I need some help with my program..

Edit: So, seems I do have a problem somewhere in my coding. Whenever I run the program and input a variable, it always returns the same answer.."The content at location 76 is 0. I'm trying to figure out how to incorporate this idea:


const int OP_LOAD = 3;
const int OP_STORE = 4;
const int OP_ADD = 5;
...

const int OP_LOCATION_MULTIPLIER = 100;

mem[0] = OP_LOAD * OP_LOCATION_MULTIPLIER + ...;
mem[1] = OP_ADD * OP_LOCATION_MULTIPLIER + ...;

operand = memory[ j ] % OP_LOCATION_MULTIPLIER;
operation = memory[ j ] / OP_LOCATION_MULTIPLIER;



Okay guys, I posted on here a few days ago about a question but it was just a compilation error, so if this looks familiar, thats why. I will reiterate, I'm new to programming, I'm not the best, so I'm going for simplicity. Also this is an SML program. Anyway, this IS a homework assignment and I'm wanting a good grade on this. So I was looking for input and making sure this program will do what I'm hoping they are looking for. Anyway, here are the instructions: Write SML (Simpletron Machine language) programs to accomplish each of the following task:

A) Use a sentinel-controlled loop to read positive number s and compute and print their sum. Terminate input when a neg number is entered. B) Use a counter-controlled loop to read seven numbers, some positive and some negative, and compute + print the avg. C) Read a series of numbers, and determine and print the largest number. The first number read indicates how many numbers should be processed.

Without further a due, here is my program. All together.

Code:

---

int main()
{
        const int READ = 10;
        const int WRITE = 11;
        const int LOAD = 20;
        const int STORE = 21;
        const int ADD = 30;
        const int SUBTRACT = 31;
        const int DIVIDE = 32;
        const int MULTIPLY = 33;
        const int BRANCH = 40;
        const int BRANCHNEG = 41;
        const int BRANCHZERO = 41;
        const int HALT = 43;

        int mem[100] = {0}; //Making it 100, since simpletron contains a 100 word mem.
       
        int operation; //taking the rest of these variables straight out of the book seeing as how they were italisized.
       
        int operand;
       
        int accum = 0; // the special register is starting at 0
       
        int j;
       
       
       
       
        // This is for part a, it will take in positive variables in a sent-controlled loop and compute + print their sum. Variables from example in text.
        memory [00] = 1010;
       
        memory [01] = 2009;
       
        memory [02] = 3008;
       
        memory [03] = 2109;
       
        memory [04] = 1109;
       
        memory [05] = 4300;
       
        memory [06] = 1009;

        j = 0; //Makes the variable j start at 0.

        while ( true )
        {
               
                operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100)
                operation = memory[ j ]/100;

                //using a switch loop to set up the loops for the cases
                switch ( operation ){
                        case 1: //reads a variable into a word from loc. Enter in -1 to exit
                                cout <<"\n Input a positive variable:  ";
                                cin >> memory[ operand ]; break;
               
                        case 2: // takes a word from location
                                cout << "\n\nThe content at location " << operand  << "is " << memory[operand]; break;
               
                        case 3:// loads
                                accum = memory[ operand ]; break;

                        case 4: //stores
                                memory[ operand ] = accum; break;

                        case 5: //adds
                                accum += mem[operand]; break;
                       

                        case 6: // subtracts
                                accum-= memory[ operand ]; break;

                        case 7: //divides
                                accum /=(memory[ operand ]); break;

                        case 8: // multiplies
                                accum*= memory [ operand ]; break;

                        case 9: // Branches to location
                                j = -1; break;

                        case 10: //branches if acc. is < 0
                                if (accum < 0)
                                j = 5;
                                break;
                       
                        case 11: //branches if acc = 0
                                if (accum == 0)
                                        j = 5;
                                break;

                        case 12: // Program ends
                                exit(0); break;
        }
        j++;
        }
return 0;
}

//Part b finding the sum + avg.

int main()
{
        const int READ = 10;
        const int WRITE = 11;
        const int LOAD = 20;
        const int STORE = 21;
        const int ADD = 30;
        const int SUBTRACT = 31;
        const int DIVIDE = 32;
        const int MULTIPLY = 33;
        const int BRANCH = 40;
        const int BRANCHNEG = 41;
        const int BRANCHZERO = 41;
        const int HALT = 43;

        int mem[100] = {0};
        int operation;
        int operand;
        int accum = 0;
        int pos = 0;

        int j;
       

        mem[22] = 7; // loop 7 times
        mem[25] = 1; // increment by 1

        mem[00] = 4306;
       
        mem[01] = 2303;
       
        mem[02] = 3402;
       
        mem[03] = 6410;
       
        mem[04] = 3412;
       
        mem[05] = 2111;

        mem[06] = 2002;
       
        mem[07] = 2312;
       
        mem[08] = 4210;
       
        mem[09] = 2109;
       
        mem[10] = 4001;

        mem[11] = 2015;
       
        mem[12] = 3212;
       
        mem[13] = 2116;

        mem[14] = 1101;
       
        mem[15] = 1116;

        mem[16] = 4300;

        j = 0;

        while ( true )
        {
               
                operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100)
                operation = memory[ j ]/100;

                //using a switch loop to set up the loops for the cases
                switch ( operation ){
                        case 1: //reads a variable into a word from loc. Enter in -1 to exit
                                cout <<"\n enter #:  ";
                                cin >> memory[ operand ]; break;
               
                        case 2: // takes a word from location
                                cout << "\n\nThe content at location " << operand  << "is " << memory[operand]; break;
               
                        case 3:// loads
                                accum = memory[ operand ]; break;

                        case 4: //stores
                                memory[ operand ] = accum; break;

                        case 5: //adds
                                accum += mem[operand];; break;
                       

                        case 6: // subtracts
                                accum-= memory[ operand ]; break;

                        case 7: //divides
                                accum /=(memory[ operand ]); break;

                        case 8: // multiplies
                                accum*= memory [ operand ]; break;

                        case 9: // Branches to location
                                j = operand; break;

                        case 10: //branches if acc. is < 0
                               
                                break;
                       
                        case 11: //branches if acc = 0
                                if (accum == 0)
                                        j = operand;
                                break;

                        case 12: // Program ends
                                exit(0); break;
        }
        j++;
        }
return 0;
}



///Part c
int main()
{
        const int READ = 10;
        const int WRITE = 11;
        const int LOAD = 20;
        const int STORE = 21;
        const int ADD = 30;
        const int SUBTRACT = 31;
        const int DIVIDE = 32;
        const int MULTIPLY = 33;
        const int BRANCH = 40;
        const int BRANCHNEG = 41;
        const int BRANCHZERO = 41;
        const int HALT = 43;

        int mem[100] = {0};
        int operation;
        int operand;
        int accum = 0;


        int j;
       
       
        mem[23] = 1; //decrements 1 place in mem
       
        mem[0] = 1030; // Takes in # of values to be stored.
       
        mem[01] = 4123; // These 4 memory slots check for the largest variable then store
        mem[02] = 4134;
        mem[03] = 1011;
        mem[04] = 3204;

        mem[05] = 4005; // These 5 decrement the count+ store + branch.
        mem[06] = 4006;
        mem[07] = 4007;
        mem[08] = 4008;
        mem[09] = 4009;

        mem[10] = 4010;
        mem[11] = 4311; // exits

        j = 0; // this is the starting value..

        while ( true )
        {
               
                operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100)
                operation = memory[ j ]/100;

                //using a switch loop to set up the loops for the cases
                switch ( operation ){
                        case 1: //reads a variable into a word from loc. Enter in -1 to exit
                                cout <<"\n enter #:  ";
                                cin >> memory[ operand ]; break;
               
                        case 2: // takes a word from location
                                cout << "\n\nThe content at location " << operand  << "is " << memory[operand]; break;
               
                        case 3:// loads
                                accum = memory[ operand ]; break;

                        case 4: //stores
                                memory[ operand ] = accum; break;

                        case 5: //adds
                                accum += mem[operand]; break;
                       

                        case 6: // subtracts
                                accum-= memory[ operand ]; break;

                        case 7: //divides
                                accum /=(memory[ operand ]); break;

                        case 8: // multiplies
                                accum*= memory [ operand ]; break;

                        case 9: // Branches to location
                                j = operand; break;

                        case 10: //branches if acc. is < 0
                               
                                break;
                       
                        case 11: //branches if acc = 0
                                if (accum == 0)
                                        j = operand;
                                break;

                        case 12: // Program ends
                                exit(0); break;
                        case 13: // checks > than
                                if (accum < mem[operand])
                                        accum = mem[operand];
                                break;
                }
        j++;
        }
return 0;
}

---

Which program are you running, a, b or c? Part a looks like it will go into an infinite loop.

Also, remember that 08 and 8 are not the same thing in C++. 08 isn't really a valid number, I'm not sure what the compiler will do with it. The initial 0 indicates that you're using an octal number, and octal digits go from 0 to 7 only. I'd get rid of the initial 0 in those numbers.