Swapping two initialized arrays in C++ using Void and Pointers - c++

Question

I need to write a C++ program where it swaps between two 1-dimensional
arrays using pointers and functions. Firstly, a void function named showValues to display both arrays before swapping takes and also a void function named swap to swap the elements between both arrays.
My question is: I'm supposed to swap the function but for some reason it wont run and I am not sure where is the error in my code
#include <iostream>
#include <iomanip>
using namespace std;
const int SIZE = 5;
void showValues(int[],int[]);
void swap(int[],int[]);
int main() {
int array1[SIZE] = {10,20,30,40,50};
int array2[SIZE] = {60,70,80,90,100};
showValues (array1, array2);
swap(array1, array2);
return 0;
}
void showValues(int array1[], int array2[]){
cout<<"The original arrays are as shown below: " << endl;
cout << " Array 1 is: ";
for (int i = 0; i < 5; ++i) {
cout << array1[i] << " ";
}
cout << "\n Array 2 is: ";
for (int i = 0; i < 5; ++i) {
cout << array2[i] << " ";
}
}
void swap(int array1[], int array2[])
{
int temp,i;
for(i=0; i<5; ++i)
{
temp = array1[SIZE];
array1[SIZE] = array2[SIZE];
array2[SIZE] = temp;
}
cout << "\nThe swapped arrays are as shown below: " << endl;
cout << " Array 1 is: ";
for (int i = 0; i < 5; ++i) {
cout << array1[i] << " ";
}
cout << "\n Array 2 is: ";
for (int i = 0; i < 5; ++i) {
cout << array2[i] << " ";
}
}

Answer 1

This part of your code doesn't make sense:
temp = array1[SIZE];
array1[SIZE] = array2[SIZE];
array2[SIZE] = temp;
SIZE is 5. So, you are accessing array1[5] and array2[5], i.e. the 6th element of the array. Yet, your arrays have only 5 elements to begin with (array1[0] to array1[4], same for array2), so you are accessing elements beyond the end of the array, which is undefined behavior that is probably just corrupting memory somewhere!
You probably meant to use i here, not SIZE, then the code makes sense. Instead, it would be useful to replace the "magic number" 5 with SIZE:
for(i = 0; i < SIZE; ++i)
{
temp = array1[i];
array1[i] = array2[i];
array2[i] = temp;
}

Answer 2

The void swap(int array1[], int array2[]) function is where you are having trouble. You actually don't even need to have another function for the swapping. You could just use std::swap() which is defined in the #include <utility> header. Since both arrays have the same size.
For example you could do something along these lines:
#include <iostream>
#include <iomanip>
#include <utility>
const int SIZE = 5;
void showValues(int[], int[]);
void swap(int[], int[]);
int main() {
int array1[SIZE] = { 10,20,30,40,50 };
int array2[SIZE] = { 60,70,80,90,100 };
int n = sizeof(array1) / sizeof(array2[0]);
showValues(array1, array2);
std::swap(array1, array2);
std::cout << "\n\nThe swapped arrays are as shown below:\n ";
std::cout << "\nArray 1 is: ";
for (int i = 0; i < n; i++)
std::cout << array1[i] << ", ";
std::cout << "\nArray 2 is: ";
for (int i = 0; i < n; i++)
std::cout << array2[i] << ", ";
return 0;
}
void showValues(int array1[], int array2[]) {
std::cout << "The original arrays are as shown below: " << std::endl;
std::cout << "\nArray 1 is: ";
for (int i = 0; i < 5; ++i) {
std::cout << array1[i] << " ";
}
std::cout << "\nArray 2 is: ";
for (int i = 0; i < 5; ++i) {
std::cout << array2[i] << " ";
}
}
Also consider not using using namespace std;.