Swap with pointers. Is it right way to use pointers in order to swap two numbers or is there any simpler method?

Question

#include <iostream>
using namespace std;

void swap(int *c, int *d)
{
int *e, m;
e = &m;
m = *c;
*c = *d;
*d = *e;
}
int main()
{
int a, b;

cout << "Program to swap two numbers using pointers\n" << endl;
cout << "Please, input 1st number: ";
cin >> a;
cout << "Please, input 2nd number: ";
cin >> b;

swap(a, b);
if (a != b)
{
cout << "\nThe 1st swapped number: " << a << endl;
cout << "The 2nd swapped number: " << b << endl;
}
else
cout << "Numbers are equal to each other! Please, input different numbers!" << endl;

system("pause");
return 0;
}

Answer 1

#include <iostream>
using namespace std;
void swap(int* x, int* y)
{
int z = *x;
*x = *y;
*y = z;
}

int main()
{
int a, b;
cout<<"a= ";
cin>>a;
cout<<"\nb= ";
cin>>b;

swap(&a, &b);

cout << "\nAfter Swap\n";
cout << "a = " << a << " b = " << b << "\n";
return 0;
}
There is best way

Answer 2

Pointers aren't absolutely necessary for this. You don't even need references, but it would save memory to use references.

void swap(int& a, int &b) {
int tmp = a;
a = b;
b = tmp;
}