A project for C/C++ by Stephanie Williams during the Spring Semester 2012.

This project was begun on March 13 and is anticipated to take two weeks to complete. Project was completed on April 5, 2012.

I took ill during the completion of this project and was unable to do work for a couple of weeks.

The purpose of this project is to test the student's working knowledge of C and how to implement arrays and do mathematical operations on them.

In order to successfully accomplish/perform this project, the listed resources/experiences need to be consulted/achieved:

• successful completion of project #1 and solid understanding of pertinent topics
• familiarity with memory allocation via malloc(3)
• familiarity with memory, accessing data via pointer dereferencing, and address calculation
• familiarity with looking up C function parameters/information in the manual
• familiarity with functions, their parameters and return types

The idea behind this project is to play with arrays and the already predetermined sizes of integer types in C.

This project will have you implementing code to support the storage and manipulation of numbers outside of the established data types.

In C, from our first project (Project #0), we explored the various established data types, and determined their various sizes and representational ranges.

From that, we should know the largest value we can store in a variable using the biggest data type size (unsigned long long int), which is: 18,446,744,073,709,551,615

That's a 20-digit number.

But this project will have us creating the ability to store and manipulate largers much larger than that. We'll start with a target of 4 and 24 digits (if you write your code effectively, the number of digits should ultimately not matter).

Why 4? Can't we already easily store values of 4 digits?

Yes, but looking to implement the ability to store and manipulate a 4 digit number will help us to better realize the logic and code necessary to scale our solution to support any number of digits.

While there are many approaches to this problem, follow through this example to get some insight. You don't have to take this approach, but it will cover some important concepts you will need to implement in your solution, whether or not you take this approach.

Let's look at a 4 digit number not as a side-effect of being able to be stored in a quantity of appropriate size, but as 4 literal stored digits in memory. To wit:

    unsigned char *value;
    value = (unsigned char *) malloc (sizeof(unsigned char) * 4);
    *(value+0) = *(value+1) = *(value+2) = *(value+3) = 0;

What just happened here? Make sure you understand, or ask questions and get clarification before attempting to continue.

Essentially, we have just allocated 4 bytes of memory (of type unsigned char), which are located consecutively in memory. To draw a picture, we'd have this:

4 bytes of memory, each containing a single digit of our 4 digit number. Let's assume we are attacking this as a decimal (base 10) value, and we'll maintain our assumption that the left-most value is the most significant digit, and the right-most value is the least significant digit.

For example, let's say we wanted to store the 4-digit number 8192 in memory using this scheme. The code and resulting “picture” would be as follows:

    *(value+0) = 8;
    *(value+1) = 1;
    *(value+2) = 9;
    *(value+3) = 2;

Make sense?

Be aware that *(value+0), the first memory address of our sequence, is at the left side of our value… therefore it stores the most significant digit. You are free to do it the other way, just make sure that whatever approach you take, you maintain your logic.

Now, what if we wanted to perform an addition?

8192+4 = 8196

Pretty easy right?

4 in our memory scheme would be represented as “0004”, and we'd accomplish the addition as follows:

    *(value+0) = *(value+0) + 0;
    *(value+1) = *(value+1) + 0;
    *(value+2) = *(value+2) + 0;
    *(value+3) = *(value+3) + 4;

As you can see, the value of “4” was added only to the last (least significant) digit stored in our value. Displaying it should should the expected answer:

There's actually two situations that occur with adding… what we just saw was the straight “sum”. In this case, the sum was the only meaningful result generated.

But there's also another situation we can have, and that is a carry. A carry is when the result is too big to be stored in a single digit (ie a 2 digit number). So we react by storing the least significant digit and carrying the most significant digit to the next placevalue.

Let's take our 8196 and add 1024 to it. What do we get? 9220

Illustrated, we have:

So, for this project I'd like for you to write a set of functions and a test program that:

• have a function that will allocate space to store a value of desired length (at least 4 and 24, but feel free to test it with larger numbers: 32, 40, 64, etc.) and return the address (so we can assign it to one of our pointers).
• have a function that will zero your value, running through each position and setting it to 0.
• have a function that will accept as a parameter the original number and number to add, perform the operation, and place the result in the original number
• implement a function to tackle subtraction being mindful of the carry
• implement a function to perform multiplication
• implement a function to perform division
• implement a function that accepts as two arguments two of our dynamically allocated “numbers”, compares them, and returns a -1 if the left parameter is greater, 0 if they are equal, and 1 if the right parameter is greater.
• implement a sample program that:
  • prompts the user to enter a the number length (4 digits, 24 digits, 32 digits, etc.)
  • prompts the user for actual values (you'll have to rig up a way to convert the user's input into the appropriate values to place in your managed data type
  • gives the user a choice (perhaps via a menu) that lets them select from all the available functions (even resetting and starting over with new digit-lengths).

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
        long long int numb1;
        long long int numb2;
        long long int total;
        int menu;

        printf("Enter a number: ");
        scanf("%lld", &numb1);
        printf("Enter another number: ");
        scanf("%lld", &numb2);
        printf("Enter selection\n");
        printf("1=addition\n");
        printf("2=subtraction\n");
        printf("3=multiplication\n");
        printf("4=division\n");
        printf("0=exit\n");
        scanf("%d", &menu);
        
        switch(menu){
                case 1: total=numb1+numb2; break;
                case 2: total=numb1-numb2; break;
                case 3: total=numb1*numb2; break;
                case 4: total=numb1/numb2; break;
        };


        if(menu==1)
        {
                printf("%lld plus %lld is %lld\n", numb1, numb2, total);
        }
        if(menu==2)
        {
                printf("%lld minus %lld is %lld\n", numb1, numb2, total);
        }
        if(menu==3)
        {
                printf("%lld times %lld is %lld\n", numb1, numb2, total);
        }
        if(menu==4)
        {
                printf("%lld divided by %lld is %lld\n", numb1, numb2, total);
        }
         if(menu=0)
        {
                printf("Hope you enjoyed your experience\n");
        }
        return(0);
}

addition:

lab46:~/src/cprog$ ./big_num
Enter a number: 123345676789
Enter another number: 243546556
Enter selection
1=addition
2=subtraction
3=multiplication
4=division
0=exit
1
123345676789 plus 243546556 is 123589223345
lab46:~/src/cprog$

subtraction

lab46:~/src/cprog$ ./big_num
Enter a number: 13245675897
Enter another number: 56432
Enter selection
1=addition
2=subtraction
3=multiplication
4=division
0=exit
2
13245675897 minus 56432 is 13245619465
lab46:~/src/cprog$

multiplication

lab46:~/src/cprog$ ./big_num
Enter a number: 2345678976543
Enter another number: 874532
Enter selection
1=addition
2=subtraction
3=multiplication
4=division
0=exit
3
2345678976543 times 874532 is 2051371326714102876
lab46:~/src/cprog$

division

lab46:~/src/cprog$ ./big_num
Enter a number: 89764512378645
Enter another number: 5126674513
Enter selection
1=addition
2=subtraction
3=multiplication
4=division
0=exit
4
89764512378645 divided by 5126674513 is 17509
lab46:~/src/cprog$

Some comments on this project are not many. I don't think I really played with arrays the way as originally intended. I did get to play with the case switch function in a little more depth.

In performing this project, the following resources were referenced:

• the book
• class notes