Array Of Pointers To Strings

Array Of Pointers To Strings

As we know , a pointer variable always contains an address. Therefore , if we construct an array of pointers it would contain a number of addresses.Let us see how the names in the earlier example can be stored in the array of pointers.

char *names[] = { "akshay", "parag", "raman",
                  "srinivas", "gopal", "rajesh"};
 In this declaration names[] is an array of pointers.It contains base addresses of respective names.That is. base address of "akshay" is stored in names[0], base address of "parag" is stored in names[1] and so on. In the two-dimensional array of characters,the strings occupied 60 bytes.As against this ,in array of pointers,the strings
occupy only 41 bytes.
   Note that in two-dimensional array of characters ,the last name ended at location number 1060,whereas in array of pointers to strings , it ends at 1041.A substantial saving, you would agree. But realise that actually 19 bytes are not saved, since 10 bytes are sacrificed for
storing the address in array names .Thus ,one reason to store strings in an array of pointers is to make a more efficient use of available memory. Another reason to use an array of pointers to store
strings is to obtain greater ease in manipulation of the strings. This is shown by the following programs. The first one uses a two-dimensional array of characters to store the names, whereas the second uses an array of  pointers to strings .The purpose of both the programs is very simple.
We want to exchange the position of the names "raman"
and "srinivas".

Exchange names using 2-D array of characters

main()
{
char names[][10] = { "akshay", "parag", "raman",
                     "srinivas", "gopal", "rajesh"};
int i;
char t;
printf("\n original: %s%s",&names[2][0], &names[3][0]);
for(i = 0; i<= 9; i++)
{
t = names[2][i];
names[2][i] = names[3][i];
names[3][i] = t;
}
printf("\n New:%s%S", &names[2][0], &names[3][0]);
}
output:
Original: raman srinivas
New: srinivas raman

  Note that in this program to exchange the names we are required to exchange corresponding characters of the two  names. In effect , 10 exchanges are needed to interchange
two names Let us see if the number of exchanges can be reduced by using an array of pointers to strings. Here isthe program...

main()
{
char *names[] = { "akshay", "parag", "raman",
                  "srinivas", "gopal", "rajesh"};
char *temp;
printf("original: %s %s", names[2], names[3]);
temp = names[2];
names[2] = names[3];
names[3] = temp;
printf("\n New: %s %s", names[2], names[3]);
}

output:
original: raman srinivas
New: srinivas raman