Section 9.2
Pointers and Arrays in C++

A POINTER IS JUST THE ADDRESS OF SOME location in memory. In Java, pointers play an important role behind the scenes in the form of references to objects. A Java variable of object type stores a reference to an object, which is just a pointer giving the address of that object in memory. When you use an object variable in a program, the computer automatically follows the pointer stored in that variable in order to find the actual object in memory. Since all this happens automatically, behind the scenes, you really don't have to worry much about the fact that objects are referenced through pointers.

In C++, everything about pointers becomes explicit. Some variables can store pointers, but only variables that are of special pointer types. When you use a pointer variable, the computer does not follow the pointer automatically -- if you want the computer to follow the pointer, you have to use a special notation to tell it to do so. You can create a new object for a pointer variable to point to, using a "new" operator that is similar to the new operator in Java. But where Java will dispose of such objects automatically when you are done with them, in C++ you are responsible for keeping track of objects created with the new operator and disposing of them when they are no longer needed. Another operator, named delete is provided for doing this. All-in-all, working with explicit pointers is fairly difficult and error-prone, and it is generally considered to be one of the advanced aspects of programming. Unfortunately, in C++ there are some pretty basic things that you can't do without using pointers (not unless you use other advanced features of the language or use a class that someone else has written for you).

Suppose that T is the name of a type in C++. Then there is a pointer type named T* which represents variables that can hold pointers to objects of type T. (In fact, C++ allows pointers to values of any type. For example, there is a type int* that represents pointers to integers. A variable of type int* holds the address of a location in memory; an integer value is stored at that location.) Suppose that T is some type and that ptr is a variable of type T*. When you use ptr in a program, you are referring to the actual pointer, that is, the memory address that is contained in the variable ptr. To refer to the object that is stored at that address, you would use the notation *ptr. (C++ uses the predefined constant NULL -- all upper case -- as the value of a pointer that doesn't point to anything. It is an error to refer to *ptr if the value of ptr is NULL.)

The notation for defining a variable, ptr, of type T* is:

              T *ptr;

Thus, to declare iptr as a variable of type pointer-to-int, you would say:

             int *iptr;

The notation "T *ptr" is meant to suggest that the value of *ptr is of type T. But note that the variable that is being declared is ptr, not *ptr.

The type of a subroutine parameter can be a pointer type. For example, here is a subroutine that will swap the integer values stored in two memory locations. Its parameters are of type pointer-to-int:

             void swap(int *x, int *y) {
                int temp;
                temp = *x;  // copy the value pointed to by x into temp
                *x = *y;    // copy the value pointed to by y into the
                            //       location pointed to by x
                *y = temp;  // copy the value in temp into the location
                            //       pointed to by y

At the end of this subroutine, x and y still point to the same memory locations, but the values in those locations have been swapped.

This becomes more interesting once you know about the unary address operator, which is written as &. Suppose that x is any variable. Then &x is a notation that stands for the memory address of x. That is, &x is a pointer that points to x. Consider the following C++ program segment:

             int x = 7;   // x is an integer variable with initial value 7
             int *iptr;   // iptr is a variable that can hold a pointer to an int
             iptr = &x;   // iptr now holds a pointer to x, so that now
                          //     x and *iptr refer to the very same memory location
             *iptr += 3;  // adds 3 to the value stored in location *iptr
             cout << x;   // print out the value of x, which is 10

Because *iptr and x refer to the same memory location, adding 3 to *iptr is really the same as adding 3 to the value of x. So the output value of x in this example is 7 + 3.

Suppose that we combine the address operator with the swap() routine defined above. Consider this example:

             int a,b;
             a = 17;
             b = 42;
             swap(&a, &b);

The subroutine call is legal because &a and &b are pointers to integers. That is, they are of type int*, and that is the same type as the formal parameters of the subroutine. The end result of this example is that the original values of a and b are swapped. So the final value of a is 42, and the final value of b is 17. Believe it or not, using pointers and addresses like this is a very common way to write subroutines that can alter the values of variables. (In the C programming language, it's the only way. C++ provides a somewhat more sightly alternative, which is discussed below. Note that in Java, it is impossible to write a subroutine that performs the same task as swap(), since there is no such thing as a pointer-to-int in Java. Of course, you can do similar things with objects and references to objects, in Java as well as in C++.)

Pointers are probably used most often with classes, but I will put off that discussion to the next section.

References and Passing Parameters by Reference

In C++, it is possible to write a subroutine that can change the value stored in a variable which is used as an actual parameter to the subroutine. Here is a variation of the swap routine that does this:

               void swap(int& x, int& y) {
                  int temp = x;
                  x = y;
                  y = temp;

The type int& which is specified in this subroutine for the formal parameters x and y is called a reference type. If T is any type in C++, then you can form a reference type, T&. You can actually declare variables using reference types, but I have never seen any practical use for them except for formal parameters and occasionally as the return type of a function. When a formal parameter is declared using a reference type, the parameter is said to be passed by reference. The effect of passing a parameter by reference is to allow the subroutine to change the value stored in an actual parameter which is passed to the subroutine. Thus, if the subroutine swap() is defined as above, using pass-by-reference, then the subroutine call statement swap(a,b) will interchange the values stored in the variables a and b. On the other hand, consider the subroutine

                void failsToSwap(int x, int y) {
                   int temp = x;
                   x = y;
                   y = temp;

In this case, a subroutine call statement failsToSwap(a,b) will have no effect at all on the values stored in a and b. The values of the actual parameters a and b are copied into the formal parameters x and y when the subroutine is called. The subroutine interchanges the values of x and y, but this has no effect on what's stored in a and b. In the failsToSwap() subroutine, the parameters are said to be passed by value.

When a parameter is passed by reference, then what's passed to the subroutine is really a reference, or pointer, back to the actual parameter. The effect is similar to what was done in the first version of swap(), which passed pointers explicitly. But in pass-by-reference, the pointers are handled implicitly.

In Java, all parameters are passed by value, and so there is no way to write a subroutine that will swap the values stored in two integer variables. More generally, in Java, no subroutine can ever change the value stored in a variable that is passed to the subroutine as an actual parameter. This is kind of a shock to programmers who are used to other languages. (Be careful: when an object variable is passed as a parameter in Java, the value stored in the variable is a reference to an object. The values of instance variables in that object can be changed by the subroutine, even though the reference stored in the variable cannot be changed. When the subroutine ends, the variable will still refer to the same location in memory, but the values in the object stored at that location might have changed. In a sense, then, objects are passed by reference in Java, even though variables are not.)


Arrays in C++ are superficially similar to arrays in Java. An array is just an indexed list of values, where the index goes from 0 up to some maximum. But in C++, an array variable is considered to be nothing more than a pointer to the first element in the array. In fact, array variables and pointer variables can be used interchangeably. For example, if list is an array variable, then you can use either list[0] or *list to refer to the first item in the array. And you can use assignment statements to copy the value of an array variable into a pointer variable.

Arrays can be created in C++ using the new operator, just as in Java. For example, to create an array of 100 ints, you could say either

              int list[] = new int[100];


              int *list = new int[100];

(Note, however, that the notation for declaring an array variable is "int list[]", not "int[] list". In Java, either notation would be allowed, but I prefer the latter.)

It is also possible to create arrays without using the new operator:

               int list[100];  // declares an array of 100 ints

The main difference is that arrays created with the new operator should eventually be disposed of with the delete operator, in order to avoid memory leaks. Arrays created with the second notation do not have to be -- and in fact cannot be -- deleted. Most beginning C++ programmers would probably use arrays for quite a while before they even learn about the new operator.

Although an array in C++ has some fixed length, the computer does not keep track of that length in any way, and there is nothing to stop you from trying to refer to array locations that don't even exist. Thus, arrays -- like pointers in general -- are a common source of bugs in C++ programs.

Strings as Arrays of Characters

Strings in C++ are considered to be arrays of characters. A string, therefore, has type char[]. Since array types and pointer types are considered to be equivalent, we could also consider a string to be of type char*, that is pointer to char. Generally, in fact, a string is considered to be of type char*.

Since a string in C++ is just as array of characters, strings can be manipulated using array notation. For example, if str is a string variable, then str[i] refers to the i-th character in the string. You can change one of the characters in a string by using an assignment statement to the appropriate array element. For example, if str is the string "bit", then the assignment statement str[1]='e' changes the value of str to "bet". (In Java, the i-th character of a string str is referred to as str.charAt(i), and there is no way to change the characters in a string.)

In C++, there is no String class, and strings are not objects. So there are no class methods for operating on strings. Instead, a large number of subroutines for working with strings are available through a standard header file, string.h.

Since strings are arrays and since the computer does not keep track of the lengths of arrays, the computer does not know how long a string is. However, by convention, strings in C++ are assumed to end with a null character (the character with ASCII code zero). The null character is not really considered to be part of the string. It's there just as an end-of-string marker. This convention is used when a string variable is initialized. For example, if you say

               char *greeting = "Hello World";

the computer creates an array of 12 characters, not just 11, to hold the string. The characters in greeting[0] through greeting[10] are the characters 'H' through 'd' in the string that you've specified. The computer adds a null character in the last location, greeting[11], to mark the end of the string.

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