## Homework

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;
}

If I call it with this code:

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

I get this output:

%%%%%%%%%%

And, if I call it with this code:

int  x = 10, y = 42;
rep(y,x);

I get this output:

**********

Why? 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 '*'.

Finally, if I call it with this code:

    int  x = 10;
double z = 42.553;
rep(z,x);

I get this output:

**********

This is because 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 '*'.

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:

    void rep(char,int);
void rep(int,int);

int main() {
int a = 55, b = 6;
char c = 'X';

rep(c,b);  //Alternative One

rep(a,b);  //Alternative Two

return 0;
}

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;
}

What happens here? Well, Alternative One causes 6 X's 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!

int max(int a, int b) {  //method A
if (a >= b)
return a;
else
return b;
}
int max(int a, int b, int c) { //method B
return max(max(a,b),c);
}

Now, suppose in my main function I call max(2,5,1). There is no conflict here, because the compiler can look at the number of arguments and decide which function to call - it will call method B, and assign the value 2 to a, 5 to b, and 1 to c. Within that method, it then makes the function call max(a,b) - again, there's no conflict - it will call method A to resolve this expression (returning the value 5), before calling max(5,c). Again, it calls method A (because it has two arguments), and evaluates to 5, which is returned.

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. If it's mystery(95,5), it'll call the first definition. You just figure out which prototype matches.

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, suppose we have functions with the following two prototypes:

void rep(char, int);
void rep(int, int);

and we make the function call rep(42.23,10). Neither is a perfect match, so which one gets called? If we only had one of these functions, the program would just cast 42.23 to the proper type, but the ambiguity that results from having both results in a compiler error - C++ isn't going to guess which one you mean.

You can clear up the abiguity by using explicit casting, for example, I might change the call rep(42.23,10); to rep(char(42.23),10);.

## 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, and see how the call rep('\$',30) actually results in calls to both rep functions.

## Recursion ← Note: this is just a teaser trailer for recursion, it gets covered in detail later

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 about:

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. Hint: recall that char + int produces an int value!
3. Step through this program with the debugger and figure out what useful work (if any) the mystery function accoplishes.