Trying to build my own strstr() function in c?

Question

I'm trying to build my strstr() function, it's look all fine but when I execute the code it return nothing ;

This is my function :

#include <stdio.h>

char* ft_strstr(char *str, char *to_find) {
  char *a;
  char *b;

  b = to_find;
  if (!*b)
    return (char*) str;
  while (*str) {
    if (*str != *b)
      continue;
    a = str;
    while (1) {
      if (!*b)
        return (char*) str;
      if (*a++ != *b++)
        break;
    }
    str++;
    b = to_find;
  }
  return (NULL);
}

int main() {
  char string[] = "Hello from morroco";
  char find[] = "from";
  ft_strstr(string, find);
  printf("%s", find);
  return 0;
}

I don't know what. I have to try to fix it cause it looks all fine for me.

OT: Regarding return (char *)str;... If str already have the type char *, there's just no need to do an explicit conversion to the same type. — Some programmer dude
– Some programmer dude, Commented Sep 20, 2023 at 11:33
Please format your code properly. Readable code is very important, especially for beginners. — Jabberwocky
– Jabberwocky, Commented Sep 20, 2023 at 11:33
The code doesn't modify the string pointed to by the to_find argument. Why should find in the main function change? You seem to have some basic misunderstandings about both the strstr function as well as the basics of C. — Some programmer dude
– Some programmer dude, Commented Sep 20, 2023 at 11:34
At a glance you seem to have some 2-3 bugs caused by sloppy formatting alone... — Lundin
– Lundin, Commented Sep 20, 2023 at 11:36
What exactly do you expect to happen when you reach continue;? What situation will you be in when you come to if(*str != *b) again? How is it different from last time you got there? — Gerhardh
– Gerhardh, Commented Sep 20, 2023 at 11:43

Vlad from Moscow · Accepted Answer · 2023-09-20 12:05:23Z

2

There is a bug in this if statement

if(*str != *b)
    continue;

the pointer str is not increased in this case and as a result the loop will be infinite.

instead of the while loop it is better to use a for loop as

for ( ; *str; ++str )

or the while loop can look like

while (*str)
{
    if( *str == *b )
    {
        a = str;
        while(1)
        {
            if (!*b )
                return (char *)str;
            if (*a++ != *b++)
                break;
        }
    }
    str++;
    b = to_find;
}

Also it seems you mean

char *pos = ft_strstr(string , find);
if ( pos ) printf("%s", pos);

instead of

ft_strstr(string , find);
printf("%s", find);

Pay attention to that the function should be declared like

char * ft_strstr( const char *str, const char *to_find);

And using such castings like

return (char *)str;

makes sense if the parameter str has the type const char * as it should have.

edited Sep 20, 2023 at 12:05

answered Sep 20, 2023 at 11:39

Vlad from Moscow

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7 Comments

Omar Ghazi Over a year ago

Thank you vlad , but in this subject for loop not allowed , cause of that i have to use while but i incremenet str++ in the last of the first while loop and i tried to stock the function in *p and print it with condition but the output still the same ,

Omar Ghazi Over a year ago

i have to notice that i can't declare the params in the str function as a const

Vlad from Moscow Over a year ago

@OmarGhazi There s nothing prevents you to declare the parameters with the qualifer const. Also the variables a and b must be declared with the qualifier const.

Jonathan Leffler Over a year ago

@OmarGhazi — why can't you declare function parameters as const? Why is your "teacher" abusing the language so? It bodes ill for you — you should be being taught how to write good C code, and good C code uses const whenever sensible. Are you also constrained to C90? If so, you have to start worrying about the standard of education you are receiving.

chux Over a year ago

@OmarGhazi restrictions concerning for and const deserve to be in the question.

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