Thread: Recursive concept

Hi guys; sorry in advance for posting this thread but it's harassing me whenever I code in a recursive and I believe here's the place to learn and post my knowledge professionally.
Lets take an example the quickSort .. generally the recursion rule I succeed of writing it..but still not make sense for me ..
Lets say I have case that QuickSort(int p, int r) which p,r is the boundaries of the array, p is the first index, r is the last index..
Lets say I got case QuickSort(9,8) logically it's not make sense and something going wrong here over the boundaries because p=9,r=8 and that's not allowed.
So here I'm stuck in; what should I return in the case?! here's exactly my problem; how to decide here which kind of "return" to return it?

Why should I have consider this case? because lets say I have QuickSort(1,8) I may get in the recursive calls of QuickSort(1,8) a case that QuickSort(9,8).. so I should consider this case with something to be returned once occured..

Thanks for your help!

I assume I should return "trash" .. but how could I analog it to PC's concept

I don't believe that You have a knowledge professionally . . . if You want to understand "recursive programming" then start simple, and not with "Quicksort".
This program calculate the Fibonacci-Number - iterativ. Change it to a recursive program.

Code:

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

void main(void)
{
    unsigned long x,y;
    int i, n;


    x = y = i = 1;
    
    //For Linux "system(clear)"
    system("cls");
    printf("\nBerechnet iterativ die Fibonaccizahl.");
    printf("\nGeben Sie eine Zahl ein: ");
    scanf("%d", &n);
    printf("\n");
    if (n == 0 || n == 1)
    {
        printf("\nDie Fibonaccizahl von %d ist: 1", n);
    }
    //For 3 show 1,2,3 -- Um bei 3 eine Anzeige von: 1 2 3 zu erreichen
    printf("1 "); 
    if (n > 40)
    {
        //Input only to 40 - sonst Überlauf
        printf("\nNur Eingaben bis 40!\n");        
    }
    else while (i < n)
    {
        x = x + y; y = x - y; i++;
        printf("%d  ", x);
    }
    printf("\n\nDie Fibonaccizahl von %d ist: %lu\n\n", n, x);
}

Originally Posted by Kernelpanic

There was a (little) mistake when the input is over forty.

Code:

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


void main(void)
{
    unsigned long x,y;
    int i, n;


    x = y = i = 1;
    
    //For Linux "system(clear)"
    system("cls");
    printf("\nBerechnet iterativ die Fibonaccizahl.");
    printf("\nGeben Sie eine Zahl ein: ");
    scanf("%d", &n);
    printf("\n");
    if (n == 0 || n == 1)
    {
        printf("\nDie Fibonaccizahl von %d ist: 1", n);
    }    
    
    if (n > 40)
    {
        //Input only to 40 - sonst Überlauf
        printf("\nNur Eingaben bis 40!\n");
        exit(0);
    }
    else while (i < n)
    {
        x = x + y; y = x - y; i++;
        printf("%d  ", x);
    }
    printf("\n\nDie Fibonaccizahl von %d ist: %lu\n\n", n, x);
}

First, I appreciate your help to understand the concept. but what's exactly confusing me like this :
lets say I have or actually after "n" splitting of array I arrived to array {2,3} and the pivot is 3(assuming that) I call it q=2(index of 3), then I must call to quicksort(1,1) which it gives me sorted "list" because {2} is already sorted ! , but what about quicksort(3,2) - the right side ? there's nothing outter than 3 in the list {2,3} ?! here what I do and how to decide which return to return it once this case occurred?! "whenever" I arrive to case like dont care/trash/outter of the bounds/blocking ..what should I return in the function?!

God damn, I'm getting another crisis! This little program is full of pitfalls ... Last version! The Lord is with me!

Code:

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


void main(void)
{
    unsigned long x,y;
    int i, n;


    x = y = i = 1;
    
    //For Linux "system(clear)"
    system("cls");
    printf("\nBerechnet iterativ die Fibonaccizahl.");
    printf("\nGeben Sie eine Zahl ein: ");
    scanf("%d", &n);
    printf("\n");
    if (n == 0 || n == 1)
    {
        printf("\nDie Fibonaccizahl von %d ist: 1", n);
        exit(0);
    }
    else if    (n == 3)
    {
        printf("1 ");
    }
    
    if (n > 40)
    {
        //Input only to 40 - sonst Überlauf
        printf("\nNur Eingaben bis 40!\n");
        //Sonst wird unten trotzdem "printf ..." ausgeführt
        exit(0);
    }
    else while (i < n)
    {
        x = x + y; y = x - y; i++;
        printf("%d  ", x);
    }
    printf("\n\nDie Fibonaccizahl von %d ist: %lu\n\n", n, x);
}

Originally Posted by christop

Just... no. Please don't tell me you're using Turbo C or some other old compiler that allows this rubbish!

Code:

int main(void)
. . .
return(0);

Aiyo, if you are concerned about invalid arguments then just add a check for that in an initially called function that does have some way of indicating an error, e.g., with a return value or pointer parameter. It's not hard.