Limitations of array Pointers to strings

Limitations of array Pointers to strings

When we are using a two-dimensional array of characters we are at liberty to either initialize the strings where we are declaring the array, or receive the stringsusing scanf() function. However,when we are using an array of  pointers to stringswe can initialize the strings at the place where we are declaring the array,but we cannot receive the strings from keyboard using scanf().Thus the following
program would never work out.

main()
{
char *names[6];
int i;
for(i =0; i<= 5; i++)
{
printf("\n enter name");
scanf("%s",names[i]);
}
}
  The program doesn't work because; when ae are declaring the array it is containing the garbage values. And it would be definitely wrong to send these garbage values to scanf() as the address where it should keep the strings received from the keyboard.  Write a function xstrstr() that will return the position where one string is present within another string. If the second string doesn't occur in the first string xstrstr() should return a 0.

For example, in the string "somewhere over the rainbow",
"over" is present at location 11.

main()
{
static char str1[] = "somewhere over the rainbow";
static char str2[] = "over";
printf("string found at %d", xstrstr(str1, str2));
}
xstrstr(s1, s2)
char *s1, *s2;
{
int i, a, len1, len2;
len1 = strlen(s1);
len2 = strlen(s2);
for(i = 0; i<=(len-1); i++)
{
a = strncmp((s1+i) , s2, len2);
if(a == 0)
return (i+1);
}
return (0);
}

output:
string found at 11

Explanation:  The two strings have been declared as str1[] and str2[] in main from where the base addresses are sent to the function xstrstr() for searching the second string in the first one. In xstrstr(), len1 and len2 store the lengths of the 2 strings withbase addresses s1 and s2 respectively .In the for loop, i is incremented len1 number of times. As many times, the standard library function
strncmp(t,s, n) gets called.This function compares the first n elements of strings starting from t and s and returns 0 if they are equal.The first time through the for loop, i is 0. Hence strncmp() compares the first len2 (here len2 is equal to 4) elements of strings starting from (s1+0) i.e s1and s2 a collects a non-zero value, as the first four elements of the two strings are found to be different.The control therefore reverts back to the for where i is incremented to 1.So the second time through the
loop strncmp() compares first len2 elements of strings starting from s1 and s2. Literally first 4 elements of "somewhere over the rainbow" and "over" are compared .Once  again a collects a non-zero value and i is incremented a second time.this goes on similarly till i is 10, when s1+10
denotes the base address of the string "over the rainbow".This time a is assigned a 0,as both the strings have o, v, e and r as the first four elements, and control returns to main() with i+1, i.e 11.This is the position of the second string in the first one.Suppose the second string is not
present in the first string at all, then at no time a would contain 0. Thus the return statement after the loop would return 0,signifying that the second string was not found in the first one.