Question

​​​​​​​in C++

Q5-pointers: Using pointer notation and dynamic array to: input m*n array of zeros and ones then write the following functions: 1- even parity row: to find the even parity of each row 2- even Rarity.col; to find the even parity of each col. 3- print array: to print the new array (n+1)*(m+1)

Answer 1

Answer:

According to the given data, the following as the C++ code.

Code:

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

using namespace std;

//method to fill the even parity of rows
void even_parity_row(int *newArr, int m, int n)
{
for (int i = 0; i < m; i++)
{
int rowParity = 0;
int j;
for (j = 0; j < n; j++)
{
rowParity = rowParity + *(newArr + i*n + j);
}
if(rowParity%2==0)
*(newArr + i*n + j) = 1;
else
*(newArr + i*n + j) = 0;
}
}

//method to fill the even parity of columns
void even_parity_col(int *newArr, int m, int n)
{
for (int j = 0; j < m; j++)
{
int colParity = 0;
int i;
for (i = 0; i < n; i++)
{
colParity = colParity + *(newArr + i*n + j);
}
if(colParity%2==0)
*(newArr + i*n + j) = 1;
else
*(newArr + i*n + j) = 0;
}
}

//method to display the new array on the computer screen
void print_arrray(int *newArr, int m, int n)
{
for(int i=0; i<m+1; i++)
{
for(int j=0; j<n+1; j++)
{
cout<<*(newArr + i*n + j)<<" ";
}
cout<<endl;
}
}

int main()
{
int m = 3, n = 3;
int* arr = new int[m*n];

//initialize array with 0 and 1
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
*(arr + i*n + j) = rand() % 2;
}
}
  
int* newArr = new int[(m+1)*(n+1)];
  
  
// here initialize new array
for (int i = 0; i < m; i++)
{
int rowParity = 0;
int j;
for (j = 0; j < n; j++)
{
rowParity = rowParity + *(arr + i*n + j);
*(newArr + i*n + j) = *(arr + i*n + j);
}
}
  
//method calling
even_parity_row(newArr, m, n);
even_parity_col(newArr, m, n);
print_arrray(newArr, m, n);

return 0;
}

OUTPUT:

1 0 1 1
1 1 1 0
0 0 1 0
1 0 0 0

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