I need some assistance with creating a function - c++

Question

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