Assignment No. 5

Write 8086 ALP to perform string manipulation. The strings to be accepted from the user is to be stored in data segment of program_l and write FAR PROCEDURES in code segment program_2 for following operations on the string: (a)Concatenation of two strings (b) Number of occurrences of a sub-string in the given string Use PUBLIC and EXTERN directive. Create .OBJ files of both the modules and link them to create an EXE file.

In this program create 3 files as follows
1. file1.asm
2. file2.asm
3. macro.asm

1. file1.asm
——————————————————————————————–
;Assignment Name :X86/64 Assembly language program (ALP)
;String Manipulation : Concatenation of 2 strings &
;Find no. of occurences of substring.
;Accept string from user.
;————————————————————————

extern    con_proc        ; [ FAR PROCRDURE
extern    sub_proc        ;   USING EXTERN DIRECTIVE ]

global    str1,str1_size, str2, str2_size

%include    “macro.asm”

;————————————————————————
section .data
nline        db    10,10
nline_len:    equ    $-nline

msg        db     10,10,10,10,”MIL assignment 07 : String Manipulation”
db    10,”————————————-“,10
msg_len:    equ    $-msg

s1msg        db    10,”Enter string 1    : ”
s1msg_len:    equ    $-s1msg

s2msg        db    10,”Enter string 2    : ”
s2msg_len:    equ    $-s2msg

menu        db    10,”———–MENU———————”
db    10,”1.String Concatenation”
db    10,”2.No. of occurences of substring”
db    10,”3.Exit”
db    10,”Enter choice    : ”
menu_len:    equ    $-menu

emsg        db    10,”INVALID CHOICE!!!!! Please try again! “,10,10,10
emsg_len:    equ    $-emsg

;—————————————————————————
section .bss
buf        resb    2
buf_len:    equ    $-buf

str1        resb    20
str1_len:    equ    $-str1

str2        resb    20
str2_len:    equ    $-str2

str1_size    resw    1
str2_size    resw    1

;————————————————————————–
section .text
global _start

_start:
print    msg,msg_len        ;assignment no.

print    s1msg,s1msg_len
read     str1,str1_len
dec    eax
mov    [str1_size],ax

print    s2msg,s2msg_len
read     str2,str2_len
dec    eax
mov    [str2_size],ax

Disp_Menu:
print    menu,menu_len
read    buf,2
mov    al,[buf]
SUB    AL,30H

C1:      CMP    al,1
JNE    C2
call    con_proc
JMP    Disp_Menu

C2:      CMP    al,2
JNE    C3
call    sub_proc
JMP    Disp_Menu

C3:    CMP    al,3
JNE    err
exit

err:    print    emsg,emsg_len
JMP    Disp_Menu
;——————————————————————————–
2. file2.asm
;———————————————————————
section .data
nline        db    10,10
nline_len:    equ    $-nline

cmsg        db    10,”The Concatenated String is    : ”
cmsg_len:    equ    $-cmsg

ymsg        db    10,”The substring is Present.”,10
db    10,”No. of occurences of substring    : ”
ymsg_len:    equ    $-ymsg

nmsg        db    10,”The substring is not Present.”,10
db    10,”No. of occurences of substring    : ”
nmsg_len:    equ    $-nmsg
;———————————————————————

section .bss
str3        resb    40
str3_len:    equ    $-str3

str3_size    resw    1

sscount        resw    1
cur_add        resq    1
end_add        resq    1

char_ans    resb    4
;———————————————————————

extern    str1,str1_size, str2, str2_size
global    con_proc, sub_proc

%include    “macro.asm”
;———————————————————————
section .text
global    _main
_main:
;    global    con_proc, sub_proc

con_proc:                  ;STRING CONCATINATION PROCEDURE
cld

mov    ecx,0
mov    cx,[str1_size]
mov    esi, str1        ; 1st string

mov    edi, str3        ; concatenated string
rep     movsb

mov    ecx,0
mov    cx,[str2_size]
mov    esi, str2        ; 2nd string

rep     movsb

mov    cx,[str1_size]
add    cx,[str2_size]
mov    [str3_size],cx

print cmsg,cmsg_len
print str3,str3_len

ret
;———————————————————————
sub_proc:                ;substring procedure

mov    esi,str1

CLD

mov    [cur_add],esi        ; store starting address of string1

mov    ecx,esi            ; calculate end address of string1
add    cx,[str1_size]
dec    ecx
mov    [end_add],ecx

back:
mov    edi,str2
xor    ecx,ecx
mov    cx,[str2_size]

repe     cmpsb

jnz     conti
inc     word[sscount]

conti:
inc     word[cur_add]
mov     esi,[cur_add]
cmp      esi,[end_add]
jbe     back

cmp     word[sscount],00
je     no

print     ymsg, ymsg_len
jmp     last

no:    print     nmsg, nmsg_len

last:
mov     ax,[sscount]
call     display_16

ret
;——————————————————————
display_16:
mov     esi,char_ans+3    ; load last byte address of char_ans in rsi
mov     ecx,4            ; number of digits

cnt:    mov     edx,0            ; make rdx=0 (as in div instruction rdx:rax/rbx)
mov     ebx,16        ; divisor=16 for hex
div     ebx
cmp     dl, 09h        ; check for remainder in RDX
jbe      add30
add      dl, 07h
add30:
add     dl,30h        ; calculate ASCII code
mov     [esi],dl        ; store it in buffer
dec     esi            ; point to one byte back

dec     ecx            ; decrement count
jnz     cnt            ; if not zero repeat

print char_ans,4        ; display result on screen
ret
;—————————————————————-

3. macro.asm

————————————————————————————
;macro.asm
;macros as per 64 bit conventions

%macro print 2
mov    eax,4        ;print/write
mov    ebx,1        ;stdout/screen
mov    ecx,%1    ;msg
mov    edx,%2    ;msg_len
int 80h
%endmacro

%macro read 2
mov    eax,3        ;read
mov    ebx,0        ;stdin/keyboard
mov    ecx,%1    ;buf
mov    edx,%2    ;buf_len
int 80h
%endmacro

%macro exit 0
print    nline,nline_len
mov    eax,1    ;exit
int 80h
%endmacro
———————————————————————————————————————-
Steps to run the program as follows

dhokane@dhokane-ThinkCentre-A70:~/MIT IN LINUX/far procedure$ nasm -f elf file1.asm
dhokane@dhokane-ThinkCentre-A70:~/MIT IN LINUX/far procedure$ nasm -f elf file2.asm
dhokane@dhokane-ThinkCentre-A70:~/MIT IN LINUX/far procedure$ ld -m elf_i386 -s -o file1 file1.o -o file2 file2.o
dhokane@dhokane-ThinkCentre-A70:~/MIT IN LINUX/far procedure$ ./file2

MIL assignment 07 : String Manipulation
————————————-

Enter string 1    : abcabcabcabcabc

Enter string 2    : abc

———–MENU———————
1.String Concatenation
2.No. of occurences of substring
3.Exit
Enter choice    : 1

The Concatenated String is    : abcabcabcabcabcabc
———–MENU———————
1.String Concatenation
2.No. of occurences of substring
3.Exit
Enter choice    : 2

The substring is Present.

No. of occurences of substring    : 0005
———–MENU———————
1.String Concatenation
2.No. of occurences of substring
3.Exit
Enter choice    : 3

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