Strings in C (strstr function) - c

Question

I'm trying to reproduce the behavior of the strstr() function, which tries to find a substring in a string and for that I created the following function and compared it to the original. I did the iterations all on paper to understand what was going on in the function, but I don't understand why the command return (&str[i]); prints ab and not just a. When the function enters in if (to_find[j] == '\0'), the values of i and j are 2 and 2, so it should just print &str[2], which is a. Why is printing ab instead of just a?
#include <stdio.h>
#include <unistd.h>
#include <string.h>
char *ft_strstr(char *str, char *to_find)
{
int i;
int j;
i = 0;
if (*to_find == '\0')
return (str);
while (str[i] != '\0')
{
j = 0;
while (str[i + j] == to_find[j])
{
//printf("%d", i);
//printf("%d\n", j);
j++;
if (to_find[j] == '\0')
return (&str[i]);
}
i++;
}
return (0);
}
int main()
{
char i[] = "ab";
char dest[] = "a ab";
printf("%s", ft_strstr(dest, i));
//printf("%s", strstr(dest, i));
}

Answer 1

return (&str[i]); What this does is:
str[i] this is the same as *(str+i). It means take the address where str points to, add i to it and get its value.
&x means get the address of x
(&str[i]) Is a combination of the 2 above. It means take the address where str points to, add i to it and get its value and then get the address of that value. Which means the last 2 steps cancel each other out and you get: take the address where str points to, add i to it. It is the same as (str+i).
The return statement just means return this pointer.
Now after you called ft_strstr(dest, i) you get this pointer. In your case this pointer points to the second a in the string "a ab". You give this pointer to printf() and with the "%s" you told printf() to print the string where this pointer points to till there is a '\0'-byte. 'b' is not a '\0'-byte so it will also be printed.
The returned pointer points to here when you call printf():
V
+---+---+---+---+---+
| a | | a | b |\0 |
+---+---+---+---+---+
printf() will then check if it points to a '\0'-byte, which is false so the byte ('a') is printed and the pointer is incremented. Then the same is done for the byte ('b') and then it points to '\0', which means printf() stops here.

Answer 2

return (&str[i]); does not print anything. It just returns a value. And the value is the address to the i:th element of str.
The printing happens in printf("%s", ft_strstr(dest, i)); and what happens here is that you start with a format string, and the %s is a specifier that basically says "print characters until hitting the zero terminator, and start with the address specified".

Answer 3

When you print the return value from strstr using the specifier %s it will print "the needle" (aka the string to find) and all of the original string after the needle.
In your case there is nothing after the needle so your output is just "ab". Had your original string been "a abHelloWorld" your output would have been "abHelloWorld".
If you just want the first character of the needle do:
printf("%c", *ft_strstr(dest, i));
Also you can try this to get a better understanding:
char str[] = "Hello World";
printf("%s\n", str);
printf("%s\n", &str[0]);
printf("%s\n", &str[1]);
printf("%s\n", &str[2]);
printf("%s\n", &str[3]);
...
which will give you output
Hello World
Hello World
ello World
llo World
lo World
...
As you can see, printing from &str[n] will skip the first N characters and print the rest of the original string.
BTW:
char *ft_strstr(char *str, char *to_find)
should be
char *ft_strstr(const char *str, const char *to_find)

Answer 4

printf("%s", ...) prints a string, not a single character. To print a character, use the %c format specifier and provide the first element of the array as the argument:
printf("%c", *ft_strstr(dest, i));
Another option is to use putchar(...) for printing a single character.
For more information on printf format specifiers, see here.

Answer 5

Within this while loop
while (str[i + j] == to_find[j])
{
//printf("%d", i);
//printf("%d\n", j);
j++;
if (to_find[j] == '\0')
return (&str[i]);
}
the variable i is not being changed. It stays unchanged. It is the variable j that is being changed.
On the other hand the return statement
return (&str[i]);
returns a pointer to the symbol at the position i that is to the beginning of the substring equal to the string to_find.
{ay attention to that the array dest declared like
char dest[]="a ab";
in fact has the following content
char dest[] = { 'a', ' ', 'a', 'b', '\0' };
and the function returns a pointer to its substring
char dest[] = { 'a', ' ', 'a', 'b', '\0' };
^
|
{ 'a', 'b', '\0' }
that is a pointer to the first character of the string "ab" that is a substring of the string stored in the array dest.
Bear in main that when you write for example
printf( "%s\n", dest );
when this call is equivalent to
int i = 0;
printf( "%s\n", &dest[i] );