在C中使用递归的strlen函数

Question

I'm kida new to the recursion subject and i've been trying to write the "strlen" function using recurion, thats what i tried:

int strlen ( char str[], int i)
{
    if ( str[i] == 0) return i+1;
    return strlen(str,i++);
}

and i tried something very similiar

int strlen( char str[], int i)
{
    if ( str[i] == 0) return 1;
    return strlen(str,i++) + 1;
}

and in my main function

int main()
{
     char word[MAX_DIGITS];
     scanf("%s",word);
     printf("%d", strlen(word,0));
     return 0;
}

but my program would crash whenever i run it, what am I missing? (I'm using C90 btw)

Answer 1

Your problem starts here:

i++

This is called a postfix. Just use ++i or i + 1

Postfix sends the value and just then increments the variable. It's like writing this:

return strlen(str,i);
i = i + 1;

You have to use Prefix, which increments the variable and then sends the value. A prefix ( ++i ) will act like that:

i = i + 1;
return strlen(str,i);

Or just send the value without changing the variable:

return strlen(str, i + 1);

Which, in my opinion, is the simplest way to do that.

Answer 2

size_t strlen (char* str) {
    if (*str == 0) {
        return 0;
    }

    return strlen (str+1) +1;
}

So :

• strlen ("") == 0
• strlen ("a") -> strln("") + 1 == 1
• strlen ("he") -> strln("e") + 1) = (strln("") + 1) + 1 == 2

etc

Answer 3

return strlen(str,i++);

You are using the wrong increment operator. i++ means the original value of i is passed as argument, and then it's incremented. That means infinite recursion.

You should try ++i instead, or better, i + 1 .

Answer 4

If you want to keep the same prototype as strlen does. This is how i see a strlen with recursion.

size_t strlen(char *str)
{
    static int i = 0;

    if (*str != '\0')
    {
        i++;
        return ft_strlen(++str);
    }
    return i;
}

I know it's not the best way to do it. Just my implementation.