Adding Binary Numbers (as arrays) C++

Question

I have to figure out a way to add two binary numbers that are stored in arrays, and the binary numbers can be of any size; i.e.:

Sample input and output:  
101 + 11 = 1000  
111 + 111 = 1110  
1010 + 1010 = 10100  
11101 + 1010 = 100111  
11111 + 11111 = 111110

This is way over my head but this is what I have so far:

#include <iostream>

using namespace std;

void addition (int a[], int b[], int sizea, int sizeb, int result[]){

int carry = 0;

    for(int i = 0; i < sizeof(a); i++)
    {
    if(a[i] + b[i] + carry == 0)
    {
        result[i] = 0;
        carry = 0;
    }
    else if(a[i]+b[i]+carry==1)
    {
        result[i]=1;
        carry=0;
    }
    else if(a[i] + b[i] + carry == 2)
    {
        result[i] = 0;
        carry = 1;
    }
    else if(a[i] + b[i] + carry > 2)
    {
        result[i] = 1;
        carry = 1;
    }
}
    result[0] = 0;

    for (int i = 10; i > 0; i--){
        cout << result[i];
    }
    cout << endl;

}

void initarray(int &sizea, int &sizeb, int a[], int b[]){

    cout << "Enter size of a" << endl;
    cin >> sizea;
    cout << "Enter size of b" << endl;
    cin >> sizeb;

    cout << "enter contents of a " << sizea << endl;
    for (int i = 0; i < sizea; i++){
        cin >> a[i];

    }

    cout << "enter contents of b  "<< sizeb << endl;
    for (int z = 0; z < sizeb; z++){
        cin >> b[z];
    }

}

int main() {

    int sizea, sizeb;

    int* a = new int[sizea];
    int* b = new int[sizeb];
    int* result = new int [10];

    initarray(sizea, sizeb, a, b);
    addition(a, b, sizea, sizeb, result);


}

Please feel free to rip me apart, I'm really having trouble with this and I think I have the logic down, I just can't figure out how to translate it into code.

Right now, if I enter in the first example, I get:

Enter size of a 3 Enter size of b 2 enter contents of a 3 1 0 1 enter contents of b 2 1 1 -18174002763276720728465360000100

So obviously there's a problem here. Can someone help?

Is this homework? I ask because you could change a lot of things to make it easier, but that won't help you if it has to be done this way. — Weak to Enuma Elish
– Weak to Enuma Elish, Commented Feb 4, 2016 at 5:18
Hey James, This is homework. What are your suggestions? The bit nmber has to be stored in arrays, that's the only requirement. — hbazzi234
– hbazzi234, Commented Feb 4, 2016 at 5:22
Also in your main, sizea and sizeb are uninitialized and using them is undefined behavior. — crashmstr
– crashmstr, Commented Feb 4, 2016 at 6:14

Rabbid76 · Accepted Answer · 2016-02-04 05:43:01Z

1

Adapt your code like this:

void addition (int a[], int b[], int sizea, int sizeb, int result[]){

    int maxSize = sizea > sizeb ? sizea : sizeb;  // number of bits is maximum of siza and sizeb

    int carry = 0;
    for( int i = 0; i < maxSize; i++ )
    {
        int bitA = i < sizea && a[i] ? 1 : 0;     // test if bit in array a is set
        int bitB = i < sizeb && b[i] ? 1 : 0;     // test if bit in array b is set
        int sum = bitA + bitB + carry;            // calculate sum of all bits
        result[i] = sum == 1 || sum == 3 ? 1 : 0; // result bit is set if sum is equal 1 or 3
        carry = sum > 1 ? 1 : 0;                  // carry bit is set if sum is eaul 2 or 3
    }
    result[ maxSize ] = carry;                    // highest bit of result is carry bit

    for (int i = 0; i <= maxSize; i++){
        cout << result[maxSize-i];
    }
    cout << endl;
}

edited Feb 4, 2016 at 5:43

answered Feb 4, 2016 at 5:36

Rabbid76

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2 Comments

hbazzi234 Over a year ago

Thanks for the help. Looking into this now, but it looks like i'm still not getting correct output:

hbazzi234 Over a year ago

a as "1 1 1" and b as "1 1 1" returns 1100 instead of 1110

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