The Art of ASSEMBLY LANGUAGE PROGRAMMING Chapter Nine (Part 3) Table of Content Chapter Nine (Part 5)
 CHAPTER NINE: ARITHMETIC AND LOGICAL OPERATIONS (Part 4) 9.3.4 - Extended Precision Multiplication 9.3.5 - Extended Precision Division

### 9.3.4 Extended Precision Multiplication

Although a 16x16 or 32x32 multiply is usually sufficient there are times when you may want to multiply larger values together. You will use the 80x86 single operand `mul `and `imul `instructions for extended precision multiplication.

Not surprisingly (in view of how `adc `and ```sbb ```work) you use the same techniques to perform extended precision multiplication on the 80x86 that you employ when manually multiplying two values.

Consider a simplified form of the way you perform multi-digit multiplication by hand:

```1) Multiply the first two               2) Multiply 5*2:
digits together (5*3):

123                                     123
45                                      45
---                                      ---
15                                      15
10

3) Multiply 5*1:                       4) 4*3:

123                                     123
45                                      45
---                                      ---
15                                      15
10                                      10
5                                       5
12

5) Multiply 4*2:                       6) 4*1:

123                                     123
45                                      45
---                                     ---
15                                      15
10                                      10
5                                       5
12                                      12
8                                       8
4

7) Add all the partial products together:

123
45
---
15
10
5
12
8
4
------
5535```

The 80x86 does extended precision multiplication in the same manner except that it works with bytes words and double words rather than digits. The figure below shows how this works.

Probably the most important thing to remember when performing an extended precision multiplication is that you must also perform a multiple precision addition at the same time. Adding up all the partial products requires several additions that will produce the result. The following listing demonstrates the proper way to multiply two 32 bit values on a 16 bit processor:

Note: `Multiplier` and `Multiplicand` are 32 bit variables declared in the data segment via the `dword `directive. Product is a 64 bit variable declared in the data segment via the `qword `directive.

```Multiply        proc    near
push    ax
push    dx
push    cx
push    bx

; Multiply the L.O. word of Multiplier times Multiplicand:

mov     ax
word ptr Multiplier
mov     bx
ax                          ;Save Multiplier val
mul     word ptr Multiplicand           ;Multiply L.O. words
mov     word ptr Product
ax            ;Save partial product
mov     cx
dx                          ;Save H.O. word

mov     ax
bx                          ;Get Multiplier in BX
mul     word ptr Multiplicand+2         ;Multiply L.O. * H.O.
0                           ;Don't forget carry!
mov     bx
ax                          ;Save partial product
mov     cx
dx                          ; for now.

; Multiply the H.O. word of Multiplier times Multiplicand:

mov     ax
word ptr Multiplier+2       ;Get H.O. Multiplier
mul     word ptr Multiplicand           ;Times L.O. word
mov     word ptr product+2
ax          ;Save partial product

mov     ax
word ptr Multiplier+2       ;Multiply the H.O.
mul     word ptr Multiplicand+2         ; words together.
0                           ;Don't forget carry!
mov     word ptr Product+4
ax          ;Save partial product
mov     word ptr Product+6
dx

pop     bx
pop     cx
pop     dx
pop     ax
ret
Multiply        endp```

One thing you must keep in mind concerning this code it only works for unsigned operands. Multiplication of signed operands appears in the exercises.

9.3.5 Extended Precision Division

You cannot synthesize a general n-bit/m-bit division operation using the `div `and `idiv` instructions. Such an operation must be performed using a sequence of shift and subtract instructions. Such an operation is extremely messy. A less general operation dividing an n bit quantity by a 32 bit (on the 80386 or later) or 16 bit quantity is easily synthesized using the `div `instruction. The following code demonstrates how to divide a 64 bit quantity by a 16 bit divisor producing a 64 bit quotient and a 16 bit remainder:

```dseg            segment para public 'DATA'
dividend        dword   0FFFFFFFFh
12345678h
divisor         word    16
Quotient        dword   0
0
Modulo          word    0
dseg            ends

cseg            segment para public 'CODE'
assume  cs:cseg
ds:dseg

; Divide a 64 bit quantity by a 16 bit quantity:

Divide64        proc    near

mov     ax
word ptr dividend+6
sub     dx
dx
div     divisor
mov     word ptr Quotient+6
ax
mov     ax
word ptr dividend+4
div     divisor
mov     word ptr Quotient+4
ax
mov     ax
word ptr dividend+2
div     divisor
mov     word ptr Quotient+2
ax
mov     ax
word ptr dividend
div     divisor
mov     word ptr Quotient
ax
mov     Modulo
dx
ret
Divide64        endp
cseg            ends```

This code can be extended to any number of bits by simply adding additional `mov / div / mov `instructions at the beginning of the sequence. Of course on the 80386 and later processors you can divide by a 32 bit value by using `edx `and `eax `in the above sequence (with a few other appropriate adjustments).

If you need to use a divisor larger than 16 bits (32 bits on an 80386 or later) you're going to have to implement the division using a shift and subtract strategy. Unfortunately such algorithms are very slow. In this section we'll develop two division algorithms that operate on an arbitrary number of bits. The first is slow but easier to understand the second is quite a bit faster (in general).

As for multiplication the best way to understand how the computer performs division is to study how you were taught to perform long division by hand. Consider the operation 3456/12 and the steps you would take to manually perform this operation:

This algorithm is actually easier in binary since at each step you do not have to guess how many times 12 goes into the remainder nor do you have to multiply 12 by your guess to obtain the amount to subtract. At each step in the binary algorithm the divisor goes into the remainder exactly zero or one times. As an example consider the division of 27 (11011) by three (11):

```

```

There is a novel way to implement this binary division algorithm that computes the quotient and the remainder at the same time. The algorithm is the following:

```Quotient := Dividend;
Remainder := 0;
for i:= 1 to NumberBits do

Remainder:Quotient := Remainder:Quotient SHL 1;
if Remainder >= Divisor then

Remainder := Remainder - Divisor;
Quotient := Quotient + 1;

endif
endfor```

`NumberBits` is the number of bits in the ```Remainder Quotient Divisor ``` and `Dividend` variables. Note that the ```Quotient := Quotient + 1 ```statement sets the L.O. bit of `Quotient` to one since this algorithm previously shifts `Quotient` one bit to the left. The 80x86 code to implement this algorithm is

```; Assume Dividend (and Quotient) is DX:AX
Divisor is in CX:BX

; and Remainder is in SI:DI.

mov     bp
32          ;Count off 32 bits in BP
sub     si
si          ;Set remainder to zero
sub     di
di
BitLoop:        shl     ax
1           ;See the section on shifts
rcl     dx
1           ; that describes how this
rcl     di
1           ; 64 bit SHL operation works
rcl     si
1
cmp     si
cx          ;Compare H.O. words of Rem

ja      GoesInto        ; Divisor.
jb      TryNext
cmp     di
bx          ;Compare L.O. words.
jb      TryNext

GoesInto:       sub     di
bx          ;Remainder := Remainder -
sbb     si
cx          ;                Divisor
inc     ax              ;Set L.O. bit of AX
TryNext:        dec     bp              ;Repeat 32 times.
jne     BitLoop```

This code looks short and simple but there are a few problems with it. First it does not check for division by zero (it will produce the value 0FFFFFFFFh if you attempt to divide by zero) it only handles unsigned values and it is very slow. Handling division by zero is very simple just check the divisor against zero prior to running this code and return an appropriate error code if the divisor is zero. Dealing with signed values is equally simple you'll see how to do that in a little bit. The performance of this algorithm however leaves a lot to be desired. Assuming one pass through the loop takes about 30 clock cycles[2] this algorithm would require almost 1 000 clock cycles to complete! That's an order of magnitude worse than the DIV/IDIV instructions on the 80x86 that are among the slowest instructions on the 80x86.

There is a technique you can use to boost the performance of this division by a fair amount: check to see if the divisor variable really uses 32 bits. Often even though the divisor is a 32 bit variable the value itself fits just fine into 16 bits (i.e. the H.O. word of `Divisor` is zero). In this special case that occurs frequently you can use the DIV instruction which is much faster.

[2] This will vary depending upon your choice of processor.

 Chapter Nine (Part 3) Table of Content Chapter Nine (Part 5)

Chapter Nine: Arithmetic And Logical Operations (Part 4)
27 SEP 1996