IC210

Class 19: Functions IV

Reading

Review - Sections 3.1-3.3 of Absolute C++

Read - Section 4.2 of Absolute C++

 

Lecture

Most of this lecture involves function overloading. This concept/feature may not seem too important right now, but as we progress in our understanding of programming and in our knowledge of C++, the importance of overloading will become more apparent. When you look at templates in higher level classes, it will be more and more apparent. We're talking about it now, because it should help you understand the role of types in C++ even better, and why I say that understanding types is one of the most important things you can get out of this class!

 

Implicit type conversion and function calls

Consider the following bit of review. Suppose I have a function with prototype: void rep(char,int);, and with the following definition:

void rep(char c, int k)
{
for(int i = 0; i < k; i++)
††† cout << c;
}
††† 

Code

Resulting Output

Explanation

char g = '%';
intx = 10;
rep(g,x);
%%%%%%%%%%

Clear from the function definition.

intx = 10, y = 42;
rep(y,x);
**********

rep expects a char as its first argument, but the first argument in the function call, y, is an int. Thus, y gets converted to a char to match the prototype, and the char interpretation of 42 (as we see from the ASCII table) is '*'.

intx = 10;
double z = 42.553;
rep(z,x);
**********

rep expects a char as its first argument, but the first argument in the function call, z, is a double. Thus, z gets converted to a char to match the prototype, and the char interpretation of 42.553 is ASCII value 42, which is '*'.

 

Overloading Names

Now, what if I add another prototype and definition for the function rep, but with different types of arguments? For example, what if I have the following program

Prototypes

Definitions

main

void rep(char,int);
void rep(int,int);
void rep(char c, int k)
{
for(int i = 0; i < k; i++)
††† cout << c;
}
void rep(int n, int k);
{
for(int i = 0; i < k; i++)
††† cout << n;
}
int main()
{
int a = 55, b = 6;
char c = 'X';
 
rep(c,b);Alternative One
 
rep(a,b);Alternative Two
 
return 0;
}

What happens here? Well, Alternative One causes 6 Xís to be printed, and Alternative Two causes

555555555555

... to be printed. Why the difference? Which function gets used? Well, in Alternative One we used an object of type char as the first argument, so it used the version of rep with an object of type char as a first argument. In Alternative Two we used an object of type int as the first argument, so it used the version of rep with an object of type int as a first argument. This is called overloading of function names. The idea is, that the name of a function isn't just rep, the name of the function is really rep(int,int) or rep(char,int). So:

Overloading of a function name occurs when two or more definitions are given for functions with the same name, but different number or types of arguments.

Note: the return type of a function is not considered in function overloading. Thus, it is illegal to have two functions whose prototype differs only in the return type. For example: double f(int); and char f(int); cannot coexist!

 

Examples of Overloading

Definition 1

Definition 2

Discussion

int max(int a, int b)
{
if (a >= b)
††† return a;
else
††† return b;
}
int max(int a, int b, int c)
{
return max(max(a,b),c);
}

No conflict here, because I can look at the number of arguments and decide which function to call

void mystery(int k, char c)
{
for(int i = 0 i < k; i++)
{
††† for(int j = 0; j < k; j++)
††††† cout << c;
††† cout << endl;
}
}
char mystery(char c, int k)
{
return ((c - 'A' + k) % 26) + 'A';
}

No mystery here! If you see a call like mystery('X',13) you'd know that mystery(char,int) (i.e. the second definition) would be called. You just figure out which prototype matches.

 

Implicit Conversions and Overloading

When you make a function call with arguments whose types and number do not match any prototype, the compiler will try to use implicit type conversion to match a prototype. This is pretty straightforward (remember the example at the top of the page?) when function name overloading is not used. With oveloading, however, things can get complicated. For example:

Definition 1

Definition 2

Function Call

Discussion

void rep(char c, int k)
{
for(int i = 0; i < k; i++)
††† cout << c;
}
void rep(int n, int k);
{
for(int i = 0; i < k; i++)
††† cout << n;
}
rep(42.23,10);

Error! This function call is ambiguous! The compiler has no way of knowing whether the double 42.23 is supposed to be implicitly converted to the char * so we can use void rep(char,int), or whether it should be implicitly converted to the int 42 so we can use void rep(int,int). This results in a compiler error!

How is the compiler supposed to decide which implicit conversion is best? When ambiguities like this arise, simply use explicit conversion to disambiguate. For the previous example, I might change the call rep(42.23,10); to rep(char(42.23),10);, thus removing any ambiguity.

 

A Common Use

One common use of overloading would go something like this: I've defined the function void rep(char c, int k, ostream& OUT);, which writes the character c k times to the output stream OUT. This allows me to write to any output stream, of course, but I typically only want to write to the screen, and I don't like writing rep('$',30,cout), because I am evidently quite lazy, and instead I'd rather write rep('$',30) with the understanding that if no output stream is specified, rep should write to cout. One way to accomplish this is to simply overload rep by adding this function definition:

void rep(char c, int k)
{
rep(c,k,cout);
}

With this definition, I have my wish. It probably looks strange to have the function rep calling the function rep, but remember --- really we have the function rep(char,int) calling the function rep(char,int,ostream&). Try stepping through this program using the debugger, and watch how the call rep('$',30) actually results in calls to both rep functions.

 

Recursion

Let's consider the following function, which is a variation on our old familiar example: A function line(int k) that prints a line of k asterisks. And just to be on the safe side, let's make sure our function still works properly when negative k's are passed to it. "Works properly" in this case means, does nothing!

void line(int k)
{
// Takes care of bad arguments!
if (k < 0)
††† return;
 
// Takes care of everything else!
for(int i = 0; i < k; i++)
††† cout << '*';
cout << endl;
}
††† 

If you call this function with an argument like 35, you get:

***********************************

Now, suppose I want to modify this so that it prints a second line below of length k-1, so that calling this function with an argument like 35 would give me

***********************************
**********************************

Well, I might decide to try the following idea. Since line already does all the work of printing out a line for me, why don't I just add a line(k-1) at the end of my function definition? In other words, how aobut:

void line(int k)
{
// Takes care of bad arguments!
if (k < 0)
††† return;
 
// Takes care of everything else!
for(int i = 0; i < k; i++)
††† cout << '*';
cout << endl;
 
// Print next line?
line(k-1);
}
††† 

Well, if I compile and run this with something like line(10), here's what I get:

**********
*********
********
*******
******
*****
****
***
**
*

So what happened? A function that "calls itself" as this one does is called recursive, and the best way to understand what's going on here is to step through it using the debugger. So, step through this program using the debugger and see what's really going on here!

 

Problems

1.      Look at this program and explain what happens, i.e. answer the question posed up in the comment block.

2.      Look at this program, similar to the one above, and explain what happens, i.e. answer the question posed up in the comment block.

3.      Step through this program with the debugger and figure out what useful work (if any) the mystery function accomplishes.


Assoc Prof Christopher Brown

Last modified: LT M. Johnson 10/03/2007 01:57:23 PM