A little something fun — making your new type act like the builtin types

C++ allows the programmer to make new types that really act like the builtin types, by which I mean that the usual operators, like *, <, ++, work with the new types, and I/O with << and >>, and so on.

There's actually some controversy about this, i.e. about whether or not it's a good idea to let programmers do this. Some people feel that it leads to hard-to-follow code.

In any event, it's not going to be a major theme for this course, but I'd like to show it to you so that you understand that we really can build new types in C++ if we want to. Also, I should note, this is why + and >> and so on work with C++ string objects. The implementers of the string library defined all these operators for their nice string type.

Simple program

Let's write a program that sort of plots a course this way.
  1. You start out at (0,0).
  2. Then, you can enter in "moves". A move of (-2,4) means move down 2 units and to the right 4 units from wherever your current position is.
  3. The user enters moves continually until he finally enters a q to quit.
First solution. My first solution is a pretty simple program.
The meat the program is:

Operator overloading. While it may be simple, it would be nice to be able to write the function as if point were a built-in type, meaning that I could add points p and m by saying p + m.

// Initialization
ofstream OUT("out.txt");
point p,m;
p.x = p.y = m.x = m.y = 0;

// Get moves & write moves
do {
  // Compute new position p from move m 
  p.x = p.x + m.x;
  p.y = p.y + m.y; 
  // Write move
  OUT << p.x << '\t' << p.y << endl;

// Initialization
ofstream OUT("out.txt");
point p,m;
p.x = p.y = m.x = m.y = 0;

// Get moves & write moves
do {
  // Compute new position p from move m 
  p = p + m;  // *** LOOK HERE ***

  // Write move
  OUT << p.x << '\t' << p.y << endl;
This wouldn't change what I could accomplish, but it would add a little "syntactic sugar" to sweeten the program a bit.
To do this, we need to be able to tell the compiler what it means to + to point objects. Doing this is quite easy once you understand the following:

a + b is just the same as the function call operator+(a,b) in C++.

So if you want to tell the compiler what + means for two point objects, you need to define the function operator+(point a,point b) --- i.e. overload the + operator for points. The prototype is clear:

point operator+(point a, point b);
... at least I hope it's clear that we should return a point when we add two points. The function definition is ... just like any other function definition:

point operator+(point a, point b)
  point S;
  S.x = a.x + b.x;
  S.y = a.y + b.y;
  return S;


"Operator overloading" is the term used for defining versions of the C++ operators for the new structs we define. In general, if you have an expression A Π B, where "Π" stands for some operator, then that is equivalent to a function call operatorΠ(A,B). So, to subtract two points we'd define
point operator-(point A, point B);
... and to compare two points with less than we'd define
bool operator<(point A, point B);
... or to multiply a point (on the left) by a real number (on the right) we'd define
point operator*(point A, double w);
So with that addition, here's my second verison of the program. Now, the syntactic sugar may not seem worth the effort here, but you'll probably be using the point struct over and over, and you'll like being able to add points. Wouldn't it be nice to define the midpoint function like this:

point midpoint(point a, point b)
  return (a + b)/2.0;
Now, in addition to defining operator+ for two point objects, what else would you need? Well, (a + b) is an object of type point, and I'm dividing it by an object of type double, so I need to define operator/(point,double). What type of object should be returned here?

point operator/(point P, double z)
  point Q;
  Q.x = P.x / z;
  Q.y = P.y / z;
  return Q;

I/O and overloading

Just to round out our examples, here's how to overload << and >>.

istream& operator>>(istream &in, point &A)
  char c;
  return in >> c >> A.x >> c >> A.y >> c;

ostream& operator<<(ostream &out, point A)
  return out << '(' << A.x << ',' << A.y << ')';
The prototypes of these two should actually make some sense. For example, we've talked before about how cin >> x actually evaluates back to cin.
What is new, however, is the "return by reference".
Normally, when a function returns an object X, the returned object in the calling function is a copy of that X. By using return by reference, we're saying "no, I want the calling function to get the exact same object I returned, not a copy". Of course, this can only be done with something that doesn't go out of scope and die at the end of the function call. To wrap all of this function overloading stuff up, consider this example: point.h and point.cpp.

Static Arrays

Dynamic arrays vs static arrays

There are two types of arrays in C++. What we've seen are called dynamic arrays, because the size of the arrays can change from run to run of the program. If your program creates and array to store data in a file, and the file (rather than you the programmer) is what determines how many elements ought to be in the array, the size of the array is bound to change from run to run of the program, because the file may change.

The other kind of array is static A static array has its size determined when the program is compiled, so from run to run of the program the array's size is always the same. A static array wouldn't work for a program that reads in data from a file that contains an unknown amount of information. How big should you the programmer make the array? You don't know.

Dynamic arrays are more general - anything you can do with static arrays you can do with dynamic arrays, and then some. However, in some instances static arrays are simple - for example if you wanted to hardcode the names of the days of the week into a program. You know the size will be 7, so there's no point in making a dynamic array.


The syntax for creating a static array is a little bit different than for creating a dynamic array.

Creating an Array of 6 ints
Static ArrayDynamic Array

int A[6]

int *A = new int[6]
Using the array after its been created is pretty much the same for either.

Consider a program that uses an array to store the vertices of a quadrilateral. Since we know that a quadrilateral always has exactly four vertices, we could use static arrays.

struct Quad
  char label;
  point vert[4];

Difference between dynamic and static arrays

One difference is that if A is a static array, the pointer A cannot be changed. The contents of the array to which it points can, of course, be changed. But not the pointer itself. Other differences are best illustrated by an example. To understand the difference between this version of Quad and the previous version, consider this picture:

You see that in the static array version the array of vertices is embedded in the Quad object. In the dynamic version, the pointer is embedded in the object, while the array is outside of the object, somewhere else in memory. Compare the static array version of main with the dynamic array version of main.

The above picture really tells you all you need to know to understand the difference between using static and dynamic arrays ... when you really can use static arrays.

Let's look at one example to see what consequences arise from this picture.

Quad S;
...       // S has 'Q' and (0,0), (1,0), (1,1), and (0,1) 
print(S); // It will print Q (0,0) (1,0) (1,1) (0,1)

Quad R;
R = S;    // Copying will have different meanings!

R.label = 'P';    
for(int i = 0; i < 4; i++)


Q: Suppose that I have a Quad object S that contains the label 'Q' and the vertices (0,0) (1,0) (1,1) (0,1). I then print out S and then R. What will I get?

A: It depends whether I'm using the static version of Quad or the dynamic version. (Drag your mouse for answers)

Static VersionDynamic Version
Q (0,0) (1,0) (1,1) (0,1)
Q (0,0) (1,0) (1,1) (0,1)
P (1,0) (2,0) (2,1) (1,1)
Q (0,0) (1,0) (1,1) (0,1)
Q (1,0) (2,0) (2,1) (1,1)
P (1,0) (2,0) (2,1) (1,1)

Why the difference? Look at the picture!

Dynamic VersionStatic Version

This is not a reason to use static over dynamic, but it is a good example of how and why they behave differently.

Here's another example:

Q: How does swap(A,B) behave differently for two Quads, A and B, with the dynamic versus static array versions of Quad?

A: Once again, the picture should tell you that while the result it the same, a lot more work gets done in the static case, where the entire contents of the arrays are swapped, rather than simply the pointers.


  1. Write a program that reads a date in mm/dd/yyyy format and prints it out in "dd monthname yyyy" format. It might be helpful to know that a static array can be initialized with a list of values in { }'s. For example, an array of the first 10 prime numbers can be constructed like this:
    int prime[10] = {2,3,5,7,11,13,17,19,23,29};
    Note: this is purely about static arrays, it doesn't concern structs at all. Here's my solution.