Question

Write an assembly intel 8088 16 bit code MASM,(The 8088 includes has four 16-bit data registers (AX, BX, CX and DX). BX can also be used as an address register for indirect addressing. The most/least significant byte of each register can also be addressed directly)

using push, pop, and stack

print the array 8,9,10,1,2,3,4,5,6,7

please provide the full assembly code.

Answer 1

Registers

The 8088 includes has four 16-bit data registers(ax,bx,cx and dx) bx can also be used as an addressregister fr indirect addressing . the most/least significant byte of each register can also be addressed directly (e.g AL is the LS byte of AX,CH is MS byte of CX.

three bits in a 16 bit program status word(psw)are used to indicate whether the resultof the previous arithmetic/logical instrunction was zero, negative,or generated a carry. An interrupt enable bit controls whether interrup requests on the IRQ pin are recognized

the address of the next instructin to be executed is held in a 16-bit instrunction pointer(IP) regsiter (the program counter)

exeecie: how many bytes can be addressed by a 16- bit value?

a 16- bit stack pointer (SP) is used to implement a stack to support subroutine calls and interupts/ exceptions.

there are also three segment registers(CS,DS,SS) which are used to allow the code. data and stack to be located in any three 64 kbyte" segments" within a 1 megabyte(20-bit) address space a sdescribed below

instruction set

Data Transfer

Transfer of 8nand 16-bit data is done using the MOV instruction. Either the source or destination has to be a register. The other operand can come from another register, from memory,from immediate data (a value encoded in the instrction ) or from a memory location " pointed at " by register BX . For example . if COUNT is the label of a memory location the followingare possibe

: register : move contents of BX to AX

MOV AX,BX

:direct : move contents of AX to memory

MOV COUNT,AX

: immediate: load CX with the value 240

MOV CX,OFOH

:register indirect:move contents of AL

:to memory location in BX

16-bit regsiters can be pushed (SP is first decremented by two and then the value stored at SP) or popped (the value is restored from memory at SP and then SP is incremented by 2 )

for example:

PUSH AX : push contents of AX

POP BX : restore into BX

8088 ASSEMBLY LANGUAGE:

The operand format is destination .source,eg.MOV AX,13 H loads AX with 19

The assembler keeps track of the type of the each symbol and uses it select immediate or direct addressing . This can be changed by using the OFFSET operator to convert a memory reference to a 16-bit value

EXAMPLE:

this is a simpole program that demostrates the mainfeatures of the 8088 instruction set. It uses the INT operatation of invoke MS-DOS to writecharaters to the screen.

: sample 8088 assembly language program. this programprints the printable charaters in a null-teminated string (similar to the unix(" strings" program)

: There is only one "segment" call "code" and the linker can assume DS and CS will be set to the right values for "code". The code begins at offset 100h with in the segment"code" (MS-DOS.COM files).

code segment public

assume cs:code,ds:code

org 100h

start:

mov bx, offset msg ;bx points to string

loop:

mov al, [bx] ;load a charater into al

cmp al,0 ; see if it's a zer0

jz done ; quit if so

cmp al,31 ;see if it's printable

jl noprt ;don't print if not

call printc ;otherwise print it

nonprt:

inc bx ;point to next charater

jmp loop ;and loop back

done:

int 20h :;return to DOS

;subroutine to print the byte in al

printc:

push ax ;push ax and dx

push dx ;

mov dl,al ;use DOS to

mov ah,02H ;print character

int 21H

pop dx ; restore ax and dx

pop ax

ret

msg db 'this ',9,31,32,'is',20H,'a string .',0

;example of how to reserve memory (not used above):

buf db 128 dup(?) ; 128 uninitialized bytes

code ends

end start

MOV [BX] ,AL