|
Table of Content | |
|
Table of Content | |
| CHAPTER
NINE: ARITHMETIC AND LOGICAL OPERATIONS (Part 8) |
|
| 9.9 -
Sample Programs 9.9.1 - Converting Arithmetic Expressions to Assembly Language 9.9.2 - Boolean Operations Example |
9.9.3 -
64-bit Integer I/O 9.9.4 - Packing and Unpacking Date Data Types |
| 9.9 Sample Programs | |
This chapter's sample programs demonstrate several important concepts including extended precision arithmetic and logical operations, arithmetic expression evaluation, boolean expression evaluation, and packing/unpacking data.
9.9.1 Converting Arithmetic Expressions to Assembly Language
The following sample program (Pgm9_1.asm on the companion CD-ROM) provides some examples of converting arithmetic expressions into assembly language:
; Pgm9_1.ASM
;
; Several examples demonstrating how to convert various
; arithmetic expressions into assembly language.
.xlist
include stdlib.a
includelib stdlib.lib
.list
dseg segment para public 'data'
; Arbitrary variables this program uses.
u word ?
v word ?
w word ?
x word ?
y word ?
dseg ends
cseg segment para public 'code'
assume cs:cseg, ds:dseg
; GETI- Reads an integer variable from the user and returns its
; its value in the AX register.
geti textequ <call _geti>
_geti proc
push es
push di
getsm
atoi
free
pop di
pop es
ret
_geti endp
Main proc
mov ax, dseg
mov ds, ax
mov es, ax
meminit
print
byte "Abitrary expression program",cr,lf
byte "---------------------------",cr,lf
byte lf
byte "Enter a value for u: ",0
geti
mov u, ax
print
byte "Enter a value for v: ",0
geti
mov v, ax
print
byte "Enter a value for w: ",0
geti
mov w, ax
print
byte "Enter a non-zero value for x: ",0
geti
mov x, ax
print
byte "Enter a non-zero value for y: ",0
geti
mov y, ax
; Okay, compute Z := (X+Y)*(U+V*W)/X and print the result.
print
byte cr,lf
byte "(X+Y) * (U+V*W)/X is ",0
mov ax, v ;Compute V*W
imul w ; and then add in
add ax, u ; U.
mov bx, ax ;Save in a temp location for now.
mov ax, x ;Compute X+Y, multiply this
add ax, y ; sum by the result above,
imul bx ; and then divide the whole
idiv x ; thing by X.
puti
putcr
; Compute ((X-Y*U) + (U*V) - W)/(X*Y)
print
byte "((X-Y*U) + (U*V) - W)/(X*Y) = ",0
mov ax, y ;Compute y*u first
imul u
mov dx, X ;Now compute X-Y*U
sub dx, ax
mov cx, dx ;Save in temp
mov ax, u ;Compute U*V
imul V
add cx, ax ;Compute (X-Y*U) + (U*V)
sub cx, w ;Compute ((X-Y*U) + (U*V) - W)
mov ax, x ;Compute (X*Y)
imul y
xchg ax, cx
cwd ;Compute NUMERATOR/(X*Y)
idiv cx
puti
putcr
Quit: ExitPgm ;DOS macro to quit program.
Main endp
cseg ends
sseg segment para stack 'stack'
stk byte 1024 dup ("stack ")
sseg ends
zzzzzzseg segment para public 'zzzzzz'
LastBytes byte 16 dup (?)
zzzzzzseg ends
end Main
The following sample program (Pgm9_2.asm on the companion CD-ROM) demonstrates how to manipulate boolean values in assembly language. It also provides an example of Demorgan's Theorems in operation.
; Pgm9_2.ASM
;
; This program demonstrates DeMorgan's theorems and
; various other logical computations.
.xlist
include stdlib.a
includelib stdlib.lib
.list
dseg segment para public 'data'
; Boolean input variables for the various functions
; we are going to test.
a byte 0
b byte 0
dseg ends
cseg segment para public 'code'
assume cs:cseg, ds:dseg
; Get0or1- Reads a "0" or "1" from the user and returns its
; its value in the AX register.
get0or1 textequ <call _get0or1>
_get0or1 proc
push es
push di
getsm
atoi
free
pop di
pop es
ret
_get0or1 endp
Main proc
mov ax, dseg
mov ds, ax
mov es, ax
meminit
print
byte "Demorgan's Theorems",cr,lf
byte "-------------------",cr,lf
byte lf
byte "According to Demorgan's theorems, all results "
byte "between the dashed lines",cr,lf
byte "should be equal.",cr,lf
byte lf
byte "Enter a value for a: ",0
get0or1
mov a, al
print
byte "Enter a value for b: ",0
get0or1
mov b, al
print
byte "---------------------------------",cr,lf
byte "Computing not (A and B): ",0
mov ah, 0
mov al, a
and al, b
xor al, 1 ;Logical NOT operation.
puti
putcr
print
byte "Computing (not A) OR (not B): ",0
mov al, a
xor al, 1
mov bl, b
xor bl, 1
or al, bl
puti
print
byte cr,lf
byte "---------------------------------",cr,lf
byte "Computing (not A) OR B: ",0
mov al, a
xor al, 1
or al, b
puti
print
byte cr,lf
byte "Computing not (A AND (not B)): ",0
mov al, b
xor al, 1
and al, a
xor al, 1
puti
print
byte cr,lf
byte "---------------------------------",cr,lf
byte "Computing (not A) OR B: ",0
mov al, a
xor al, 1
or al, b
puti
print
byte cr,lf
byte "Computing not (A AND (not B)): ",0
mov al, b
xor al, 1
and al, a
xor al, 1
puti
print
byte cr,lf
byte "---------------------------------",cr,lf
byte "Computing not (A OR B): ",0
mov al, a
or al, b
xor al, 1
puti
print
byte cr,lf
byte "Computing (not A) AND (not B): ",0
mov al, a
xor al, 1
and bl, b
xor bl, 1
and al, bl
puti
print
byte cr,lf
byte "---------------------------------",cr,lf
byte 0
Quit: ExitPgm ;DOS macro to quit program.
Main endp
cseg ends
sseg segment para stack 'stack'
stk byte 1024 dup ("stack ")
sseg ends
zzzzzzseg segment para public 'zzzzzz'
LastBytes byte 16 dup (?)
zzzzzzseg ends
end Main
This sample program (Pgm9_3.asm on the companion CD-ROM) shows how to read and write 64-bit integers. It provides the ATOU64 and PUTU64 routines that let you convert a string of digits to a 64-bit unsigned integer and output a 64-bit unsigned integer as a decimal string to the display.
; Pgm9_3.ASM
;
; This sample program provides two procedures that read and write
; 64-bit unsigned integer values on an 80386 or later processor.
.xlist
include stdlib.a
includelib stdlib.lib
.list
.386
option segment:use16
dp textequ <dword ptr>
byp textequ <byte ptr>
dseg segment para public 'data'
; Acc64 is a 64 bit value that the ATOU64 routine uses to input
; a 64-bit value.
Acc64 qword 0
; Quotient holds the result of dividing the current PUTU value by
; ten.
Quotient qword 0
; NumOut holds the string of digits created by the PUTU64 routine.
NumOut byte 32 dup (0)
; A sample test string for the ATOI64 routine:
LongNumber byte "123456789012345678",0
dseg ends
cseg segment para public 'code'
assume cs:cseg, ds:dseg
; ATOU64- On entry, ES:DI point at a string containing a
; sequence of digits. This routine converts that
; string to a 64-bit integer and returns that
; unsigned integer value in EDX:EAX.
;
; This routine uses the algorithm:
;
; Acc := 0
; while digits left
;
; Acc := (Acc * 10) + (Current Digit - '0')
; Move on to next digit
;
; endwhile
ATOU64 proc near
push di ;Save because we modify it.
mov dp Acc64, 0 ;Initialize our accumulator.
mov dp Acc64+4, 0
; While we've got some decimal digits, process the input string:
sub eax, eax ;Zero out eax's H.O. 3 bytes.
WhileDigits: mov al, es:[di]
xor al, '0' ;Translates '0'..'9' -> 0..9
cmp al, 10 ; and everything else is > 9.
ja NotADigit
; Multiply Acc64 by ten. Use shifts and adds to accomplish this:
shl dp Acc64, 1 ;Compute Acc64*2
rcl dp Acc64+4, 1
push dp Acc64+4 ;Save Acc64*2
push dp Acc64
shl dp Acc64, 1 ;Compute Acc64*4
rcl dp Acc64+4, 1
shl dp Acc64, 1 ;Compute Acc64*8
rcl dp Acc64+4, 1
pop edx ;Compute Acc64*10 as
add dp Acc64, edx ; Acc64*2 + Acc64*8
pop edx
adc dp Acc64+4, edx
; Add in the numeric equivalent of the current digit.
; Remember, the H.O. three words of eax contain zero.
add dp Acc64, eax ;Add in this digit
inc di ;Move on to next char.
jmp WhileDigits ;Repeat for all digits.
; Okay, return the 64-bit integer value in eax.
NotADigit: mov eax, dp Acc64
mov edx, dp Acc64+4
pop di
ret
ATOU64 endp
; PUTU64- On entry, EDX:EAX contain a 64-bit unsigned value.
; Output a string of decimal digits providing the
; decimal representation of that value.
;
; This code uses the following algorithm:
;
; di := 30;
; while edx:eax <> 0 do
;
; OutputNumber[di] := digit;
; edx:eax := edx:eax div 10
; di := di - 1;
;
; endwhile
; Output digits from OutNumber[di+1]
; through OutputNumber[30]
PUTU64 proc
push es
push eax
push ecx
push edx
push di
pushf
mov di, dseg ;This is where the output
mov es, di ; string will go.
lea di, NumOut+30 ;Store characters in string
std ; backwards.
mov byp es:[di+1],0 ;Output zero terminating byte.
; Save the value to print so we can divide it by ten using an
; extended precision division operation.
mov dp Quotient, eax
mov dp Quotient+4, edx
; Okay, begin converting the number into a string of digits.
mov ecx, 10 ;Value to divide by.
DivideLoop: mov eax, dp Quotient+4 ;Do a 64-bit by
sub edx, edx ; 32-bit division
div ecx ; (see the text
mov dp Quotient+4, eax ; for details).
mov eax, dp Quotient
div ecx
mov dp Quotient, eax
; At this time edx (dl, actually) contains the remainder of the
; above division by ten, so dl is in the range 0..9. Convert
; this to an ASCII character and save it away.
mov al, dl
or al, '0'
stosb
; Now check to see if the result is zero. When it is, we can
; quit.
mov eax, dp Quotient
or eax, dp Quotient+4
jnz DivideLoop
OutputNumber: inc di
puts
popf
pop di
pop edx
pop ecx
pop eax
pop es
ret
PUTU64 endp
; The main program provides a simple test of the two routines
; above.
Main proc
mov ax, dseg
mov ds, ax
mov es, ax
meminit
lesi LongNumber
call ATOU64
call PutU64
printf
byte cr,lf
byte "%x %x %x %x",cr,lf,0
dword Acc64+6, Acc64+4, Acc64+2, Acc64
Quit: ExitPgm ;DOS macro to quit program.
Main endp
cseg ends
sseg segment para stack 'stack'
stk byte 1024 dup ("stack ")
sseg ends
zzzzzzseg segment para public 'zzzzzz'
LastBytes byte 16 dup (?)
zzzzzzseg ends
end Main
This sample program demonstrates how to pack and unpack data using the Date data type introduced in Chapter One.
; Pgm9_4.ASM
;
; This program demonstrates how to pack and unpack
; data types. It reads in a month, day, and year value.
; It then packs these values into the format the textbook
; presents in chapter two. Finally, it unpacks this data
; and calls the stdlib DTOA routine to print it as text.
.xlist
include stdlib.a
includelib stdlib.lib
.list
dseg segment para public 'data'
Month byte ? ;Holds month value (1-12)
Day byte ? ;Holds day value (1-31)
Year byte ? ;Holds year value (80-99)
Date word ? ;Packed data goes in here.
dseg ends
cseg segment para public 'code'
assume cs:cseg, ds:dseg
; GETI- Reads an integer variable from the user and returns its
; its value in the AX register.
geti textequ <call _geti>
_geti proc
push es
push di
getsm
atoi
free
pop di
pop es
ret
_geti endp
Main proc
mov ax, dseg
mov ds, ax
mov es, ax
meminit
print
byte "Date Conversion Program",cr,lf
byte "-----------------------",cr,lf
byte lf,0
; Get the month value from the user.
; Do a simple check to make sure this value is in the range
; 1-12. Make the user reenter the month if it is not.
GetMonth: print
byte "Enter the month (1-12): ",0
geti
mov Month, al
cmp ax, 0
je BadMonth
cmp ax, 12
jbe GoodMonth
BadMonth: print
byte "Illegal month value, please re-enter",cr,lf,0
jmp GetMonth
GoodMonth:
; Okay, read the day from the user. Again, do a simple
; check to see if the date is valid. Note that this code
; only checks to see if the day value is in the range 1-31.
; It does not check those months that have 28, 29, or 30
; day months.
GetDay: print
byte "Enter the day (1-31): ",0
geti
mov Day, al
cmp ax, 0
je BadDay
cmp ax, 31
jbe GoodDay
BadDay: print
byte "Illegal day value, please re-enter",cr,lf,0
jmp GetDay
GoodDay:
; Okay, get the year from the user.
; This check is slightly more sophisticated. If the user
; enters a year in the range 1980-1999, it will automatically
; convert it to 80-99. All other dates outside the range
; 80-99 are illegal.
GetYear: print
byte "Enter the year (80-99): ",0
geti
cmp ax, 1980
jb TestYear
cmp ax, 1999
ja BadYear
sub dx, dx ;Zero extend year to 32 bits.
mov bx, 100
div bx ;Compute year mod 100.
mov ax, dx
jmp GoodYear
TestYear: cmp ax, 80
jb BadYear
cmp ax, 99
jbe GoodYear
BadYear: print
byte "Illegal year value. Please re-enter",cr,lf,0
jmp GetYear
GoodYear: mov Year, al
; Okay, take these input values and pack them into the following
; 16-bit format:
;
; bit 15 8 7 0
; | | | |
; MMMMDDDD DYYYYYYY
mov ah, 0
mov bh, ah
mov al, Month ;Put Month into bit positions
mov cl, 4 ; 12..15
ror ax, cl
mov bl, Day ;Put Day into bit positions
mov cl, 7 ; 7..11.
shl bx, cl
or ax, bx ;Create MMMMDDDD D0000000
or al, Year ;Create MMMMDDDD DYYYYYYY
mov Date, ax ;Save away packed date.
; Print out the packed date (in hex):
print
byte "Packed date = ",0
putw
putcr
; Okay, the following code demonstrates how to unpack this date
; and put it in a form the standard library's LDTOAM routine can
; use.
mov ax, Date ;First, extract Month
mov cl, 4
shr ah, cl
mov dh, ah ;LDTOAM needs month in DH.
mov ax, Date ;Next get the day.
shl ax, 1
and ah, 11111b
mov dl, ah ;Day needs to be in DL.
mov cx, Date ;Now process the year.
and cx, 7fh ;Strip all but year bits.
print
byte "Date: ",0
LDTOAM ;Convert to a string
puts
free
putcr
Quit: ExitPgm ;DOS macro to quit program.
Main endp
cseg ends
sseg segment para stack 'stack'
stk byte 1024 dup ("stack ")
sseg ends
zzzzzzseg segment para public 'zzzzzz'
LastBytes byte 16 dup (?)
zzzzzzseg ends
end Main
|
Table of Content | |
Chapter Nine: Arithmetic And Logical
Operations (Part 8)
27 SEP 1996