implementation of strstr() function - c

Question

The code says at many places "invalid indirection".Please help.
int main()
{
char *s1,*s2,*position;
printf("Enter string:\n");
gets(s1);
printf("Enter word to find:\n");
gets(s2);
*position=ststr(*s1,*s1);
if(*position)
printf("word is found at %c loc\n",*position);
else
printf("word not found");
getch();
return 0;
}
char *strstr(char *s1,char *s2)
{
int flag=1;
char i,j;
for(i=0; ;i++)
{
if(*s1[i]==*s2[0])
for(j=i;*s2;j++)
{
if(*s1[j]!=*s2[j])
flag=0;
}
}
if(flag)
return i;
else
return 0;
}

Answer 1

First, s1 and s2 in main have not been initialized to point anywhere meaningful. Either declare them as static arrays, or allocate memory to them at runtime using malloc() or calloc():
#define SIZE 20 // or some number big enough to hold your input
...
char s1[SIZE], s2[SIZE], *position; // s1 and s2 declared statically
Second, NEVER NEVER NEVER NEVER NEVER use gets(); it will introduce a point of failure in your program. Use fgets() instead:
if (fgets(s1, sizeof s1, stdin) != NULL)
// process s1
else
// check for EOF or error on read
EDIT
And like everyone else has pointed out, your comparison in the strstr() function needs to be either
*s1 == *s2
or
s1[i] == s2[i]
but first you need to deal with allocating your buffers in main properly.

Answer 2

One of the problems I'm noticing is whenever you do *s1[j]. The asterisk is dereferencing the array, and so is the [] notation.
s[i] really means *(s + i), so you don't have to dereference it again. The way you have it would read **(s + i), and since it's a single pointer you can't do that.

Answer 3

if(*s1[i]==*s2[0])
is such an example where my gcc complains:
error: invalid type argument of ‘unary *’ (have ‘int’)
if s1 is a pointer to char, s1[i] is a char. So you can't dereference it any more (with the *), i.e. s1[i] does not point to anything any more.
Try
if(s1[i]==s2[0])
instead.
You should also change the return value of strstr: you return an integer where you declare to return a pointer to a character. So try returning s1+i instead.
This here:
for(j=i;*s2;j++)
probably does not what you want. You're not advancing the pointer s2 anywhere in the loop, in fact you're just testing whether s2[0] (which is the same as *s2) is zero for each iteration. If s2 isn't the empty string, this loop will never terminate.

Answer 4

#include "stdio.h"
char *strstr(char *str, char *substr)
{
int len = strlen(substr);
char *ref = substr;
while(*str && *ref)
{
if (*str++ == *ref)
{
ref++;
}
if(!*ref)
{
return (str - len);
}
if (len == (ref - substr))
{
ref = substr;
}
}
return NULL;
}
int main(int argc, char *argv[])
{
printf("%s \n", strstr("TEST IS NOT DONE", "IS NOT"));
}

Answer 5

if(*s1[j]!=*s2[j])
*s1 means "the character where s1 is pointing".
s1[j] means "*(s1+j)" or "the character j positions after where s1 is pointing"
You have to use one or the other; not both.

Answer 6

#include <stdio.h>
char* my_strstr(char *s2, char *s1)
{
int i, j;
int flag = 0;
if ((s2 == NULL || s1 == NULL)) return NULL;
for( i = 0; s2[i] != '\0'; i++)
{
if (s2[i] == s1[0])
{
for (j = i; ; j++)
{
if (s1[j-i] == '\0'){ flag = 1; break;}
if (s2[j] == s1[j-i]) continue;
else break;
}
}
if (flag == 1) break;
}
if (flag) return (s2+i);
else return NULL;
}
int main()
{
char s2[] = "This is the statement";
char s1[] = "the";
char *temp;
temp = my_strstr(s2,s1);
printf("%s\n",temp);
return 0;
}