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

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

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

    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.

Modify the example program above by changing

    ptr = &my_array[0];     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.  We said that

my_array is 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 define 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 which lies on 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.

    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[0] 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.

==================================================================

CHAPTER 3:  Pointers and Strings

    The study of strings is useful to further tie in the

relationship between pointers and arrays.  It also makes it easy

to illustrate how some of the standard C string functions can be

implemented. Finally it illustrates how and when pointers can and

should be passed to functions.

    In C, strings are arrays of characters.  This is not

necessarily true in other languages.  In Pascal or (most versions

of) Basic, strings are treated differently from arrays.  To start

off our discussion we will write some code which, while preferred

for illustrative purposes, you would probably never write in an

actual program.  Consider, for example:

    char my_string[40];

    my_string[0] = 'T';

    my_string[1] = 'e';

    my_string[2] = 'd':

    my_string[3] = '\0';

    While one would never build a string like this, the end

result is a string in that it is an array of characters

_terminated_with_a_nul_character_.  By definition, in C, a string

is an array of characters terminated with the nul character. Note

that "nul" is _not_ the same as "NULL".  The nul refers to a zero

as is defined by the escape sequence '\0'.  That is it occupies

one byte of memory.  The NULL, on the other hand, is the value of

an uninitialized pointer and pointers require more than one byte

of storage.  NULL is #defined in a header file in your C

compiler, nul may not be #defined at all.

    Since writing the above code would be very time consuming, C

permits two alternate ways of achieving the same thing.  First,

one might write:

    char my_string[40] = {'T', 'e', 'd', '\0',};

    But this also takes more typing than is convenient.  So, C

permits:

    char my_string[40] = "Ted";

    When the double quotes are used, instead of the single quotes

as was done in the previous examples, the nul character ( '\0' )

is automatically appended to the end of the string.

    In all of the above cases, the same thing happens.  The

compiler sets aside an contiguous block of memory 40 bytes long

to hold characters and initialized it such that the first 4

characters are Ted\0.

    Now, consider the following program:

------------------program 3.1-------------------------------------

#include

char strA[80] = "A string to be used for demonstration purposes";

char strB[80];

int main(void)

{

   char *pA;     /* a pointer to type character */

   char *pB;     /* another pointer to type character */

   puts(strA);   /* show string A */

   pA = strA;    /* point pA at string A */

   puts(pA);     /* show what pA is pointing to */

   pB = strB;    /* point pB at string B */

   putchar('\n');       /* move down one line on the screen */

   while(*pA != '\0')   /* line A (see text) */

   {

     *pB++ = *pA++;     /* line B (see text) */

   }

   *pB = '\0';          /* line C (see text) */

   puts(strB);          /* show strB on screen */

   return 0;

}

--------- end program 3.1 -------------------------------------

    In the above we start out by defining two character arrays of

80 characters each.  Since these are globally defined, they are

initialized to all '\0's first.  Then, strA has the first 42

characters initialized to the string in quotes.

    Now, moving into the code, we define two character pointers

and show the string on the screen.  We then "point" the ponter pA

at strA.  That is, by means of the assignment statement we copy

the address of strA[0] into our variable pA.  We now use puts()

to show that which is pointed to by pA on the screen.  Consider

here that the function prototype for puts() is:

    int puts(const char *s);

    For the moment, ignore the "const".  The parameter passed to

puts is a pointer, that is the _value_ of a pointer (since all

parameters in C are passed by value), and the value of a pointer

is the address to which it points, or, simply, an address.  Thus

when we write:

    puts(strA);        as we have seen, we are passing the

address of strA[0].  Similarly, when we write:

    puts(pA);          we are passing the same address, since

we have set pA = strA;

    Given that, follow the code down to the while() statement on

line A.  Line A states:

    While the character pointed to by pA (i.e. *pA) is not a nul

character (i.e. the terminating '\0'), do the following:

    line B states:  copy the character pointed to by pA to the

space pointed to by pB, then increment pA so it points to the

next character and pB so it points to the next space.

    Note that when we have copied the last character, pA now

points to the terminating nul character and the loop ends.

However, we have not copied the nul character.  And, by

definition a string in C _must_ be nul terminated.  So, we add

the nul character with line C.

    It is very educational to run this program with your debugger

while watching strA, strB, pA and pB and single stepping through

the program.  It is even more educational if instead of simply

defining strB[] as has been done above, initialize it also with

something like:

strB[80] = "12345678901234567890123456789012345678901234567890"

where the number of digits used is greater than the length of

strA and then repeat the single stepping procedure while watching

the above variables.  Give these things a try!

    Of course, what the above program illustrates is a simple way

of copying a string.  After playing with the above until you have

a good understanding of what is happening, we can proceed to

creating our own replacement for the standard strcpy() that comes

with C.  It might look like:

   char *my_strcpy(char *destination, char *source)

   {

        char *p = destination

        while (*source != '\0')

        {

           *p++ = *source++;

        }

        *p = '\0';

        return destination.

   }

    In this case, I have followed the practice used in the

standard routine of returning a pointer to the destination.

    Again, the function is designed to accept the values of two

character pointers, i.e. addresses, and thus in the previous

program we could write:

int main(void)

{

  my_strcpy(strB, strA);

  puts(strB);

}

    I have deviated slightly from the form used in standard C

which would have the prototype:

    char *my_strcpy(char *destination, const char *source);

    Here the "const" modifier is used to assure the user that the

function will not modify the contents pointed to by the source

pointer.  You can prove this by modifying the function above, and

its prototype, to include the "const" modifier as shown.  Then,

within the function you can add a statement which attempts to

change the contents of that which is pointed to by source, such

as:

    *source = 'X';

which would normally change the first character of the string to

an X.  The const modifier should cause your compiler to catch

this as an error.  Try it and see.

    Now, let's consider some of the things the above examples

have shown us.  First off, consider the fact that *ptr++ is to be

interpreted as returning the value pointed to by ptr and then

incrementing the pointer value.  On the other hand, note that

this has to do with the precedence of the operators.  Were we to

write (*ptr)++ we would increment, not the pointer, but that

which the pointer points to!  i.e. if used on the first character

of the above example string the 'T' would be incremented to a

'U'.  You can write some simple example code to illustrate this.

    Recall again that a string is nothing more than an array

of characters.  What we have done above is deal with copying

an array.  It happens to be an array of characters but the

technique could be applied to an array of integers, doubles,

etc.  In those cases, however, we would not be dealing with

strings and hence the end of the array would not be

_automatically_ marked with a special value like the nul

character.  We could implement a version that relied on a

special value to identify the end. For example, we could

copy an array of postive integers by marking the end with a

negative integer.  On the other hand, it is more usual that

when we write a function to copy an array of items other

than strings we pass the function the number of items to be

copied as well as the address of the array, e.g. something

like the following prototype might indicate:

    void int_copy(int *ptrA, int *ptrB, int nbr);

where nbr is the number of integers to be copied.  You might want

to play with this idea and create an array of integers and see if

you can write the function int_copy() and make it work.

    Note that this permits using functions to manipulate very

large arrays.  For example, if we have an array of 5000 integers

that we want to manipulate with a function, we need only pass to

that function the address of the array (and any auxiliary

information such as nbr above, depending on what we are doing).

The array itself does _not_ get passed, i.e. the whole array is

not copied and put on the stack before calling the function, only

its address is sent.

    Note that this is different from passing, say an integer, to

a function.  When we pass an integer we make a copy of the

integer, i.e. get its value and put it on the stack.  Within the

function any manipulation of the value passed can in no way

effect the original integer.  But, with arrays and pointers we

can pass the address of the variable and hence manipulate the

values of of the original variables.

==================================================================

CHAPTER 4: More on Strings

    Well, we have progressed quite aways in a short time!  Let's

back up a little and look at what was done in Chapter 3 on

copying of strings but in a different light.  Consider the

following function:

   char *my_strcpy(char dest[], char source[])

   {

        int i = 0;

        while (source[i] != '\0')

        {

           dest[i] = source[i];

           i++;

        }

        dest[i] = '\0';

        return dest;

   }

    Recall that strings are arrays of characters.  Here we have

chosen to use array notation instead of pointer notation to do

the actual copying.  The results are the same, i.e. the string

gets copied using this notation just as accurately as it did

before.  This raises some interesting points which we will

discuss.

    Since parameters are passed by value, in both the passing of

a character pointer or the name of the array as above, what

actually gets passed is the address of the first element of each

array.  Thus, the numerical value of the parameter passed is the

same whether we use a character pointer or an array name as a

parameter.  This would tend to imply that somehow:

        source[i]  is the same as  *(p+i);

In fact, this is true, i.e wherever one writes   a[i]  it can be

replaced with  *(a + i) without any problems.  In fact, the

compiler will create the same code in either case.   Now, looking

at this last expression, part of it..  (a + i)  is a simple

addition using the + operator and the rules of c state that such

an expression is commutative.  That is   (a + i) is identical to

(i + a).  Thus we could write *(i + a) just as easily as

*(a + i).

    But *(i + a) could have come from i[a] !  From all of this

comes the curious truth that if:

    char a[20];

    int i;

    writing    a[3] = 'x';   is the same as writing

               3[a] = 'x';

    Try it!  Set up an array of characters, integers or longs,

etc. and assigned the 3rd or 4th element a value using the

conventional approach and then print out that value to be sure

you have that working.  Then reverse the array notation as I have

done above.  A good compiler will not balk and the results will

be identical.   A curiosity... nothing more!

    Now, looking at our function above, when we write:

        dest[i] = source[i];

    this gets interpreted by C to read:

        *(dest + i) = *(source + i);

    But, this takes 2 additions for each value taken on by i.

Additions, generally speaking, take more time than

incrementations (such as those done using the ++ operator as in

i++).  This may not be true in modern optimizing compilers, but

one can never be sure.  Thus, the pointer version may be a bit

faster than the array version.

    Another way to speed up the pointer version would be to

change:

    while (*source != '\0')     to simply    while (*source)

since the value within the parenthesis will go to zero (FALSE) at

the same time in either case.  

    At this point you might want to experiment a bit with writing

some of your own programs using pointers.  Manipulating strings

is a good place to experiment.  You might want to write your own

versions of such standard functions as:

            strlen();

            strcat();

            strchr();

and any others you might have on your system.

    We will come back to strings and their manipulation through

pointers in a future chapter.  For now, let's move on and discuss

structures for a bit.

==================================================================

CHAPTER 5: Pointers and Structures

    As you may know, we can declare the form of a block of data

containing different data types by means of a structure

declaration.  For example, a personnel file might contain

structures which look something like:

  struct tag{

       char lname[20];        /* last name */

       char fname[20];        /* first name */

       int age;               /* age */

       float rate;            /* e.g. 12.75 per hour */

       };

    Let's say we have an bunch of these structures in a disk file

and we want to read each one out and print out the first and last

name of each one so that we can have a list of the people in our

files.  The remaining information will not be printed out.  We

will want to do this printing with a function call and pass to

that function a pointer to the structure at hand.  For

demonstration purposes I will use only one structure for now. But

realize the goal is the writing of the function, not the reading

of the file which, presumably, we know how to do.

    For review, recall that we can access structure members with

the dot operator as in:

--------------- program 5.1 ------------------

#include

#include

struct tag{

       char lname[20];      /* last name */

       char fname[20];      /* first name */

       int age;             /* age */

       float rate;          /* e.g. 12.75 per hour */

       };

struct tag my_struct;       /* declare the structure m_struct */

int main(void)

{

  strcpy(my_struct.lname,"Jensen");

  strcpy(my_struct.fname,"Ted");

  printf("\n%s ",my_struct.fname);

  printf("%s\n",my_struct.lname);

  return 0;

}

-------------- end of program 5.1 --------------

    Now, this particular structure is rather small compared to

many used in C programs.  To the above we might want to add:

  date_of_hire;

  date_of_last_raise;

  last_percent_increase;

  emergency_phone;

  medical_plan;

  Social_S_Nbr;

  etc.....

    Now, if we have a large number of employees, what we want to

do manipulate the data in these structures by means of functions.

For example we might want a function print out the name of any

structure passed to it.  However, in the original C (Kernighan &

Ritchie) it was not possible to pass a structure, only a pointer

to a structure could be passed.  In ANSI C, it is now permissible

to pass the complete structure.  But, since our goal here is to

learn more about pointers, we won't pursue that.

    Anyway, if we pass the whole structure it means there must be

enough room on the stack to hold it.  With large structures this

could prove to be a problem.  However, passing a pointer uses a

minimum amount of stack space.

    In any case, since this is a discussion of pointers, we will

discuss how we go about passing a pointer to a structure and then

using it within the function.

    Consider the case described, i.e. we want a function that

will accept as a parameter a pointer to a structure and from

within that function we want to access members of the structure.

For example we want to print out the name of the employee in our

example structure.

    Okay, so we know that our pointer is going to point to a

structure declared using struct tag.  We define such a pointer

with the definition:

    struct tag *st_ptr;

and we point it to our example structure with:

    st_ptr = &my_struct;

    Now, we can access a given member by de-referencing the

pointer. But, how do we de-reference the pointer to a structure?

Well, consider the fact that we might want to use the pointer to

set the age of the employee.  We would write:

    (*st_ptr).age = 63;

    Look at this carefully.  It says, replace that within the

parenthesis with that which st_ptr points to, which is the

structure my_struct.  Thus, this breaks down to the same as

my_struct.age.

    However, this is a fairly often used expression and the

designers of C have created an alternate syntax with the same

meaning which is:

    st_ptr->age = 63;

    With that in mind, look at the following program:

------------ program 5.2 --------------

#include

#include

struct tag{                   /* the structure type */

       char lname[20];        /* last name */

       char fname[20];        /* first name */

       int age;               /* age */

       float rate;            /* e.g. 12.75 per hour */

       };

struct tag my_struct;         /* define the structure */

void show_name(struct tag *p);    /* function prototype */

int main(void)

{

  struct tag *st_ptr;         /* a pointer to a structure */

  st_ptr = &my_struct;        /* point the pointer to my_struct */

  strcpy(my_struct.lname,"Jensen");

  strcpy(my_struct.fname,"Ted");

  printf("\n%s ",my_struct.fname);

  printf("%s\n",my_struct.lname);

  my_struct.age = 63;

  show_name(st_ptr);          /* pass the pointer */

  return 0;

}

void show_name(struct tag *p)

{

  printf("\n%s ", p->fname);     /* p points to a structure */

  printf("%s ", p->lname);

  printf("%d\n", p->age);

}

-------------------- end of program 5.2 ----------------

    Again, this is a lot of information to absorb at one time.

The reader should compile and run the various code snippets and

using a debugger monitor things like my_struct and p while single

stepping through the main and following the code down into the

function to see what is happening.

==================================================================

CHAPTER 6:  Some more on Strings, and Arrays of Strings

   Well, let's go back to strings for a bit.  In the following

all assignments are to be understood as being global, i.e. made

outside of any function, including main.

   We pointed out in an earlier chapter that we could write:

   char my_string[40] = "Ted";

which would allocate space for a 40 byte array and put the string

in the first 4 bytes (three for the characters in the quotes and

a 4th to handle the terminating '\0'.

    Actually, if all we wanted to do was store the name "Ted" we

could write:

      char my_name[] = "Ted";

and the compiler would count the characters, leave room for the

nul character and store the total of the four characters in memory

the location of which would be returned by the array name, in this

case my_string.

    In some code, instead of the above, you might see:

     char *my_name = "Ted";

which is an alternate approach.  Is there a difference between

these?  The answer is.. yes.  Using the array notation 4 bytes of

storage in the static memory block are taken up, one for each

character and one for the nul character.  But, in the pointer

notation the same 4 bytes required, _plus_ N bytes to store the

pointer variable my_name (where N depends on the system but is

usually a minimum of 2 bytes and can be 4 or more).

    In the array notation, my_name is a constant (not a

variable).  In the pointer notation my_name is a variable.  As to

which is the _better_ method, that depends on what you are going

to do within the rest of the program.

    Let's now go one step further and consider what happens if

each of these definitions are done within a function as opposed

to globally outside the bounds of any function.

void my_function_A(char *ptr)

{

  char a[] = "ABCDE";

  .

  .

}

void my_function_B(char *ptr)

{

  char *cp = "ABCDE";

  .

  .

}

    Here we are dealing with automatic variables in both cases.

In my_function_A the automatic variable is the character array

a[]. In my_function_B it is the pointer cp.  While C is designed

in such a way that a stack is not required on those processors

which don't use them, my particular processor (80286) has a

stack.  I wrote a simple program incorporating functions similar

to those above and found that in my_function_A the 5 characters

in the string were all stored on the stack.  On the other hand,

in my_function_B, the 5 characters were stored in the data space

and the pointer was stored on the stack.

    By making a[] static I could force the compiler to place the

5 characters in the data space as opposed to the stack.  I did

this exercise to point out just one more difference between

dealing with arrays and dealing with pointers.  By the way, array

initialization of automatic variables as I have done in

my_function_A was illegal in the older K&R C and only "came of

age" in the newer ANSI C.  A fact that may be important when one

is considering portabilty and backwards compatability.

    As long as we are discussing the relationship/differences

between pointers and arrays, let's move on to multi-dimensional

arrays.  Consider, for example the array:

    char multi[5][10];

    Just what does this mean?   Well, let's consider it in the

following light.

        char multi[5][10];

        ^^^^^^^^^^^^^

    If we take the first, underlined, part above and consider it

to be a variable in its own right, we have an array of 10

characters with the "name"  multi[5].  But this name, in itself,

implies an array of 5 somethings.  In fact, it means an array of

five 10 character arrays.  Hence we have an array of arrays.  In

memory we might think of this as looking like:

      multi[0] = "0123456789"

      multi[1] = "abcdefghij"

      multi[2] = "ABCDEFGHIJ"

      multi[3] = "9876543210"

      multi[4] = "JIHGFEDCBA"

with individual elements being, for example:

      multi[0][3] = '3'

      multi[1][7] = 'h'

      multi[4][0] = 'J'

    Since arrays are to be contiguous, our actual memory block

for the above should look like:

    "0123456789abcdefghijABCDEFGHIJ9876543210JIHGFEDCBA"

    Now, the compiler knows how many columns are present in the

array so it can interpret multi + 1 as the address of the 'a' in

the 2nd row above.  That is, it adds 10, the number of columns,

to get this location.  If we were dealing with integers and an

array with the same dimension the compiler would add

10*sizeof(int) which, on my machine, would be 20.  Thus, the

address of the "9" in the 4th row above would be &multi[3][0] or

*(multi + 3) in pointer notation.  To get to the content of the

2nd element in row 3 we add 1 to this address and dereference the

result as in

    *(*(multi + 3) + 1)

    With a little thought we can see that:

    *(*(multi + row) + col)    and

    multi[row][col]            yield the same results.

    The following program illustrates this using integer arrays

instead of character arrays.

------------------- program 6.1 ----------------------

#include

#define ROWS 5

#define COLS 10

int multi[ROWS][COLS];

int main(void)

{

  int row, col;

  for (row = 0; row < ROWS; row++)

    for(col = 0; col < COLS; col++)

      multi[row][col] = row*col;

  for (row = 0; row < ROWS; row++)

    for(col = 0; col < COLS; col++)

    {

      printf("\n%d  ",multi[row][col]);

      printf("%d ",*(*(multi + row) + col));

    }

  return 0;

}

----------------- end of program 6.1 ---------------------

    Because of the double de-referencing required in the pointer

version, the name of a 2 dimensional array is said to be a

pointer to a pointer.  With a three dimensional array we would be

dealing with an array of arrays of arrays and a pointer to a

pointer to a pointer.  Note, however, that here we have initially

set aside the block of memory for the array by defining it using

array notation.  Hence, we are dealing with an constant, not a

variable.  That is we are talking about a fixed pointer not a

variable pointer.  The dereferencing function used above permits

us to access any element in the array of arrays without the need

of changing the value of that pointer (the address of multi[0][0]

as given by the symbol "multi").

EPILOG:

    I have written the preceding material to provide an

introduction to pointers for newcomers to C.  In C, the more one

understands about pointers the greater flexibility one has in the

writing of code.  The above has just scratched the surface of the

subject. In time I hope to expand on this material.  Therefore,

if you have questions, comments, criticisms, etc. concerning that

which has been presented, I would greatly appreciate your

contacting me using one of the mail addresses cited in the

Introduction.

可以说是深入浅出吧，本人是个菜鸟，希望与各位菜鸟朋友共勉