Bubble sort array of pointers

Question

So I'm trying to sort an array of pointers as seen below. The problem I have is that the array contains one null element. I have to dereference all the elements except NULL other wise i get an error of course, but this causes my sort to not properly sort any elements after NULL appears. I can create a specific exception for a NULL case, but is there anyway to avoid this, and treat NULL at 0 while I still dereference everything else? right now i tell the sort to ignore NULL. This is just a place holder as I have been unable to find a solution to my problem.

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

void arr(int ar[], int ele);

int main(){
    int a=0, b=9, x=3, p=2, *ar[]={&a, &b, &x, NULL, &p}, i=0, ele=(sizeof(ar)/sizeof(ar[0]));
    arr(ar, ele);
    printf("\n\n");
    for(;i<ele;i++){
        if(ar[i]==NULL){
            printf("");
        }else{
     printf("%i", *ar[i]);
        }
    }
}

void arr(int *ar[], int ele){
    int i=ele-1, c=0;
    for(;i>0; i--){
        for(;c<i; c++){
            if((ar[c]!=NULL && ar[c+1]!=NULL) && *ar[c]>*ar[c+1]){
                int t=*ar[c+1];
                *ar[c+1]=*ar[c];
                *ar[c]=t;
            }
        }
    }
}

just remove the null, sort, and then put it back at whatever position you want. much simpler than trying to make an otherwise simple algorithm work with a null. — goat
– goat, Commented May 7, 2012 at 2:24

JustinDanielson · Accepted Answer · 2012-05-07 02:47:49Z

2

Change this

if((ar[c]!=NULL && ar[c+1]!=NULL) && *ar[c]>*ar[c+1]){

To

//If the pointer is NULL, it will have a value of 0, so the conditional will be false.
x = (ar[c]) ? *ar[c] : 0;
y = (ar[c+1]) ? *ar[c+1] : 0;
if(x > y){

Add int x,y; to the top of the function as well.

edit: added dereferencing pointers. lol

edited May 7, 2012 at 2:47

answered May 7, 2012 at 2:32

JustinDanielson

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5 Comments

Jonathan Leffler Over a year ago

Are the values in the array all positive (or non-negative)? You're treating NULL pointers as if they pointed at a 0. It is at least deterministic, though.

Painguy Over a year ago

ok so make an exception i guess. i made a change in ur code. afte checking ar[c] or ar[c+1] would u not have to dereference the ar[c] in the case that it is not NULL?

Seth Carnegie Over a year ago

x = ar[c] ? ar[c] : 0 is the same as x = ar[c]. Same for the line after that.

JustinDanielson Over a year ago

I added an edit to fix that Nima. @Jonathan Leffler, the question stated to "treat NULL at 0." So I was just working from that.

jdizzle Over a year ago

Agreed. Just treat null pointers as if they're the minimal (or maximal?) value. Then they'll be clumped together at the end of the list after sorting.

cloudygoose · Accepted Answer · 2012-05-07 02:25:13Z

1

How about you let

Int *ptNull = new int;
*ptNull = -100(the smallest);

Then you first find that NULL in the array, and set it to be ptNull.
And then you can sort as if there is no NULL in the array.

answered May 7, 2012 at 2:25

cloudygoose

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Comments

Jonathan Leffler · Accepted Answer · 2012-05-07 02:33:50Z

1

Should the NULL be sorted first or last? Decide. The decision controls your comparison code:

if (compare(ar[c], ar[c+1]) < 0)
{
    int t=*ar[c+1];
    *ar[c+1]=*ar[c];
    *ar[c]=t;
}

Where:

static int compare(int const *v1, int const *v2)
{
    if (v1 == NULL)
        return -1;
    if (v2 == NULL)
        return +1;
    if (*v1 < *v2)
        return -1;
    if (*v1 > *v2)
        return +1;
    return 0;
}

This sorts NULL before any valid value.

You have another problem:

void arr(int ar[], int ele);

vs

void arr(int *ar[], int ele){

These are not the same signature; your code should not be compiling.

answered May 7, 2012 at 2:33

Jonathan Leffler

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Comments

Kaz · Accepted Answer · 2012-05-07 02:35:42Z

1

    for(;c<i; c++){
        int left = ar[c] != NULL ? ar[c] : 0;
        int right = ar[c+1] != NULL ? ar[c+1] : 0;

        if (left > right){
            /* swap the pointers, not what they point to! */
            int *t = ar[c+1];
            ar[c+1] = ar[c];
            ar[c] = t;
        }
    }

answered May 7, 2012 at 2:35

Kaz

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