Recursive Function with a pointer array

Question

I am trying to create a recursive function with the output:

Once upon a time, there was a child who couldnt sleep, so the child's mother told her a story about a mouse who couldnt sleep, so the mouse's mother told her a story about a turtle who couldnt sleep, so the turtle's mother told her a story about a frog who couldnt sleep, so the frog's mother told her a story about a bear who went to sleep.

Then the frog went to sleep. Then the turtle went to sleep. Then the mouse went to sleep. Then the child went to sleep.

I have the first paragraph done but when I tried to re loop back around and print out the next paragraph it turns into an infinite loop. Do I need another recursive function to complete this? Any help would be appreciated.

#include <stdio.h>
#include <stdlib.h>

void story(char **string1, int n)
{
    if ( n > 0 )
    {
        printf("a %s who couldn't sleep,\n", *(string1 + n));
        printf("so the %s's mother told her a story about\n", *(string1 + n));
        story(string1, n - 1);
    }
    else
    {
        printf("a %s who went to sleep.", *string1);
        if ( n < 4)
        {
            story(string1, n + 1);
            printf("Then the %s went to sleep.", *(string1 + n));
        }
    }
}

int main()
{
    char* animals[] = {"bear", "frog", "turtle", "mouse", "child"};
    int start = 4;

    printf("Once upon a time, there was ");
    story(animals, start);

    printf("%s", *animals);
    return 0;
}

Amit · Accepted Answer · 2015-08-05 22:06:17Z

1

The problem is that your recursion goes down (n - 1) when n > 0, and when it hits 0 it calls itself again with 1 (n + 1) which again would call with 0 (n - 1) indefinitely.

What you probably wanted to do is:

void story(char **string1, int n)
{
    if ( n > 0 )
    {
        printf("a %s who couldn't sleep,\n", *(string1 + n));
        printf("so the %s's mother told her a story about\n", *(string1 + n));
        story(string1, n - 1);
        printf("Then the %s went to sleep.", *(string1 + n));
    }
    else
    {
        printf("a %s who went to sleep.", *string1);
    }
}

So each iteration before the last (n == 0) prints it's part and calls the recursion to fill the "middle" till it's time to "go to sleep". The ending rule, n == 0, only prints "... who went to sleep".

answered Aug 5, 2015 at 22:06

Amit

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3 Comments

Jack Swanson Over a year ago

Thanks! I wasn't exactly sure on how to print out the last part but exactly what you did makes perfect sense!

Jack Swanson Over a year ago

Is the else or where n= 0 considered the base case? @Amit

Amit Over a year ago

Yeah, you could say that. I call it the ending rule in this scenario since this type of recursion doesn't use the result of inner iterations to calculate current result.

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