MASM

MASM 8086 Assembler in Ubuntu or Windows 7(x64) using DOSBox

Run MASM 8086 Assembler in Ubuntu or Windows 7(x64) using DOSBox

Here’s how to run 16 bit DOS executables like the MASM assembler or Turbo C compiler in Ubuntu (GNU/Linux) or in 64 bit editions of Windows 7 using DOSBox, a DOS environment emulator. DOSBox is available for Linux as well as Windows.

DOSBox installation

For Ubuntu users (using repository)
Open the terminal and type in the following commands to download and install DOSbox in Ubuntu
sudo apt-get update
sudo apt-get install dosbox
You will find it installed under Applications_Menu->Games->DOSBox Emulator

For other GNU/Linux users
Download DOSbox from below link.
Open terminal and cd to the directory containing the downloaded tar.gz file. Type in the following commands to build and install :
tar -xzvf dosbox-0.74.tar.gz
cd dosbox-0.74
./configure
make
Check the src subdir for the binary.

For Windows users
Download DOSbox from below.
Run the downloaded .exe file and install it like any other software.

Now that you’ve installed DOSBox, you’ll be able to run any 16bit or 32bit DOS executable inside it.

Download the 8086 MASM Assembler from below. The zip file contains the following files :
masm.exe, tasm.exe, link.exe, bin2hex.exe, exe2bin.exe, td.exe, edit.com and debug.exe

Windows users extract the .zip file into C:\ so that the path C:\8086 contains all the above mentioned files. GNU/Linux users can extract it and place it in say /home/imran/8086

Launch DOSBox and type the following commands :
For Linux users : 
mount c /home/imran/8086
c:
For Windows : 
mount c c:\8086
c:

DOSBox running in Ubuntu

Now the contents of the folder /home/imran/8086 or c:\8086 is mounted as c: drive inside the DOS emulator. You can assemble programs inside DOSBox as you do in your Microprocessor Lab under Windows XP; i.e your usual sequence of commands -
edit file.asm
masm file.asm
link file
debug file.exe

When you are done, type exit to quit DOSBox.

P.S : For GNU/Linux users, there’s an alternative assembler known as the NASM. NASM is considered to be one of the most popular assemblers for GNU/Linux.

Click to Downloads :
DOSBox for Linux : dosbox-0.74.tar.gz
DOSBox for Windows : DOSBox0.74-win32-installer.exe
8086 Assembler : 8086_Assembler.zip

MASM Programs

8086 ASSEMBLY PROGRAM FOR ADDITION OF TWO 8 BIT NUMBERS

data segment

a db 09h

b db 02h

c dw ?

data ends

code segment

assume cs:code,ds:data

start:

mov ax,data

mov ds,ax

mov al,a

mov bl,b

add al,bl

mov c,ax

int 3

code ends

end start

C:\TASM>masm an8add.asm Microsoft (R) Macro Assembler Version 5.00 Copyright (C) Microsoft Corp 1981-1985, 1987.  All rights reserved. Object filename [an8add.OBJ]: Source listing  [NUL.LST]: Cross-reference [NUL.CRF]:   50402 + 450254 Bytes symbol space free       0 Warning Errors       0 Severe  Errors C:\TASM>link an8add.obj Microsoft (R) Overlay Linker  Version 3.60 Copyright (C) Microsoft Corp 1983-1987.  All rights reserved. Run File [AN8ADD.EXE]: List File [NUL.MAP]: Libraries [.LIB]: LINK : warning L4021: no stack segment C:\TASM>debug an8add.exe -g AX=0B0B  BX=0002  CX=0022  DX=0000  SP=0000  BP=0000  SI=0000  DI=0000 DS=0B97  ES=0B87  SS=0B97  CS=0B98  IP=0011   NV UP EI PL NZ NA PO NC 0B98:0011 CC            INT     3 -d 0B97:0000 0B97:0000  09 02 0B 0B 00 00 00 00-00 00 00 00 00 00 00 00   ................ 0B97:0010  B8 97 0B 8E D8 A0 00 00-8A 1E 01 00 02 C3 A3 02   ................ 0B97:0020  00 CC 86 72 FF 77 15 8A-86 70 FF 2A E4 50 B8 FD   ...r.w...p.*.P.. 0B97:0030  05 50 FF 36 24 21 E8 77-63 83 C4 06 FF 36 24 21   .P.6$!.wc....6$! 0B97:0040  B8 0A 00 50 E8 47 5E 83-C4 04 5E 8B E5 5D C3 90   ...P.G^...^..].. 0B97:0050  55 8B EC 81 EC 84 00 C4-5E 04 26 80 7F 0A 00 74   U.......^.&....t 0B97:0060  3E 8B 46 08 8B 56 0A 89-46 FC 89 56 FE C4 5E FC   >.F..V..F..V..^. 0B97:0070  26 8A 47 0C 2A E4 40 50-8B C3 05 0C 00 52 50 E8   &.G.*.@P.....RP. -q C:\TASM>

CENG 222 – Computer Organization Lab Work 9

Standard I/O In 8086

Standard input output operations  are performed INT 21h interrupt in assembly.

Int 21h acts like a function. The operation to perform in interrupt no 21h is determined by the contents of the AH register.

Read the entry regarding Int 21h interrupt in “Supported Interrupt Functions” section of Emu8086 help (“help”->”Documentation and Tutorials”->“Supported Interrupt Functions”)

Pay attention to Int 21h functions where the value of AH is 1,2,9,0Ah

Exercises

1) Write an assembly program to read a character and print the same character back to screen

2) The user will enter a 2 digit decimal number write an assembly program to store the number in CX register.

3) Write an assembly program to print the English alphabet (capital letters only)

4) design an assembly function (that uses a variable for return value) that reads 5 characters from the keyboard. Call the function in main to get the characters and write the program to print the characters.

CENG 222 – Computer Organization Lab Work 8

Variables

Syntax for a variable declaration:

name DB value

name DW value

DB - stays for Define Byte.

DW - stays for Define Word.

name - can be any letter or digit combination, though it should start with a letter. It's possible to declare unnamed variables by not specifying the name (this variable will have an address but no name).

value - can be any numeric value in any supported numbering system (hexadecimal, binary, or decimal), or "?" symbol for variables that are not initialized.

Reference Address:

http://www.yecd.com/os/8086%20assembler%20tutorial%20for%20beginners%20(part%203).htm

.model small .stack 100h .data VAR1 DB 5 var2 DW 1A2Bh .code main proc      mov ax,@data      mov ds,ax      MOV AL, var1      MOV BX, var2      mov ax, 4c00h      int 21h main endp end main

Some previous Examples to remember...

MOV AX, [1BFFH]    ; Copy 2-bytes (word) data stored at address 1BFF to AX

MOV [BX], 20h          ; Copy the word (2-bytes) data to the memory

                        ; block started at BX. (*BX = 0020h)

MOV WORD PTR [BX], 20h ; Exactly the same as above

MOV BYTE PTR [BX], 20h                ; Copy 1 byte (*BX = 20h)

Offset command

When used with a variable offset command gives the address of the variable.

MOV BX, offset var1               ; Copy address of var1 to BX

MOV DL, [offset var2]            ; Copy 1-byte data at var2 to DL

LEA (Load effective address) instruction

If we use “offset”, we need the following instruction to get the address of a data:

MOV DX, offset var1

Here, msg is defined as name of the data itself. Another way to do this without using offset:

LEA DX, var1

Both instructions above, loads the address of var1 into DX.

Experiment 1

Analyze the assembly code below:

title Register test program .model small .stack 100h .data msg db "AB",0dh,0ah,'$' .code main proc      mov ax,@data      mov ds,ax ; Add your code here ; ------------------ ; ------------------      mov ah,9      mov dx,offset msg      int 21h      mov ax, 4c00h      int 21h main endp end main

The program above prints “AB” to the console. The message “AB” is stored at msg in data segment. Therefore, “offset msg” is the starting address of the string. Modify the code according to the following experiments:

1-      Copy the second byte stored at “msg” (which is ‘B’ currently) to the first byte. The output should be “BB”. In order to do that, load 2-bytes data stored at “msg” into a 16-bits register. Copy one part of the register to the other part. Then, write your register back to msg.

2-      Copy the first byte stored at “msg” (which is ‘A’ currently) to the second byte. The output should be “AA”. To achieve this, load the first byte stored at “msg” into one 8-bits register (high or low part) and copy it onto the other part and write the data stored in register back to msg.

3-      Load the data stored at “msg” into one register and swap the high and low parts of the register using a temporary 8-bits register. Then write back your “swapped” register onto msg. The output should be “BA”.

Experiment 2

title Logical .model small .stack 100h .data varA db ? varB db ? varC db ? varX db ? .code main proc        mov ax,@data        mov ds,ax        ; Add your code below main endp end main

In the assembly code above, four variables are defined without initialized. Use debug mode to test your program and see registers’ and memory’s status after you make necessary modifications to compute some logical operations:

1 – Assign values to all variables except varX.

            varA = 10101010b

            varB = 11001100b

            varC = 11110000b

2 – Compute the following logical operations:

            a)         varX = varA OR (varB AND varC)

            b)         varX = varA OR varB OR (NOT varC)

Experiment 3

Define variables facparam and facreturn. Write an assembly function that calculates the factorial of the number stored in facparam variable and write the result to facreturn variable. The registers should remain unchanged at the end of the function call.

Use the function to calculate 3! and 5!.

CENG 222 – Computer Organization

Lab Work 7

CALL and RET instructions

CALL instruction is used for calling a subroutine/procedure/function.

Usage:

CALL <address or procedure name>

RET instruction is used for returning back to caller procedure from the called procedure. RET has no arguments. Analyze the code given in experiment.

Using stack

Stack is used for temporarily storing data to memory. It is also used by CALL and RET to store caller instruction address. Hence, it is critically important to use stack properly:

Leave stack as how you find!

It means that you need to get all the data out which you store in the stack. Do not forget any data in the stack. You can use it temporarily!

PUSH instruction stores a data in the stack.

PUSH <word data or word register>

POP instruction retrieves a data from the stack.

POP <word memory block or word register>

Examples:

PUSH AX

POP BX

EXPERIMENTS

1)

Write the following C code in assembly using no jump instructions (use loop command instead) You should not use other registers to store the value of CX temporarily. You should use the stack properly to store the value of CX properly.

Hint: Don't forget that loop command only operates on CX and to use a inner loop you must remember the value of CX for the outer loop at each iteration.

int b =0;

for(int i=0; i<5;i++)
            for(int j=0; j<10;j++)

                        b = I + 2*j;

2)

Write an assembly function that calculates factorial. Use the function to calculate 3! and 5!.

3)

Write an assembly program that calculates first 10 Fibonacci numbers. Numbers should be observed in dx consequently (since we did not learn how to print at the moment we can simulate it by observing the register in emulator).

Hint: Use the stack to reverse the order. Once you have pushed all the items, pop command will give you the last item.

CENG 222 – Computer Organization - Lab Work 6

Exercises

Re-write the following C codes in assembly.

1. While loop

   int a = 125;
   int b = 0;
   while(a>0){
     b += a;
     a -=2;
   }

2. While loop with a condition

   int a = 125;
   int b = 0;
   while(a>0){
     b += a;
     if(a>10)
       a -= 2;
     else
       a--;
   }

3. For loop

   b=0;
   for(i=0;i<10;i++)
     b = b+ i*2;
   

4. Double for loop

   b=0;
   for(i=0;i<5;i++)
     for(j=0;j<5;j++)
       b = b+ j*2;

5. Do-while loop with a condition

   int a = 125;
   int b = 0;
   do{
     b += a;
     if(a>10)
        a -= 2;
     else
       a--;
   }while(a>0)

6. Collatz conjecture

   int a = 21;
   while(a != 1){
     if(a%2 == 1)
       a = 3*a+1;
     else
       a = a/2;
   }