?? begpoint.txt
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UNDERSTANDING POINTERS (for beginners)
by Ted Jensen
Version 0.0
This material is hereby placed in the public domain.
September 5, 1993
TABLE OF CONTENTS
INTRODUCTION;
CHAPTER 1: What is a pointer?
CHAPTER 2: Pointer types and Arrays
CHAPTER 3: Pointers and Strings
CHAPTER 4: More on Strings
CHAPTER 5: Pointers and Structures
CHAPTER 6: Some more on Strings, and Arrays of Strings
EPILOG:
==================================================================
INTRODUCTION:
Over a period of several years of monitoring various
telecommunication conferences on C I have noticed that one of the
most difficult problems for beginners was the understanding of
pointers. After writing dozens of short messages in attempts to
clear up various fuzzy aspects of dealing with pointers, I set up
a series of messages arranged in "chapters" which I could draw
from or email to various individuals who appeared to need help in
this area.
Recently, I posted all of this material in the FidoNet CECHO
conference. It received such a good acceptance, I decided to
clean it up a little and submit it for inclusion in Bob Stout's
SNIPPETS file.
It is my hope that I can find the time to expand on this text
in the future. To that end, I am hoping that those who read this
and find where it is lacking, or in error, or unclear, would
notify me of same so the next version, should there be one, I can
correct these deficiencys.
It is impossible to acknowledge all those whose messages on
pointers in various nets contributed to my knowledge in this
area. So, I will just say Thanks to All.
I frequent the CECHO on FidoNet via RBBSNet and can be
contacted via the echo itself or by email at:
RBBSNet address 8:916/1.
I can also be reached via
Internet email at ted.jensen@spacebbs.com
Or Ted Jensen
P.O. Box 324
Redwood City, CA 94064
==================================================================
CHAPTER 1: What is a pointer?
One of the things beginners in C find most difficult to
understand is the concept of pointers. The purpose of this
document is to provide an introduction to pointers and their use
to these beginners.
I have found that often the main reason beginners have a
problem with pointers is that they have a weak or minimal feeling
for variables, (as they are used in C). Thus we start with a
discussion of C variables in general.
A variable in a program is something with a name, the value
of which can vary. The way the compiler and linker handles this
is that it assigns a specific block of memory within the computer
to hold the value of that variable. The size of that block
depends on the range over which the variable is allowed to vary.
For example, on PC's the size of an integer variable is 2 bytes,
and that of a long integer is 4 bytes. In C the size of a
variable type such as an integer need not be the same on all
types of machines.
When we declare a variable we inform the compiler of two
things, the name of the variable and the type of the variable.
For example, we declare a variable of type integer with the name
k by writing:
int k;
On seeing the "int" part of this statement the compiler sets
aside 2 bytes (on a PC) of memory to hold the value of the
integer. It also sets up a symbol table. And in that table it
adds the symbol k and the address in memory where those 2 bytes
were set aside.
Thus, later if we write:
k = 2;
at run time we expect that the value 2 will be placed in that
memory location reserved for the storage of the value of k.
In a sense there are two "values" associated with k, one
being the value of the integer stored there (2 in the above
example) and the other being the "value" of the memory location
where it is stored, i.e. the address of k. Some texts refer to
these two values with the nomenclature rvalue (right value,
pronounced "are value") and lvalue (left value, pronunced "el
value") respectively.
The lvalue is the value permitted on the left side of the
assignment operator '=' (i.e. the address where the result of
evaluation of the right side ends up). The rvalue is that which
is on the right side of the assignment statment, the '2' above.
Note that rvalues cannot be used on the left side of the
assignment statement. Thus: 2 = k; is illegal.
Okay, now consider:
int j, k;
k = 2;
j = 7; <-- line 1
k = j; <-- line 2
In the above, the compiler interprets the j in line 1 as the
address of the variable j (its lvalue) and creates code to copy
the value 7 to that address. In line 2, however, the j is
interpreted as its rvalue (since it is on the right hand side of
the assignment operator '='). That is, here the j refers to the
value _stored_ at the memory location set aside for j, in this
case 7. So, the 7 is copied to the address designated by the
lvalue of k.
In all of these examples, we are using 2 byte integers so all
copying of rvalues from one storage location to the other is done
by copying 2 bytes. Had we been using long integers, we would be
copying 4 bytes.
Now, let's say that we have a reason for wanting a variable
designed to hold an lvalue (an address). The size required to
hold such a value depends on the system. On older desk top
computers with 64K of memory total, the address of any point in
memory can be contained in 2 bytes. Computers with more memory
would require more bytes to hold an address. Some computers,
such as the IBM PC might require special handling to hold a
segment and offset under certain circumstances. The actual size
required is not too important so long as we have a way of
informing the compiler that what we want to store is an address.
Such a variable is called a "pointer variable" (for reasons
which will hopefully become clearer a little later). In C when
we define a pointer variable we do so by preceding its name with
an asterisk. In C we also give our pointer a type which, in this
case, refers to the type of data stored at the address we will be
storing in our pointer. For example, consider the variable
definition:
int *ptr;
ptr is the _name_ of our variable (just as 'k' was the name
of our integer variable). The '*' informs the compiler that we
want a pointer variable, i.e. to set aside however many bytes is
required to store an address in memory. The "int" says that we
intend to use our pointer variable to store the address of an
integer. Such a pointer is said to "point to" an integer. Note,
however, that when we wrote "int k;" we did not give k a value.
If this definiton was made outside of any function many compilers
will initialize it to zero. Simlarly, ptr has no value, that is
we haven't stored an address in it in the above definition. In
this case, again if the definition is outside of any function, it
is intialized to a value #defined by your compiler as NULL. It
is called a NULL pointer. While in most cases NULL is #defined
as zero, it need not be. That is, different compilers handle
this differently. Also note that while zero is an integer, NULL
need not be.
But, back to using our new variable ptr. Suppose now that we
want to store in ptr the address of our integer variable k. To
do this we use the unary '&' operator and write:
ptr = &k;
What the '&' operator does is retrieve the lvalue (address)
of k, even though k is on the right hand side of the assignment
operator '=', and copies that to the contents of our pointer ptr.
Now, ptr is said to "point to" k. Bear with us now, there is
only one more operator we need to discuss.
The "dereferencing operator" is the asterisk and it is used
as follows:
*ptr = 7;
will copy 7 to the address pointed to by ptr. Thus if ptr
"points to" (contains the address of) k, the above statement will
set the value of k to 7. That is, when we use the '*' this way
we are refering to the value of that which ptr is pointing
at, not the value of the pointer itself.
Similarly, we could write:
printf("%d\n",*ptr);
to print to the screen the integer value stored at the address
pointed to by "ptr".
One way to see how all this stuff fits together would be to
run the following program and then review the code and the output
carefully.
-------------------------------------------------
#include <stdio.h>
int j, k;
int *ptr;
int main(void)
{
j = 1;
k = 2;
ptr = &k;
printf("\n");
printf("j has the value %d and is stored at %p\n",j,&j);
printf("k has the value %d and is stored at %p\n",k,&k);
printf("ptr has the value %p and is stored at %p\n",ptr,&ptr);
printf("The value of the integer pointed to by ptr is %d\n",
*ptr);
return 0;
}
---------------------------------------
To review:
A variable is defined by giving it a type and a name (e.g.
int k;)
A pointer variable is defined by giving it a type and a name
(e.g. int *ptr) where the asterisk tells the compiler that
the variable named ptr is a pointer variable and the type
tells the compiler what type the pointer is to point to
(integer in this case).
Once a variable is defined, we can get its address by
preceding its name with the unary '&' operator, as in &k.
We can "dereference" a pointer, i.e. refer to the value of
that which it points to, by using the unary '*' operator as
in *ptr.
An "lvalue" of a variable is the value of its address, i.e.
where it is stored in memory. The "rvalue" of a variable is
the value stored in that variable (at that address).
==================================================================
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 defined 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 integer.
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
current address, 100, is the address of an integer), it adds 2 to
ptr instead of 1, so the pointer "points to" the _next_
_integer_, at memory location 102. Similarly, were the ptr
defined 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.
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) (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[0] through my_array[5]. But, we could
alternatively access them via a pointer as follows:
int *ptr;
ptr = &my_array[0]; /* 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[0]; /* point our pointer to 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 note how we dereferenced our pointer in line B, i.e.
we first added i to it and then dereferenced the 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[0] we can replace that with var_name, thus in our code
where we wrote:
ptr = &my_array[0];
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 a _constant_ pointer". Many beginners
(including myself when I was learning) forget that _constant_
qualifier. In my opinon this leads to some confusion. For
example, while we can write ptr = my_array; we cannot write
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