When I write my own strchr function, there is a part in the return step that I don't understand

I'm trying to write my own strchr function for school, but there is something I don't understand.

char    *ft_strchr(const char *s, int c)
{
    size_t  i;
    size_t  len;

    i = -1;
    len = ft_strlen(s);
    while (++i < len + 1)
    {
        if (s[i] == (char)c)
        {
            return (((void *)&((char *)s)[i]));
        }
    }
    return (NULL);
}

I have completed the code's general structure and testing stages. But I couldn't understand the return step.

return (((void *)&((char *)s)[i]));

Can you briefly explain to me what's going on here?

Solution 1:

That line is overcomplicating things and not really best possible solution.

Let's take a closer look:

char    *ft_strchr(const char *s, int c)
{
...
   return (((void *)&((char *)s)[i]));
...
}

First of all, return is not a function and therefore does not need brackets around the expression. (And there are even 2 levels of unnecessary brackets) That line is exactly the same as this one:

   return (void *)&((char *)s)[i];
...

That line takes some value and casts it to void*. Casting a pointer is normally done to interpret an address as a pointer to another type than the pointer was before. Casting to void* is normally not needed as any pointer to a data object can automatically be converted to void* if needed.

But look at the function return type. It is char*, not void*. Casting the return value to void* does not make much sense here. Instead it should be a cast to char* if necessary. But the value already is of type char*. Therefore just drop that thing.

   return &((char *)s)[i];

That is what you actually need. The whole point of the cast is that you have a pointer to const char and you need a pointer to char.

You could write it a bit more readable like this:

   return (char *)&s[i];

But in the end both version will return same value with same type.

I assume with that reduced expression it is easier for you to understand what is needed in that return statement.