Thread: Passing arguments by value

Since the parameter is passed by value, the function cannot modify it. The function modifies a copy, then returns the result. So do this instead:

Code:

int n=1;
n = add(n); // Assign n to what the function returns

Because the add function isn't modifying the passed value. If you want the add function to modify the value that is passed in, then do this:

Code:

int add (int& i)
{
	i=i+100;
	return i;
}

This way i is passed by reference instead of by value. Now any changes add makes will change the caller's variable.

Code:

int n = 1;
add(n);

In the case where the variable is passed by value, the above code will change the local version of the variable n (called i within the function itself) however, since this new value (although returned to the caller) is not assigned back to n in main, it (n) maintains its old value of 1.

Code:

int n = 1;
n=add(n);

In this case, the returned value 101 is assigned back to n overwritting what was there (a 1). Thus n will now contain 101 after the function call.

Code:

int n = 1;
cout << "n:" << add(n) << endl;

This will print the value returned by the function (101) but will not update n, n will still remain 1 after the call to add because it hasn't been changed.