[ fromfile: functions.xml id: valueparm ]

By default, C++ parameters are passed by value. When a function is called, a temporary (local) copy of each argument object is made and placed on the program stack. Only the local copy is manipulated inside the function, and the argument objects from the calling block are not affected by these manipulations. The temporary stack variables are all destroyed when the function returns. A useful way to think of value parameters is this: Value parameters are merely local variables initialized by copies of the corresponding argument objects specified when the function is called. Example 5.11 gives a short demonstration.

#include <iostream>

int sumit(int num) {
    int sum = 0;
    for (; num ; --num)  
        sum += num;
    return sum;
}


int main() {
    using namespace std;
    int n = 10;
    cout << n  << endl;
    cout << sumit(n) << endl;
    cout << n << endl;  
    return 0;
}
Output:

  10
  55
  10

If a pointer is passed to a function, a temporary copy of that pointer is placed on the stack. Changes to that pointer will have no effect on the pointer in the calling block. For example, the temporary pointer could be assigned a different value (see Example 5.12).

#include <iostream>
using namespace std;

void messAround(int* ptr) {
    *ptr = 34;       
    ptr = 0;         
}


int main() {
    int n(12);       
    int* pn(&n);     
    cout << "n = " << n << "\tpn = " << pn << endl;
    messAround(pn);  
    cout << "n = " << n << "\tpn = " << pn << endl;
    return 0;
}
Output:

n = 12  pn = 0xbffff524
n = 34  pn = 0xbffff524

The output displays the hexadecimal value of the pointer pn and the value of n so that there can be no doubt about what was changed by the action of the function.

	Note
	To summarize: When an object is passed by value to a function, a copy is made of that object. The copy is treated as a local variable by the function. The copy is destroyed when the function returns.