The arguments passed in to the function are local to the function. Changes
made to the arguments do not affect the values in the calling function. This is
known as passing by value, which means a local copy of each argument is made in
the function. These local copies are treated just like any other local
variables. Program below illustrates this point.
A demonstration of
passing by value.
1: // Demonstrates passing by value
3: #include <iostream.h>
5: void swap(int x, int y);
7: int main()
9: int x = 5, y = 10;
11: cout << "Main. Before swap, x: " << x << " y: " << y << "\n";
13: cout << "Main. After swap, x: " << x << " y: " << y << "\n";
14: return 0;
17: void swap (int x, int y)
19: int temp;
21: cout << "Swap. Before swap, x: " << x << " y: " << y << "\n";
23: temp = x;
24: x = y;
25: y = temp;
27: cout << "Swap. After swap, x: " << x << " y: " << y << "\n";
Output: Main. Before swap, x: 5 y: 10
Swap. Before swap, x: 5 y: 10
Swap. After swap, x: 10 y: 5
Main. After swap, x: 5 y: 10
Analysis: This program initializes two
variables in main() and then passes them to the swap()
function, which appears to swap them. When they are examined again in main(),
however, they are unchanged!
The variables are initialized on line 9, and their values are displayed on line
11. swap() is called, and the variables are passed in.
Execution of the program switches to the swap() function, where on
line 21 the values are printed again. They are in the same order as they were in
main(), as expected. On lines 23 to 25 the values are swapped, and this
action is confirmed by the printout on line 27. Indeed, while in the swap()
function, the values are swapped.
Execution then returns to line 13, back in main(), where the values
are no longer swapped.
As you've figured out, the values passed in to the swap() function
are passed by value, meaning that copies of the values are made that are local
to swap(). These local variables are swapped in lines 23 to 25, but the
variables back in main() are unaffected.