imulinstructions for extended precision multiplication.
sbbwork), you use the same techniques to perform extended precision multiplication on the 80x86 that you employ when manually multiplying two values.
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 ------ 5535The 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:
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
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. add ax, cx ;Add partial product adc dx, 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 add ax, bx ;Add partial product mov word ptr product+2, ax ;Save partial product adc cx, dx ;Add in carry/H.O.! mov ax, word ptr Multiplier+2 ;Multiply the H.O. mul word ptr Multiplicand+2 ; words together. add ax, cx ;Add partial product adc dx, 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 endpOne thing you must keep in mind concerning this code, it only works for unsigned operands. Multiplication of signed operands appears in the exercises.
idivinstructions. 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
divinstruction. 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 endsThis code can be extended to any number of bits by simply adding additional
mov / div / movinstructions at the beginning of the sequence. Of course, on the 80386 and later processors you can divide by a 32 bit value by using
eaxin the above sequence (with a few other appropriate adjustments).
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
NumberBitsis the number of bits in the
Remainder, Quotient, Divisor,and
Dividendvariables. Note that the
Quotient := Quotient + 1statement sets the L.O. bit of
Quotientto one since this algorithm previously shifts
Quotientone 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 BitLoopThis 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, 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.
Divisoris zero). In this special case, that occurs frequently, you can use the DIV instruction which is much faster.