Lab Description:

This lab required us to create a decimal to Roman numeral converting program in HLA. This specific program used one of the simplest methods which is utilizing only subtraction.

Code:

program decimaltoroman;
    #include("stdlib.hhf")

// Yevgeniy Babkin   --- CET3510
//
// this program provides a simple way to convert
// Decimal numbers to Roman numerals

static
// declare a 16-bit variable thats greater than 0.

    x: int32:=999;

begin decimaltoroman;
// console.cls(); clears the original cmd window for a cleaner look.

console.cls();

// main while loop that establishes an exit strategy;

        while((x != 0)&&(x != 4000)) do

      stdout.put(nl,"Please input a number [1-3999] : ");

//the following commands "cin" the user input number
// and move it into the 32 bit register

       stdin.get(x);
       mov(x, EAX);
//next line is the main if loop that see's if the number is within 1-3999 range
//prgrm will not run right without it!!!

       if(EAX<4000 && EAX>0)then

//spacing is added to the 1st stdout to center the wording for 
//easier readability

        stdout.put(nl "                       Your number: ",x, nl, "          Roman Numeral equivalent: ");

//the program is fairly basic as it subtracts known roman numeral values
//to figure out which and how many roman numearals to display.

        if(EAX>=1000 && EAX<4000) then             
                        while(EAX>=1000) do
                sub(1000, EAX);
                stdout.put("M");
            endwhile;
        endif;
        if(EAX>=900) then
            sub(900, EAX);
            stdout.put("CM");
        endif;
        if(EAX>=500) then
            sub(500, EAX);
            stdout.put("D");
        endif;

        if(EAX>=400) then
            sub(400,EAX);
            stdout.put("CD");
        endif;
        if(EAX>=100 && EAX<400) then             
                        while(EAX>=100) do
                sub(100, EAX);
                stdout.put("C");
            endwhile;
        endif;
        if(EAX>=90) then
            sub(90, EAX);
            stdout.put("XC");
        endif;
        if(EAX>=50) then
            sub(50, EAX);
            stdout.put("L");
        endif;
        if(EAX>=40) then
            sub(40,EAX);
            stdout.put("XL");
        endif;
        if(EAX>=10 && EAX<40) then             
                        while(EAX>=10) do
                sub(10, EAX);
                stdout.put("X");
            endwhile;
        endif;
        if(EAX>=9) then
            sub(9, EAX);
            stdout.put("IX");
        endif;
        if(EAX>=5) then
            sub(5, EAX);
            stdout.put("V");
        endif;
        if(EAX>=4) then
            sub(4,EAX);
            stdout.put("IV");
        endif;
        if(EAX>=1 && EAX<4) then             
                        while(EAX>=1) do
                sub(1, EAX);
                stdout.put("I");
            endwhile;

                 endif;

// last stdout lets user know how to exit the program
// by entering 0, even though a user can enter; either 0 or
// a number greater than 3999 and still exit the program
// but for simplicity and a cleaner look a 0 is enough.

                stdout.put(nl, nl, " TO END PROGRAM ENTER [0]",nl);
        stdout.put(" Otherwise keep converting!!!" ,nl,nl);
    endif;

        endwhile;
end decimaltoroman;

Screenshot: