I need some assistance with creating a function

Question

I am new, not that good with functions, and I am trying to solve this question:

16. Suppose A, B, C are arrays of integers of size [M], [N], and [M][N], respectively. The user will enter the values for the array A and B. Write a user defined function in C++ to calculate the third array C by adding the elements of A and B. If the elements have the same index number, they will be multiplied. C is calculated as the following: -

    Use A, B and C as arguments in the function.

Below is my attempt at the problem.

     #include<iostream>
using namespace std;

void Mix(int(&A)[], int(&B)[], int(&C)[][100], int N, int M);
//dont understand why you used Q

int main()
{
    //variable declaration
    int A[100], B[100], C[100][100], n, m, l = 0;


    //input of size of elements for first ararys
    cout << "Enter number of elements you want to insert in first array: ";
    cin >> n;
    cout << "-----------------" << endl;
    cout << "-----------------" << endl;
    cout << "Enter your elements in ascending order" << endl;
    //input the elements of the array
    for (int i = 0; i < n; i++)
    {
        cout << "Enter element " << i + 1 << ":";
        cin >> A[i];
    }
    cout << endl << endl;
    //input of size of elements for first ararys
    cout << "Enter number of elements you want to insert in second array: ";
    cin >> m;
    cout << "-----------------" << endl;
    cout << "-----------------" << endl;
    cout << "Enter your elements in descending order" << endl;
    //input the elements of the array
    for (int i = 0; i < m; i++)
    {
        cout << "Enter element " << i + 1 << ":";
        cin >> B[i];
    }

    Mix(A, B, C, n, m);

    cout << "\nThe Merged Array in Ascending Order" << endl;


    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++)
        {
            cout << C[i][j] << " ";
        }
        cout << "\n"; //endline never use endl its 10 times slower
    }

    system("pause");
    return 0;
}

void Mix(int(&A)[], int(&B)[], int(&C)[][100], int N, int M)
{
    // rows is the index for the B array, cols is index for A array
    int rows = 0;
    int cols = 0;
    while (rows < M) {
        while (cols < N) {
            if (rows == cols) { // remember ==
                C[rows][cols] = B[rows] * A[cols];
            }
            else {
                C[rows][cols] = B[rows] + A[cols];
            }
            cols++; // increment here
        }
        rows++; // increment here
    }
    return;
}

Here is an example of the output:

enter image description here

Answer 1

In order to make the C array two-dimensional, it needs to be expressed as C[100][100], instead of C[200]. That is the first step. Next, in your Mix() function, you need to cycle through each element of both A and B (ex. two for loops). Your rows change as B changes, and your columns change as A changes. Include a check for identical indices that will determine whether to add or multiply the two values together.

void Mix(int A[], int B[], int C[][], int N, int M) {
    // rows is the index for the B array, cols is index for A array
    for (int rows = 0; rows < M; rows++) {
        for (int cols = 0; cols < N; cols++) {
            if (rows == cols) { // remember ==
                C[rows][cols] = B[rows] * A[cols];
            } else {
                C[rows][cols] = B[rows] + A[cols];
            }
        }
    }
}

Make sure your arrays are properly defined and print out the C array by row and column to match the specifications.

UPDATE: If you want to use while loops, I would default to deconstructing the for loops and apply the same logic:

void Mix(int A[], int B[], int C[][], int N, int M) {
    // rows is the index for the B array, cols is index for A array
    int rows = 0;
    int cols = 0;
    while (rows < M) {
        while (cols < N) {
            if (rows == cols) { // remember ==
                C[rows][cols] = B[rows] * A[cols];
            } else {
                C[rows][cols] = B[rows] + A[cols];
            }
            cols++; // increment here
        }
        rows++; // increment here
    }
}

I would definitely recommend the for loop approach, as it is more compact, yet does the exact same operations.

Answer 2

There are a lot of things wrong with your code. First off an 2D array must be declared with 2 squared brackets so C[200][200]. In the Mix function the logical operator is == not = in if (A[I] = B[J]) Anyway here's the function that you need:

#include<iostream>
using namespace std;


void Mix(int A[], int B[], int C[], int N, int M) {
    //dont understand why you used Q
    int i, j;
    for(i=0; i<N; i++) {
        for(j=0; j<M; j++) {
            if(i==j){
                C[i][j] = A[i] * B[j];
            }
            else {
                C[i][j] = A[i] + B[j];
            }
        }
    }
    return C[i][j];
}
int main()
{
    //variable declaration
    int A[100], B[100], C[200], j, i, n, m, l = 0;
    string Comma;

    //input of size of elements for first ararys
    cout << "Enter number of elements you want to insert in first array: ";
    cin >> n;
    cout << "-----------------" << endl;
    cout << "-----------------" << endl;
    cout << "Enter your elements in ascending order" << endl;
    //input the elements of the array
    for (i = 0; i < n; i++)
    {
        cout << "Enter element " << i + 1 << ":";
        cin >> A[i];
    }
    cout << endl << endl;
    //input of size of elements for first ararys
    cout << "Enter number of elements you want to insert in second array: ";
    cin >> m;
    cout << "-----------------" << endl;
    cout << "-----------------" << endl;
    cout << "Enter your elements in descending order" << endl;
    //input the elements of the array
    for (j = 0; j < m; j++)
    {
        cout << "Enter element " << j + 1 << ":";
        cin >> B[j];
    }
    C = Mix(A, B, C, n, m);

    cout << "\nThe Merged Array in Ascending Order" << endl;


    for(i=0; i<n; i++) {
        for(j=0; j<m; j++) {
           cout<<C[i][j]<<" ";
        }
        cout<<"\n" //endline never use endl its 10 times slower
    }

    system("pause");
    return 0;
}

Answer 3

Because M and N are defined at run time, you'll really want to use vectors to represent them. Additionally consider returning a 2D container so as to leverage return value optimization.

I'm going to write an example using a vector of vectors for simplicity (see What are the Issues with a vector-of-vectors? for more on why that's really just good for a toy example):

vector<vector<int>> Mix(const vector<int>& A, const vector<int>& B) {
    vector<vector<int>> result(size(B), vector<int>(size(A)));

    for(size_t i = 0U; i < size(B); ++i) {
        for(size_t j = 0U; j < size(A); ++j) {
            result[i][j] = A[j] * B[i];
        }
    }
    return result;
}

Live Example

EDIT:

If you must use arrays you'll miss out on return value optimization. I'd only choose this as a good option in the situations:

1. That you weren't returning anything, in which case your function would probably look something like:

void Mix(const int* A, const int* B, const size_t size_A, const size_t size_B) 
{
    for(size_t i = 0U; i < size_B; ++i) {
        for(size_t j = 0U; j < size_A; ++j) {
            cout << '[' << i << "][" << j << "]: " << A[j] * B[i] << '\t';
        }
        cout << endl;
    }
}

2. That you weren't calling a function and you'd already been given int A[M] and int B[N] as inputs and int C[N][M] as an output, in which case the code you'd inline would probably look something like this:

for(size_t i = 0U; i < size(B); ++i) {
    for(size_t j = 0U; j < size(A); ++j) {
        C[i][j] = A[j] * B[i];
    }
}