I want to swap 2 integers in a function, but cant do it without passing the parameters by reference, but when I use the same code inside main(), it works without using the integer’s memory address

passing parameters as reference

#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";

  return 0;
}

/*output

Before swap:
1020
After swap:
2010

*/

but when I pass the parameters as integers, the numbers don’t swap

passing parameters as integers

#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";

  swapNums(firstNum, secondNum);

  cout << "After swap: " << "n";
  cout << firstNum << secondNum << "n";

  return 0;
}

/*output

Before swap:
1020
After swap:
1020

*/

and again, if I don’t use a function and write the code inside the main block, it works without any problem.

not using function and running the code in main() block

#include <iostream>
using namespace std;

int main() {
  int firstNum = 10;
  int secondNum = 20;

  cout << "Before swap: " << "n";
  cout << firstNum << secondNum << "n";

  int z = firstNum;
  firstNum = secondNum;
  secondNum = z;

  cout << "After swap: " << "n";
  cout << firstNum << secondNum << "n";

  return 0;
}

/*output

Before swap:
1020
After swap:
2010

*/

thank you for helping me and appreciate your patience to read my first question in stack overflow