C++ Programming Tutorials

Functions (II)
Arguments passed by value and by reference.
Until now, in all the functions we have seen, the arguments passed to the functions have been passed by value. This means that when calling a function with parameters, what we have passed to the function were copies of their values but never the variables themselves. For example, suppose that we called our first function addition using the following code:

int x=5, y=3, z;
z = addition ( x , y );

What we did in this case was to call to function addition passing the values of x and y, i.e. 5 and 3 respectively, but not the variables x and y themselves.

C++ Programming Tutorials - addition function

This way, when the function addition is called, the value of its local variables a and b become 5 and 3 respectively, but any modification to either a or b within the function addition will not have any effect in the values of x and y outside it, because variables x and y were not themselves passed to the function, but only copies of their values at the moment the function was called.

But there might be some cases where you need to manipulate from inside a function the value of an external variable. For that purpose we can use arguments passed by reference, as in the function duplicate of the following example:

// passing parameters by reference
#include <iostream>
using namespace std;

void duplicate (int& a, int& b, int& c)
{
a*=2;
b*=2;
c*=2;
}

int main ()
{
int x=1, y=3, z=7;
duplicate (x, y, z);
cout << "x=" << x << ", y=" << y << ", z=" << z;
return 0;
}

x=2, y=6, z=14

The first thing that should call your attention is that in the declaration of duplicate the type of each parameter was followed by an ampersand sign (&). This ampersand is what specifies that their corresponding arguments are to be passed by reference instead of by value.

When a variable is passed by reference we are not passing a copy of its value, but we are somehow passing the variable itself to the function and any modification that we do to the local variables will have an effect in their counterpart variables passed as arguments in the call to the function.

C++ Programming Tutorials - local variables

To explain it in another way, we associate a, b and c with the arguments passed on the function call (x, y and z) and any change that we do on a within the function will affect the value of x outside it. Any change that we do on b will affect y, and the same with c and z.

That is why our program's output, that shows the values stored in x, y and z after the call to duplicate, shows the values of all the three variables of main doubled.

If when declaring the following function:

void duplicate (int& a, int& b, int& c)

we had declared it this way:

void duplicate (int a, int b, int c)

i.e., without the ampersand signs (&), we would have not passed the variables by reference, but a copy of their values instead, and therefore, the output on screen of our program would have been the values of x, y and z without having been modified.

Passing by reference is also an effective way to allow a function to return more than one value. For example, here is a function that returns the previous and next numbers of the first parameter passed.

// more than one returning value
#include <iostream>
using namespace std;

void prevnext (int x, int& prev, int& next)
{
prev = x-1;
next = x+1;
}

int main ()
{
int x=100, y, z;
prevnext (x, y, z);
cout << "Previous=" << y << ", Next=" << z;
return 0;
}

Previous=99, Next=101

Default values in parameters.
When declaring a function we can specify a default value for each parameter. This value will be used if the corresponding argument is left blank when calling to the function. To do that, we simply have to use the assignment operator and a value for the arguments in the function declaration. If a value for that parameter is not passed when the function is called, the default value is used, but if a value is specified this default value is ignored and the passed value is used instead. For example:

// default values in functions
#include <iostream>
using namespace std;

int divide (int a, int b=2)
{
int r;
r=a/b;
return (r);
}

int main ()
{
cout << divide (12);
cout << endl;
cout << divide (20,4);
return 0;
}

6
5

As we can see in the body of the program there are two calls to function divide. In the first one:

divide (12)

we have only specified one argument, but the function divide allows up to two. So the function divide has assumed that the second parameter is 2 since that is what we have specified to happen if this parameter was not passed (notice the function declaration, which finishes with int b=2, not just int b). Therefore the result of this function call is 6 (12/2).

In the second call:

divide (20,4)

there are two parameters, so the default value for b (int b=2) is ignored and b takes the value passed as argument, that is 4, making the result returned equal to 5 (20/4).

Overloaded functions.
In C++ two different functions can have the same name if their parameter types or number are different. That means that you can give the same name to more than one function if they have either a different number of parameters or different types in their parameters. For example:

// overloaded function
#include <iostream>
using namespace std;

int operate (int a, int b)
{
return (a*b);
}

float operate (float a, float b)
{
return (a/b);
}

int main ()
{
int x=5,y=2;
float n=5.0,m=2.0;
cout << operate (x,y);
cout << "\n";
cout << operate (n,m);
cout << "\n";
return 0;
}

10
2.5

In this case we have defined two functions with the same name, operate, but one of them accepts two parameters of type int and the other one accepts them of type float. The compiler knows which one to call in each case by examining the types passed as arguments when the function is called. If it is called with two ints as its arguments it calls to the function that has two int parameters in its prototype and if it is called with two floats it will call to the one which has two float parameters in its prototype.

In the first call to operate the two arguments passed are of type int, therefore, the function with the first prototype is called; This function returns the result of multiplying both parameters. While the second call passes two arguments of type float, so the function with the second prototype is called. This one has a different behavior: it divides one parameter by the other. So the behavior of a call to operate depends on the type of the arguments passed because the function has been overloaded.

Notice that a function cannot be overloaded only by its return type. At least one of its parameters must have a different type.

NEXT >> inline functions

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