Function Overloading & Type Conversion

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. We're talking about it now because it should help you understand the role of types even better, and why we say that understanding types is one of the most important things you can get out of this class!

It's also of note that C does not allow overloading of functions, and this is another one of those handy C++ things. Even if you find yourself only using C in the future, this topic is helpful in deepening your understanding of types, and understanding the different constraints under which each programming language uniquely has.

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;
}
CodeResulting OutputExplanation

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

%%%%%%%%%%
Clear from the function definition.

int  x = 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 '*'.

int  x = 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
PrototypesDefinitionsmain

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 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!

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:
Definitions 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.

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 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. Step through this program 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. Going back to the scoping issue, look at this program. What will be the output of the program?
  4. Step through this program and figure out what useful work (if any) the mystery function accomplishes.