strstr returns a pointer to the found character, so you could use pointer arithmetic: (Note: this code not tested for its ability to compile, it's one step away from pseudocode.)

Copychar * source = "test string";         /* assume source address is */
                                       /* 0x10 for example */
char * found = strstr( source, "in" ); /* should return 0x18 */
if (found != NULL)                     /* strstr returns NULL if item not found */
{
  int index = found - source;          /* index is 8 */
                                       /* source[8] gets you "i" */
}

Try the find function.

Here is the example from the article I linked:

 string str1( "Alpha Beta Gamma Delta" );
 string::size_type loc = str1.find( "Omega", 0 );
 if( loc != string::npos ) {
   cout << "Found Omega at " << loc << endl;
 } else {
   cout << "Didn't find Omega" << endl;
 }

Just subtract the string address from what strchr returns:

char *string = "qwerty";
char *e;
int index;

e = strchr(string, 'e');
index = (int)(e - string);

Note that the result is zero based, so in above example it will be 2.

scanf("%c",&c)..you are getting input a character.

Working copy would besomethign like:-

char c; // you want to find the character not some integer. 
int result;

printf("Please enter string: ");
scanf("%s", str1);

printf("Please enter char: ");
scanf("%d", &c);

result = indexof(str1, c);
printf("Result value: %d\n", result);

Working example:-

#include <stdio.h>
#include <string.h>
 int indexof(char *str, char c)
    {
        int i;
        for (i = 0; i < strlen(str); i++)
        {
            if (str[i] == c)
                return i;
        }

        return -1;
    }

int main()
{
    char c;
    int result;
    char str1[100];

        printf("Please enter string: ");
        scanf("%s", str1);

        printf("Please enter char: ");
        scanf(" %c", &c);

        result = indexof(str1, c);
        printf("Result value: %d\n", result);

}