Can't get my pointers to send a reference to the array - c++

Question

#include<stdio.h>
#include<stdlib.h>
int* getEvenNumbers(int arr[], int N)
{
int i, k = 0 , a[50], p;
for (i = 0; i < N; i++)
{
if (arr[i] % 2 == 0)
{
arr[k]=arr[i];
k++;
}
}
return arr[k];
}
int main ()
{
int i, arr[5000000], N, a[500000], k, *p;
printf("\nEnter your desired length of the array:\n\n");
scanf("%d", &N);
for (i = 0; i < N; i++)
arr[i]= rand();
getEvenNumbers (arr, N);
printf("\n\nEven numbers in the array are as follows:\n\n");
for (i = 0; i < N; i++)
{
a[i]= *(p+i);
printf("\n[%d] = %d", (i+1), a[i]);
}
}
please i know this is probably very easy for you guys but i need help figuring out how to return a pointer to the array without all my values of my array getting deleted, also i can't use global variables and it has to be a function that returns a pointer pointing to the array

Answer 1

First of all, decrease the size of those arrays, you don't need that much space. Second of all, you made your
getEvenNumbers
function return an int *, and not an int. arr[k] is not an int *. I also don't get why you are returning something if nothing is being assigned when you call the function. You can just change the return type to void.
void getEvenNumbers(int arr[], int N)
You also never allocate any memory for p. You can do
p = (int*) malloc(sizeof(int));
And since you never allocated any memory for p, the following line of code
a[i]= *(p+i);
is assigning a[i] to a random address. You should just try to rewrite this program. There a lot of errors in this program that I didn't even correct. Go on Google an look up finding even numbers in array program or something similar and look at the code of those examples.
EDIT:
I found some code examples for you to use. I hope it helps!
StackOverflow
sanfoundry.com

Answer 2

The caller already knows the address of the array, so you just need to return the new length. If we also remove the unused variables and take advantage of C++ declarations, we will have:
int getEvenNumbers(int* arr, int N)
{
for (int i = 0, k = 0; i < N; i++) {
if (arr[i] & 1 == 0) { // even if lowest bit is zero
arr[k] = arr[i];
k++;
}
}
return k;
}
Now you can print the even numbers easily:
int k = getEvenNumbers(arr, N);
printf("\n\nEven numbers in the array are as follows:\n\n");
for (i = 0; i < k; i++) {
printf("\n[%d] = %d", (i+1), arr[i]);
}

Answer 3

Dynamically allocate memory from heap.
int* a= new int [N];
//Now store the elements from index 1.
// at a[0] store the number of even number you have found in this function.
return a;
In main you know how many even numbers are there.
int *a1=getEvenNumbers(arr,n);
count_even=a1[0];
for(index=1;index<=count_even;index++)
cout<<a1[index];
The code is given here-
#include<stdio.h>
#include<stdlib.h>
int* getEvenNumbers(int arr[], int N)
{
int i, k = 1 , p;
int* a=new int[N+1];
for (i = 0; i < N; i++)
{
if (arr[i] % 2 == 0)
a[k++]=arr[i];
}
a[0]=k-1;
return a;
}
int main ()
{
int i, N;
printf("\nEnter your desired length of the array:\n\n");
scanf("%d", &N);
int arr[N];
for (i = 0; i < N; i++)
arr[i]= rand();
int *a=getEvenNumbers (arr, N);
printf("\n\nEven numbers in the array are as follows:\n\n");
for (i = 1; i <= a[0]; i++)
printf("\n[%d] = %d", (i), a[i]);
delete []a;
}
A better option is to use std::vector.You can read it here.