Доброго вечера!
Потихоньку решаю ассемблерную часть из задачника А.В. Столярова. По итогу решения задачки 3.20 «Получи на вводе два числа. Выведи их сумму, разность и произведение» родил данный код:
%include"stud_io.inc"
global _start
section .bss
FstArg resd 1
SndArg resd 1
string resb 10
summ resd 10
diff resd 10
prod resd 20
section .data
ErrMsg db "There must be two numbers divided with one space symbol"
factor db 10
divider dw 10
section .text
_start xor eax, eax; prepare char buffer
xor ebx, ebx; prepare current number holder (arg calculation)
xor ecx, ecx; prepare number counter
xor edx, edx; prepare data register
mov dword [FstArg], 0
; prepare 1st argument
mov dword [SndArg], 0
; prepare 2nd argument
NRead: GETCHAR ; read a char
cmp eax, ' '; do we have space character?
je ZChck ; then check - has anything already been printed?
jmp NxtStg ; if we dont have space char - continue alalyz char
ZChck: cmp ecx, 0 ; has anything already been printed?
je ErrPrt ; if not then print error message
jmp ArgPr1 ; prepare 1st argument
NxtStg: cmp eax, '0'; if we have ascii symbol less than '0'
jb ErrPrt ; then print error message
cmp eax, '9'; if we have ascii symbol more than '9'
ja ErrPrt ; then print error message
sub eax, '0'; convert ascii code into a number
inc ecx ; increase number counter
mov [string + ecx - 1], al
; write another number into array
jmp NRead ;
ArgPr1 xor eax, eax; prepare the accamulator
lp: mov ebx, [string + ecx - 1]
; get the current number from the array
mul byte [factor]
; multiplying accumulator value by 10
add eax, ebx; adding to accumulator current number
loop lp
add dword [FstArg], eax
; we got complete 1st argument now
mov ecx, 10 ; prepare counter for string cleanup
StrClUp:mov byte [string + ecx - 1], 0
; set current byte to zero
loop StrClUp
xor eax, eax; prepare char buffer
xor ebx, ebx; prepare current number holder (arg calculation)
xor ecx, ecx; prepare the counter
NRead2 :GETCHAR ; read a char
cmp eax, 10 ; do we have break line character?
je ZChck2 ; then check - has anything already been printed?
jmp NxtStg2 ; if we dont have space char - continue alalyz char
ZChck2: cmp ecx, 0 ; has anything already been printed?
je ErrPrt ; if not then print error message
jmp ArgPr2 ; prepare 2nd argument
NxtStg2:cmp eax, '0'; if we have ascii symbol less than '0'
jb ErrPrt ; then print error message
cmp eax, '9'; if we have ascii symbol more than '9'
ja ErrPrt ; then print error message
sub eax, '0'; convert ascii code into a number
inc ecx ; increase number counter
mov [string + ecx - 1], al
; write another number into array
jmp NRead2 ;
ErrPrt: PRINT ErrMsg
ArgPr2: xor eax, eax; prepare the accamulator
lp2: mov ebx, [string + ecx - 1]
; get the current number from the array
mul byte [factor]
; multiplying accumulator value by 10
add eax, ebx; adding to accumulator current number
loop lp2
add dword [SndArg], eax
; we got complete 2nd argument now
PRINT "SUM" ; <<<< sum opperation
PUTCHAR 10
add eax, [FstArg]
lp3: cmp eax, 0 ; have we finished?
je SumPrt
div dword [divider]
inc ecx
mov [summ + ecx - 1], edx
jmp lp3
SumPrt: xor ebx, ebx; PRINT SUM
lp4: mov ebx, [string + ecx - 1]
; copying a number from string array
add ebx,'0' ; converting a number from binary to ascii format
PUTCHAR bl ; print numbers of the string array
loop lp4 ; one by one
PUTCHAR 10
xor eax, eax
xor ebx, ebx
xor ecx, ecx
xor edx, edx
PRINT "DIFF" ; <<<< sub operation
PUTCHAR 10
mov eax, [FstArg]
mov ebx, [SndArg]
sub eax, ebx; do we have negative difference?
jns lp4 ; if not - proceed with the standart algoritm
not eax ; else take additional actions - invert
add eax, 1 ; and add 1 - so we convert neg num into positiv one
PUTCHAR "-"
lp5: cmp eax, 0 ; (have we finished?) if no start work with pos num
je SubPrt
div dword [divider]
inc ecx
mov [diff + ecx - 1], edx
jmp lp5
SubPrt: xor ebx, ebx; PRINT SUB
lp6: mov ebx, [diff + ecx - 1]
; copying a number from string array
add ebx, '0'; converting a number from binary to ascii format
PUTCHAR bl ; print numbers of the string array
loop lp6 ; one by one
PUTCHAR 10
xor eax, eax
xor ebx, ebx
xor ecx, ecx
xor edx, edx
PRINT "PROD" ; <<<< mul operation
PUTCHAR 10
mov eax, [FstArg]
mov ebx, [SndArg]
mul ebx
mov ecx, 20
lp7: cmp eax, 0 ; (have we finished?) if no start work with pos num
je MulPrt
div dword [divider]
dec ecx
mov [prod + ecx], edx
jmp lp7
cmp edx, 0 ; does the second multiplyer register is empty?
je MulPrt ; then just print the number
mov eax, edx; else start ro work with it too
lp8: cmp eax, 0 ; (have we finished?) if no start work with pos num
je MulPrt
div dword [divider]
dec ecx
mov [prod + ecx], edx
jmp lp8
MulPrt: nop
lp9: mov ebx, [prod + ecx]
; copying a number from string array
inc ecx
add ebx, '0'; converting a number from binary to ascii format
PUTCHAR bl ; print numbers of the string array
loop lp9 ; one by one
PUTCHAR 10
При запуске компоновщика получаю данный поток нецензурной брани в свой адрес:
~/assembly$ ld -m elf_i386 TwoNumSumDifMul.o -o TwoNumSumDifMul
TwoNumSumDifMul.o: в функции «..@10.str»:
TwoNumSumDifMul.asm:(.text+0xfc): перемещение обрезано по месту: R_386_8 у неопределённого символа «.data»
Перемещено hobbit из general
