How to Understand "Pass By Reference Within a Function" w3schools example? - c++

Question

I'm having a hard time understanding how this example from the w3schools tutorial works.
#include <iostream>
using namespace std;
void swapNums(int &x, int &y) {
int z = x;
x = y;
y = z;
}
int main() {
int firstNum = 10;
int secondNum = 20;
cout << "Before swap: " << "\n";
cout << firstNum << secondNum << "\n";
// Call the function, which will change the values of firstNum and secondNum
swapNums(firstNum, secondNum);
cout << "After swap: " << "\n";
cout << firstNum << secondNum << "\n";
system("pause");
return 0;
}
I think I understand the first part:
void swapNums(int &x, int &y) {
int z = x;
x = y;
y = z;
}
I'm basically referencing whatever x and y are going to be when I call the function. So x is going to be "pointing" to firstNum and y is going to be pointing to secondNum. It's going to do the switcheroo using a third variable as a placeholder.
However, after I call the function swapNums(firstNum, secondNum);, I don't understand how the function with its local variables has the ability to change the values of int firstNum = 10; and int secondNum = 20;.
My understanding is that variables within a function are "local" and the scope of said variables only extend within the function itself. How do the local variables change other variables outside their own function without any return statements?

Answer 1

Try this
#include <iostream>
void change_val_by_ref(int &x)
{
x=100
}
void change_val_by_val(int x)
{
x=50;
}
int main()
{
int whatever=0;
std::cout<<"Original value: "<<whatever<<"\n";
change_val_by_ref(whatever);
std::cout<<"After change by ref: "<<whatever<<"\n";
change_val_by_val(whatever);
std::cout<<"After change by val: "<<whatever<<"\n";
}
The output you will see is:
0
100
100
Let's see what happened
change_val_by_ref changed the original whatever, because the ref
was "pointing" to the VARIABLE.
change_val_by_val didn't change the whatever, because the argument of the function x has just copied the value of whatever, and anything that happens to x will not affect whatever, because they are not related.
That's the point of passing by ref.

Answer 2

Imagine you have banana and orange placed in two different plates. You place them into big box with a robotic hand.
You, obviously, can change their destination using you hand. But you want the robotic hand to do this task. You need to tell where the banana and orange are placed, so robotic hand can do its job. !Important robotic hand doesn't care about fruit, which is placed into these plates, it only cares about changing their positions.
By using a pointer, you just tell a function the placement of an object you want to change. And you can set your local, suitable name for this. So pointer is just the name of a plate, which contains a certain value.

Answer 3

References are syntactic sugar for passing around addresses of variables. Instead of the variable itself, you pass its memory address and then the function uses the address of the variable to change its value. Simple as that. Equivalent C code would be:
void swap(int* a, int* b) {
int tmp = *a;
*a = *b;
*b = tmp;
}