Syntax error: incomplete instruction.
.model small .stack 100 .data line1 db 03,12,15,14,01 avg db ? .code mov ax,@data mov ds,ax lea si,line1 mov cx,0005h mov al,00+1 l1:add al,[si] inc si loop l1 mov ah,00h mov bh,05h div bh mov [si],al mov ah,4ch int 21h end
No. Assembly language is a low-level symbolic language that needs to be translated (assembled) to produce the machine code. The reverse of assembly is disassembly, where machine code is disassembled to produce code that is similar to assembly but has no symbolic names or comments. Disassembly is essentially a human-readable version of machine code whereas assembly is the code written by a human in order to produce the machine code, the only language the machine actually understands.
data segment strs db 'madam',0 strr db 5 dup(0) data ends code segment assume cs:code,ds:data,es:data start:mov ax,data mov ds,ax mov es,ax mov cx,6 lea si,strs lea di,strr l1:mov al,strs[si] mov strr[di],al inc si dec di loop l1 mov cx,6 repe cmpsb jz l3 mov bx,0000h jmp l4 l3:mov bx,1111h l4:int 3h code ends end start
org 100h .data str1 db "Computer" str2 db "computer" mes1 db "string are same $" mes2 db "string are different $" .code assume cs:code,ds:data start: mov ax,@data mov ds,ax mov es,ax mov si,offset str1 mov di,offset str2 cld mov cx,8 repe cmpsb mov ah,9 jz skip lea dx,mes2 jmp over skip: lea dx,mes1 over: int 21h mov ax,4c00h int 21h end start ret
8096
Code for An Assembly Language Program to find 2's Complement of given binary number in Assembly LanguageData Segment num db 00000010B Data Ends Code Segment Assume cs:code, ds:data Begin: mov ax, data mov ds, ax mov es, ax mov ah, 0000h mov al, num NOT al mov bl, al adc al, 00000001B mov bl, al Exit: mov ax, 4c00h int 21h Code Ends End Begin
mov eax,cr0 or eax,1 mov cr0,eax jmp cs:@pm @pm:
•The mov instruction carries out assignment in 8086 assembly language. •It which allows us place a number in a register or in a memory location (a variable) i.e. it assigns a value to a register or variable. Example: Store the ASCII code for the letter A in register bx.•A has ASCII code 65D (01000001B, 41H) •The following mov instruction carries out the task:•mov bx, 65d•We could also write it as:•mov bx, 41h•or mov bx, 01000001b•or mov bx, 'A'
Code segment assume cs:code mov si,1500h mov di,1600h mov ax,[si] mov bx,[si+2] add ax,bx mov [di],ax mov [di+2],dx mov ah,4ch int 21h code ends end
.model small .stack 100 .data line1 db 03,12,15,14,01 avg db ? .code mov ax,@data mov ds,ax lea si,line1 mov cx,0005h mov al,00+1 l1:add al,[si] inc si loop l1 mov ah,00h mov bh,05h div bh mov [si],al mov ah,4ch int 21h end
.model small .stack .data m db 'the no is odd $' m1 db 'the no is even $' a db 04h b db 02h .code mov ax,@data mov ds,ax mov ah,0 mov al,a mov bl,b div bl cmp ah,00h je l1 mov dx,offset m jmp l l1: mov dx,offset m1 jmp l l: mov ah,09 int 21h mov ah,4ch int 21h end
No. Assembly language is a low-level symbolic language that needs to be translated (assembled) to produce the machine code. The reverse of assembly is disassembly, where machine code is disassembled to produce code that is similar to assembly but has no symbolic names or comments. Disassembly is essentially a human-readable version of machine code whereas assembly is the code written by a human in order to produce the machine code, the only language the machine actually understands.
Mov ax,[2020h] mov bx,[2022h] mul bl mov [2024h],ax hlt
hi frenz here is an ALP on fibbonacci series...... source code: mov si,2000(2000 is a address location of si register) mov cx,05 (cx is a count register for how many values we have to given) mov al,00 (initially we have 2 initialize al register at starting point....) mov [si],al mov bl,01 inc si mov [si],bl add al,bl mov dl,al mov al,bl mov bl,al inc si mov [si,dl dec cx jnz (address of the location add al,bl jump) hlt
data segment strs db 'madam',0 strr db 5 dup(0) data ends code segment assume cs:code,ds:data,es:data start:mov ax,data mov ds,ax mov es,ax mov cx,6 lea si,strs lea di,strr l1:mov al,strs[si] mov strr[di],al inc si dec di loop l1 mov cx,6 repe cmpsb jz l3 mov bx,0000h jmp l4 l3:mov bx,1111h l4:int 3h code ends end start
code segment assume cs:code,ds:code mov bx,1200h mov cx,[bx] mov ax,01h l1:mul cx dec cl jnz l1 mov[bx+2],ax mov ah,4ch int 21h code ends end
The airport code for Moranbah Airport is MOV.