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>>Assembly Simple Program please if you can <<< simple_program.asm If you fail to name your file this way,you will almost certainly fail the project since my evaluation program won't even be able to find your file! Technical Description and Instructions Your program needs to do only three things: 1. Ask user for input 2. Perform the equation 3. Print the result Your program will be graded on the following criteria (100 pts total) Proper indentation and commenting (20 pts) .2 screen shots (one of the console and one of the simulator 20 pts total) Correct result (40 pts) Correct submission of source code (20 pts) simple_program.asm If you fail to name your file this way,you will almost certainly fail the project since my evaluation program won't even be able to find your file! Technical Description and Instructions Your program needs to do only three things: 1. Ask user for input 2. Perform the equation 3. Print the result Your program will be graded on the following criteria (100 pts total) Proper indentation and commenting (20 pts) .2 screen shots (one of the console and one of the simulator 20 pts total) Correct result (40 pts) Correct submission of source code (20 pts)


Best answer
4.13
8

As the question is not clear, I provide here with an assembly program to find the fibonocci series up to the limit the user enter. print macro args move dx, offset args mov ah, 09h int 21h endm data segment cr equ 0dh lf equ 0ah m1 db cr, lf, "ENTER THE LIMIT:$" m2 db cr, lf, "FIBONOCCI SERIES IS:$" m3 db ",$" m dw 0001h n dw 0000h p dw 0000h outs dw 10 dup(?) data ends code segment assume cs:code, ds:data start: mov ax,data mov ds, ax mov ax, n sub ax, 0001h mov n, ax print ml call read mov p, cx print m2 L5: mov ax, m mov bx, n mov n, ax add ax, bx mov bx, n mov m, ax call disp print m3 print outs mov cx, p dec p loop L5 mov ah, 4ch int 21h read proc mov bx, 000ah mov cx, 0000h L1: mov ah, 01h int 21h cmp al, 0dh jz L2 mov ah, 00h sub al, 30h push ax mov ax, cx mul bx mov cx, ax pop ax add cx, ax jmp L1 L2: ret read endp disp proc mov bx, 000ah mov cx, 0000h mov di, offset outs mov dx, 0024h push dx inc cx L3: mov dx, 0000h div bx add dx, 0030h push dx inc cx cmp ax, 0000h jnz L3 L4: pop[di] inc di loop L4 ret disp endp code ends end start

4.13  (8 votes )
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