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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_name.

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 terminating 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 short for &myname[0] which is the address of the first element of the array. Since the location of the array is fixed during run time, this 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 declarations 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 = "FGHIJ"

In the case of my_function_A, the content, or value(s), of the array a[] is considered to be the data. The array is said to be initialized to the values ABCDE. In the case of my_function_B, the value of the pointer cp is considered to be the data. The pointer has been initialized to point to the string FGHIJ. In both my_function_A and my_function_B the definitions are local variables and thus the string ABCDE is stored on the stack, as is the value of the pointer cp. The string FGHIJ can be stored anywhere. On my system it gets stored in the data segment.

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 portability and backwards compatibility.

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];

Let’s take the underlined part to be the "name" of an array. Then prepending the char and appending the [10] we have an array of 10 characters. But, the name multi[5] is itself an array indicating that there are 5 elements each being an array of 10 characters. Hence we have an array of 5 arrays of 10 characters each..

Assume we have filled this two dimensional array with data of some kind. In memory, it might look as if it had been formed by initializing 5 separate arrays using something like:

    multi[0] = {'0','1','2','3','4','5','6','7','8','9'}
    multi[1] = {'a','b','c','d','e','f','g','h','i','j'}
    multi[2] = {'A','B','C','D','E','F','G','H','I','J'}
    multi[3] = {'9','8','7','6','5','4','3','2','1','0'}
    multi[4] = {'J','I','H','G','F','E','D','C','B','A'}

At the same time, individual elements might be addressable using syntax such as:

    multi[0][3] = '3'
    multi[1][7] = 'h'
    multi[4][0] = 'J'

Since arrays are contiguous in memory, our actual memory block for the above should look like:

    |_____ starting at the address &multi[0][0]

Note that I did not write multi[0] = "0123456789". Had I done so a terminating ‘\0′ would have been implied since whenever double quotes are used a ‘\0‘ character is appended to the characters contained within those quotes. Had that been the case I would have had to set aside room for 11 characters per row instead of 10.

My goal in the above is to illustrate how memory is laid out for 2 dimensional arrays. That is, this is a 2 dimensional array of characters, NOT an array of "strings".

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 the 4th row 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 ----------------------

/* Program 6.1 from PTRTUT10.HTM   6/13/97*/

#include <stdio.h>
#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 often said to be equivalent to a pointer to a pointer. With a three dimensional array we would be dealing with an array of arrays of arrays and some might say its name would be equivalent to a pointer to a pointer to a pointer. However, here we have initially set aside the block of memory for the array by defining it using array notation. Hence, we are dealing with a constant, not a variable. That is we are talking about a fixed address 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 address (the address of multi[0][0] as given by the symbol multi).

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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)
    printf("\n%s ",my_struct.fname);
    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)

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 */
    printf("\n%s ",my_struct.fname);
    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];
        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;


    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:


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);

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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