there are digits in the bottom number:
- Remove the rightmost digit of the bottom number.
- Multiply the top number by the digit just removed.
- Add the result to the answer, shifted one more place to the left each time.

The secret of binary multiplication is that *it works exactly like
decimal multiplication*, as long as you use the addition and
multiplication tables for binary digits:

+| 0 1 x| 0 1 --+----- --+---- 0| 0 1 0| 0 0 1| 1 10 1| 0 1

The algorithm in the binary case is slightly simpler than the decimal algorithm:

- Set the answer to 0.
- Repeat as many times as there are bits in the bottom number:
- Remove the rightmost bit of the bottom number.
- If the bit removed was 1, add the top number to the answer, shifted one more place to the left each time.

Here's an example:

110 x 101 ----- 110 110 ----- 11110

This is quite easy to turn into machine language - the only tricky thing in the code below is that instead of shifting the added number to the left, it shifts the answer one place to the right each time, catching the lost bit in a memory location. Here it is for one-byte numbers:

LDA #0 ;Initialize RESULT to 0 LDX #8 ;There are 8 bits in NUM2 L1 LSR NUM2 ;Get low bit of NUM2 BCC L2 ;0 or 1? CLC ;If 1, add NUM1 ADC NUM1 L2 ROR A ;"Stairstep" shift (catching carry from add) ROR RESULT DEX BNE L1 STA RESULT+1

Note that though we were multiplying one-byte numbers, the result requires two bytes. The general rule for multiplication is that the number of bytes in the result will be equal to the number of bytes in the first number, plus the number of bytes in the second number.

The method is easily extendable to wider numbers. Here's a routine for multiplying two-byte numbers, giving a four-byte result:

LDA #0 ;Initialize RESULT to 0 STA RESULT+2 LDX #16 ;There are 16 bits in NUM2 L1 LSR NUM2+1 ;Get low bit of NUM2 ROR NUM2 BCC L2 ;0 or 1? TAY ;If 1, add NUM1 (hi byte of RESULT is in A) CLC LDA NUM1 ADC RESULT+2 STA RESULT+2 TYA ADC NUM1+1 L2 ROR A ;"Stairstep" shift ROR RESULT+2 ROR RESULT+1 ROR RESULT DEX BNE L1 STA RESULT+3

At this point it seems to be common for people to ask for additional routines to multiply even wider numbers. Rather than bloat this article with even more routines for numbers of various sizes, I encourage readers to look back over the algorithm just described and the examples that implement it - once these are understood, writing additional routines is straightforward.

Passing over the trivial case of dividing by 1, the easiest number to divide by is 2:

LSR NUM

Or, for two-byte numbers:

LSR NUMHI ROR NUMLO

This replaces the original number with the result, and leaves the remainder in the carry flag.

As with multiplication, dividing by higher powers of two is simply a matter of repeating the above. The remainder can be found by saving the bits shifted off into the carry flag - the first bit shifted out is the lowest bit of the remainder, the second bit shifted is the next-to-lowest bit of the remainder, and so on.

Unfortunately, dividing by constants that are not powers of two is harder than multiplying by them - hard enough that it's often easiest just to go for the general divide-anything-by-anything routine, especially if you need the remainder.

As with multiplication, a reasonable division algorithm can be found by looking at the way you would write out a long division with pencil and paper. Consider the following example, written in the manner traditionally taught in U.S. schools:

184 _______ 67 ) 12345 -67 --- 564 -536 ---- 285 -268 ---- 17

Here we divided a *dividend*, 12345, by a *divisor*, 67,
and got *quotient* of 184, with a remainder of 17. Note the
procedure...at each step, we find the largest multiple of the divisor
that is less than the leftmost remaining digits of the dividend, and
subtract. Then we bring down the next digit of the dividend, and repeat
until no more digits are left.

Binary division works exactly the same way, as long as you use the rules for binary digits instead of decimal digits. For example:

10101 _________ 101 ) 1101101 -101 ---- 11 -0 -- 111 -101 ---- 100 -0 --- 1001 -101 ---- 100

Note how much easier this is than the decimal version, mostly because there are only two possible numbers to subtract: 0 or the divisor.

Again, writing code to do this isn't very hard. We will shift the bits of the the dividend, one at a time, into a work area, and then try subtracting the divisor from the work area. If the subtraction succeeded, we replace the work area with the result of the subtraction and record a 1 bit in the quotient. If the subtraction failed, we discard its result and record a 0 bit in the quotient. The process is repeated until all bits of the dividend are used up.

Here's an example that divides the two-byte number NUM1 by the two-byte number NUM2, leaving the quotient in NUM1 and the remainder in REM:

LDA #0 ;Initialize REM to 0 STA REM STA REM+1 LDX #16 ;There are 16 bits in NUM1 L1 ASL NUM1 ;Shift hi bit of NUM1 into REM ROL NUM1+1 ;(vacating the lo bit, which will be used for the quotient) ROL REM ROL REM+1 LDA REM SEC ;Trial subtraction SBC NUM2 TAY LDA REM+1 SBC NUM2+1 BCC L2 ;Did subtraction succeed? STA REM+1 ;If yes, save it STY REM INC NUM1 ;and record a 1 in the quotient L2 DEX BNE L1

Again, I'm not going to bloat this article by showing division routines for different sizes of numbers. Once the method is understood, writing additional routines is straightforward.

The routines shown above are reasonably compact, but not very speedy. If speed is important, different techniques must be used. The best way to get fast multiplication or division on the 6502 is table lookup - i.e., the programmer precomputes the answers to the multiplication or division problems and stores them in memory, and when the program needs to multiply or divide, it looks up the answer in the table.

Of course this means there's a tradeoff between speed and code size. Table lookup works best when the numbers to be multiplied or divided fall in a limited range, otherwise the tables quickly grow to unmanageable sizes - for example, a multiplication table for multiplying two arbitrary one-byte numbers requires more memory than the 6502 can directly access.

Fortunately, with a little bit of algebra, the multiplcation table can be made much smaller, at the cost of a little extra execution time. Consider these two ways of writing the binomial equation:

(a + b)^2 = a^2 + 2*a*b + b^2, (a - b)^2 = a^2 - 2*a*b + b^2.

If the bottom equation is subtracted from the top, we get

(a + b)^2 - (a - b)^2 = 4*a*b,

Or, rearranging,

a*b = ((a + b)^2 - (a - b)^2)/4 = (a + b)^2/4 - (a - b)^2/4.

Thus we can do a multiplication with an addition, two subtractions, a couple of right shifts, and a couple of lookups in a table of squares. And if we're clever about the coding, we can make most of that work trivial - Here are the tricks to make it work:

- Instead of storing a table of
`x^2`

, the table will store`x^2/4`

. This means each table entry needs only two bytes intstead of three. The two low-order bits of each entry are lost, but that's OK, since`(a+b)^2`

and`(a-b)^2`

are guaranteed to lose*exactly the same bits*, so the subtraction cancels the lost bits. - We'll actually store the table twice, once to look up
`(a+b)^2/4`

, and once to look up`(a-b)^2/4`

. The total memory cost is 2048 bytes (2K). - The
`(a-b)^2/4`

table is offset by one. This is so we can store`b`

in the accumulator, and negate it with`EOR #$FF`

without having to add 1. - The tables are page-aligned. This greatly streamlines the indexing.

A brief setup routine is needed to set up four pointers in page 0:

LDA #SSQLO/256 STA PSLO+1 LDA #SSQHI/256 STA PSHI+1 LDA #DSQLO/256 STA PDLO+1 LDA #DSQHI/256 STA PDHI+1

With that done, we can multiply two bytes by putting one in the accumulator and the other in the Y register, and calling this routine:

STA PSLO ;Index into sum table by A STA PSHI EOR #$FF STA PDLO ;Index into diff table by -A-1 STA PDHI LDA (PSLO),Y ;Get (a+y)^2/4 (lo byte) SEC SBC (PDLO),Y ;Subtract (-a+y)^2/4 (lo byte) TAX ;Save it LDA (PSHI),Y ;Get (a+y)^2/4 (hi byte) SBC (PDHI),Y ;Subtract (-a+y)^2/4 (hi byte)

This leaves the product in the accumulator (high byte) and the X register (low byte).

The tables necessary to make this work are too long to show here, so I've provided them in a separate text file. Remember, the tables have to be page-aligned, or the above code won't work.

Up to this point, all the code presented in this article has been
designed to work with unsigned (positive) numbers. The simple
multiply-by-constant routines will also work with negative numbers,
but none of the others will. Generally, if you see a `LSR`

or `ROR`

instruction in one of the above routines, it will
give wrong results for negative numbers.

The problem is that the 6502's `LSR`

and `ROR`

instructions are not sign-preserving. They could be replaced by calls
to subroutines that perform the shift in a sign-preserving manner,
but that's probably not the best way to solve the problem.

The usual way to multiply signed numbers on the 6502 is to compute the sign of the result first, then make both numbers positive, multiply them, and apply the proper sign to the result.

Computing the sign of the result is very easy: If the two original
numbers have the same sign, the result is positive, and if they have
different signs, the result is negative. Assuming the usual
two's-complement representation of negative numbers, the sign of the
result can be found by `EOR`

ing the high bytes of the two
numbers, which leaves the sign of the result in the high bit of the
accumulator and in the `N`

flag.

To make a number positive, first examine its high bit. If the high bit is 0, no action is necessary, otherwise the number will have to negated. The best way to do this depends on where the number is stored - if it's in the accumulator, the quickest way is this:

EOR #$FF CLC ADC #1

If the number to be negated is in a memory location, the quickest way to negate it is to subtract it from 0. For example, to negate a two-byte number:

LDA #0 SEC SBC NUM STA NUM LDA #0 SBC NUM+1 STA NUM+1

Putting it all together, here's a routine that multiplies two signed one-byte numbers, using the one-byte multiply routine from above:

LDA NUM1 ;Compute sign of result EOR NUM2 PHP ;Save it on the stack LDA NUM1 ;Is NUM1 negative? BPL T1 EOR #$FF ;If so, make it positive CLC ADC #1 STA NUM1 T1 LDA NUM2 ;Is NUM2 negative? BPL T2 EOR #$FF ;If so, make it positive CLC ADC #1 STA NUM1 T2 JSR MUL1BYTE ;Do the unsigned multiplication PLP ;Get sign of result BPL T3 LDA #0 ;If negative, negate result SEC SBC RESULT STA RESULT LDA #0 SBC RESULT+1 STA RESULT+1 T3 ...

Doing signed division is a bit trickier. The trick is to preserve the following relation:

dividend = divisor * quotient + remainder

In the unsigned case there's only one way to arrange this: The quotient is always less than or equal to the true mathematical quotient (which may have a fractional part), and the remainder is always positive.

In the signed case, the ability to have signed remainders makes the issue murkier. There are two conventions in common use:

*Floored division:*If the true mathematical quotient has a fractional part, the same choices are made as in the unsigned case: the quotient is always less than (more negative) or equal to the mathematical quotient, and the remainder is always positive.*Toward-zero division:*If the true mathematical quotient has a fractional part, the fractional part is truncated (i.e. the quotient is rounded toward zero), and the remainder has the sign of the dividend.

Signed division is implemented with a wrapper around an unsigned routine, as with signed multiplication, but the wrapper varies depending on which convention you choose. In either case, first compute the sign of the quotient the same way you would compute the sign of the result of a multiplication, and then make both numbers positive, and call the unsigned division routine. What happens next depends on which convention you choose.

*Floored division:*If the quotient should be negative, negate it. Leave the remainder unchanged.*Toward-zero division:*If the quotient should be negative, negate it, and then add one to it. If the quotient should be negative*and*the original dividend was negative, negate the remainder.

The code is analogous to the signed multiplication example above.

LLX > Neil Parker > Apple II > Multiplying and Dividing on the 6502

Original: August 31, 2005

Modified: June 2, 2016