# CHAPTER 2: Pointer types and Arrays

CHAPTER 2: Pointer types and Arrays

Okay, let’s move on. Let us consider why we need to identify
the “type” of variable that a pointer points to, as in:

int *ptr;

One reason for doing this is so that later, once ptr “points
to” something, if we write:

*ptr = 2;

the compiler will know how many bytes to copy into that memory
location pointed to by ptr. If ptr was declared as pointing to an
integer, 2 bytes would be copied, if a long, 4 bytes would be
copied. Similarly for floats and doubles the appropriate number
will be copied. But, defining the type that the pointer points
to permits a number of other interesting ways a compiler can
interpret code. For example, consider a block in memory
consisting if ten integers in a row. That is, 20 bytes of memory
are set aside to hold 10 integers.

Now, let’s say we point our integer pointer ptr at the first
of these integers. Furthermore lets say that integer is located
at memory location 100 (decimal). What happens when we write:

ptr + 1;

Because the compiler “knows” this is a pointer (i.e. its
value is an address) and that it points to an integer (its
ptr instead of 1, so the pointer “points to” the _next_
_integer_, at memory location 102. Similarly, were the ptr
declared as a pointer to a long, it would add 4 to it instead of
1. The same goes for other data types such as floats, doubles,
or even user defined data types such as structures. This is
obviously not the same kind of “addition” that we normally think
of. In C it is referred to as addition using “pointer
arithmetic”, a term which we will come back to later.

Similarly, since ++ptr and ptr++ are both equivalent to
ptr + 1 (though the point in the program when ptr is incremented
may be different), incrementing a pointer using the unary ++
operator, either pre- or post-, increments the address it stores
by the amount sizeof(type) where “type” is the type of the object
pointed to. (i.e. 2 for an integer, 4 for a long,
etc.).

Since a block of 10 integers located contiguously in memory
is, by definition, an array of integers, this brings up an
interesting relationship between arrays and pointers.

Consider the following:

int my_array[] = {1,23,17,4,-5,100};

Here we have an array containing 6 integers. We refer to
each of these integers by means of a subscript to my_array, i.e.
using my_array through my_array. But, we could
alternatively access them via a pointer as follows:

int *ptr;

ptr = &my_array; /* point our pointer at the first
integer in our array */

And then we could print out our array either using the array
notation or by dereferencing our pointer. The following code
illustrates this:
——————————————————
#include <stdio.h>

int my_array[] = {1,23,17,4,-5,100};
int *ptr;

int main(void)
{
int i;
ptr = &my_array; /* point our pointer to the first
element of the array */
printf(“\n\n”);
for(i = 0; i < 6; i++)
{
printf(“my_array[%d] = %d “,i,my_array[i]); /*<– A */
printf(“ptr + %d = %d\n”,i, *(ptr + i)); /*<– B */
}
return 0;
}
—————————————————-
Compile and run the above program and carefully note lines A
and B and that the program prints out the same values in either
case. Also observe how we dereferenced our pointer in line B,
i.e. we first added i to it and then dereferenced the new
pointer. Change line B to read:

printf(“ptr + %d = %d\n”,i, *ptr++);

and run it again… then change it to:

printf(“ptr + %d = %d\n”,i, *(++ptr));

and try once more. Each time try and predict the outcome and
carefully look at the actual outcome.

In C, the standard states that wherever we might use
&var_name we can replace that with var_name, thus in our code
where we wrote:

ptr = &my_array;

we can write:

ptr = my_array; to achieve the same result.

This leads many texts to state that the name of an array is a
pointer. While this is true, I prefer to mentally think “the
name of the array is the address of first element in the array”.
Many beginners (including myself when I was learning) have a
tendency to become confused by thinking of it as a pointer.
For example, while we can write ptr = my_array; we cannot write

my_array = ptr;

The reason is that the while ptr is a variable, my_array is a
constant. That is, the location at which the first element of
my_array will be stored cannot be changed once my_array[] has
been declared.

Earlier when discussing the term “lvalue” I cited K&R-2 where
it stated:

“An _object_ is a named region of storage; an _lvalue_ is an
expression referring to an object”.

This raises an interesting problem. Since my_array is a named
region of storage, why is “my_array” in the above assignment
statement not an lvalue? To resolve this problem, some refer to
“my_array” as an “unmodifiable lvalue”.

Modify the example program above by changing

ptr = &my_array; to ptr = my_array;

and run it again to verify the results are identical.

Now, let’s delve a little further into the difference between
the names “ptr” and “my_array” as used above. Some writers will
refer to an array’s name as a _constant_ pointer. What do we
mean by that? Well, to understand the term “constant” in this
sense, let’s go back to our definition of the term “variable”.
When we declare a variable we set aside a spot in memory to hold
the value of the appropriate type. Once that is done the name of
the variable can be interpreted in one of two ways. When used on
the left side of the assignment operator, the compiler interprets
it as the memory location to which to move that value resulting
from evaluation of the right side of the assignment operator.
But, when used on the right side of the assignment operator, the
name of a variable is interpreted to mean the contents stored at
that memory address set aside to hold the value of that variable.

With that in mind, let’s now consider the simplest of
constants, as in:

int i, k;
i = 2;

Here, while “i” is a variable and then occupies space in the
data portion of memory, “2” is a constant and, as such, instead
of setting aside memory in the data segment, it is imbedded
directly in the code segment of memory. That is, while writing
something like k = i; tells the compiler to create code which at
run time will look at memory location &i to determine the value
to be moved to k, code created by i = 2; simply puts the ‘2’ in
the code and there is no referencing of the data segment. That
is, both k and i are objects, but 2 is not an object.

Similarly, in the above, since “my_array” is a constant, once
the compiler establishes where the array itself is to be stored,
it “knows” the address of my_array and on seeing:

ptr = my_array;

it simply uses this address as a constant in the code segment and
there is no referencing of the data segment beyond that.

Well, that’s a lot of technical stuff to digest and I don’t
expect a beginner to understand all of it on first reading. With
time and experimentation you will want to come back and re-read
the first 2 chapters. But for now, let’s move on to the
relationship between pointers, character arrays, and strings.