Pointers to Arrays
Pointers to Arrays Pointers, of course, can be “pointed at” any type of data
object, including arrays. While that was evident when we
discussed program 3.1, it is important to expand on how we do
this when it comes to multi-dimensional arrays.
To review, in Chapter 2 we stated that given an array of
integers we could point an integer pointer at that array using:
int *ptr;
ptr = &my_array[0]; /* point our pointer at the first
integer in our array */
As we stated there, the type of the pointer variable must match
the type of the first element of the array.
In addition, we can use a pointer as a formal parameter of a
function which is designed to manipulate an array. e.g.
Given:
int array[3] = {’1′, ‘5′, ‘7′};
void a_func(int *p);
we can pass the address of the array to the function by making
the call
a_func(array);
This kind of code promotes the mis-conception that pointers and
arrays are the same thing. Of course, if you have followed this
text carefully up to this point you know the difference between a
pointer and an array. The function would be better written (in
terms of clarity) as a_func(int p[]); Note that here
we need not include the dimension since what we are passing is
the address of the array, not the array itself.
We now turn to the problem of the 2 dimensional array. As
stated in the last chapter, C interprets a 2 dimensional array as
an array of one dimensional arrays. That being the case, the
first element of a 2 dimensional array of integers is a one
dimensional array of integers. And a pointer to a two
dimensional array of integers must be a pointer to that data
type. One way of accomplishing this is through the use of the
keyword “typedef”. typedef assigns a new name to a specified
data type. For example:
typedef unsigned char byte;
provides the name “byte” to mean type “unsigned char”. Hence
byte b[10]; would be an array of unsigned characters.
Note that in the typedef declaration, the word “byte” has
replaced that which would normally be the name of our unsigned
char. That is, the rule for using typedef is that the new name
for the data type is the name used in the definition of the data
type. Thus in:
typedef int Array[10];
Array becomes a data type for an array of 10 integers. i.e.
Array my_arr;
declares my_arr as an array of 10 integers and
Array arr2d[5];
makes arr2d an array of 5 arrays of 10 integers each.
Also note that Array *p1d; makes p1d a pointer to an
array of 10 integers. Because *p1d points to the same type as
arr2, assigning the address of the two dimensional array arr2d to
p1d, the pointer to a one dimensional array of 10 integers is
acceptable. i.e. p1d = &arr2d[0]; or p1d = arr2d;
are both correct.
Since the data type we use for our pointer is an array of 10
integers we would expect that incrementing p1d by 1 would change
its value by 10*sizeof(int), which it does. That is sizeof(*p1d)
is 20. You can prove this to yourself by writing and running a
simple short program.
Now, while using typedef makes things clearer for the reader
and easier on the programmer, it is not really necessary. What
we need is a way of declaring a pointer like p1d without the need
of the typedef keyword. It turns out that this can be done and
that int (*p1d)[10]; is the proper declaration, i.e. p1d here
is a pointer to an array of 10 integers just as it was under the
declaration using the Array type. Note that this is different
than int *p1d[10]; which would make p1d the name of an
array of 10 pointers to type int.
