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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 a 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 ------------------

/* Program 5.1 from PTRTUT10.HTM     6/13/97 */


#include <stdio.h>
#include <string.h>

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 my_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;                  (data types not shown)
    date_of_last_raise;
    last_percent_increase;
    emergency_phone;
    medical_plan;
    Social_S_Nbr;
    etc.....

If we have a large number of employees, what we want to do is manipulate the data in these structures by means of functions. For example we might want a function print out the name of the employee listed in any structure passed to it. However, in the original C (Kernighan & Ritchie, 1st Edition) 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 that we must copy the contents of the structure from the calling function to the called function. In systems using stacks, this is done by pushing the contents of the structure on the stack. 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 declare such a pointer with the declaration:

    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 ---------------------

/* Program 5.2 from PTRTUT10.HTM   6/13/97 */

#include <stdio.h>
#include <string.h>

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.

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More on Strings

Well, we have progressed quite a way 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. Thus we see that pointer arithmetic is the same thing as array indexing. Either syntax produces the same result.

This is NOT saying that pointers and arrays are the same thing, they are not. We are only saying that to identify a given element of an array we have the choice of two syntaxes, one using array indexing and the other using pointer arithmetic, which yield identical results.

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];

due to the fact that array indexing and pointer arithmetic yield identical results, we can write this as:

    *(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

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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 BASIC, Pascal, Fortran and various other languages, a string has its own data type. But in C it does not. In C a string is an array of characters terminated with a binary zero character (written as ‘\0′). 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. Be aware that "nul" is not the same as "NULL". The nul refers to a zero as defined by the escape sequence ‘\0′. That is it occupies one byte of memory. NULL, on the other hand, is the name of the macro used to initialize null pointers. 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-------------------------------------

/* Program 3.1 from PTRTUT10.HTM   6/13/97 */

#include <stdio.h>

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 declare two character pointers and show the string on the screen. We then "point" the pointer 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.

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!

Getting back to the prototype for puts() for a moment, the "const" used as a parameter modifier informs the user that the function will not modify the string pointed to by s, i.e. it will treat that string as a constant.

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. 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, with the last character being a ‘\0′. 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 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 positive 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.

This permits using functions to manipulate 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.

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 the original variables.

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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 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 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[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:

-----------  Program 2.1  -----------------------------------

/* Program 2.1 from PTRTUT10.HTM   6/13/97 */

#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 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[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. 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 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[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. 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[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.

This might be a good place explain further the use of the (void *) expression used in Program 1.1 of Chapter 1. As we have seen we can have pointers of various types. So far we have discussed pointers to integers and pointers to characters. In coming chapters we will be learning about pointers to structures and even pointer to pointers.

Also we have learned that on different systems the size of a pointer can vary. As it turns out it is also possible that the size of a pointer can vary depending on the data type of the object to which it points. Thus, as with integers where you can run into trouble attempting to assign a long integer to a variable of type short integer, you can run into trouble attempting to assign the values of pointers of various types to pointer variables of other types.

To minimize this problem, C provides for a pointer of type void. We can declare such a pointer by writing:

void *vptr;

A void pointer is sort of a generic pointer. For example, while C will not permit the comparison of a pointer to type integer with a pointer to type character, for example, either of these can be compared to a void pointer. Of course, as with other variables, casts can be used to convert from one type of pointer to another under the proper circumstances. In Program 1.1. of Chapter 1 I cast the pointers to integers into void pointers to make them compatible with the %p conversion specification. In later chapters other casts will be made for reasons defined therein.

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.

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