Sort the pointers by the value they are pointing to in C

Question

I want to sort pointers that point to an array in C, but without moving index positions. Here is a picture of what I am trying to do:

Here is the code, but it doesnt work, and I dont know what the problem is.

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

    int main()
    {

      int niz[7] = { 4, 2, 5, 7, 6, 1, 3};


    int j,i;
    int *temp;

    int *nizptr = niz;
    int **niz2ptr = &nizptr;

    for(i = 0; i < 7; i++)
    {
        for(j = 0; j < 7; j++)
        {
            if( niz[i] < niz[j] )
            {
                temp = *(niz2ptr + i);
                *(niz2ptr+i) = *(niz2ptr + j);
                *(niz2ptr+j) = temp;
            }
        }
    }


    for(i = 0; i < 7; i++)
    {
        printf("%d",*(*(niz2ptr + i)));
    }
    return 0;
}

Your code doesn't have an array of pointers like your picture seems to think you do. Maybe start with making your code reflect what your picture is trying to say. — WhozCraig
– WhozCraig, Commented Sep 9, 2016 at 8:34
You don't need a pointer to pointer, once you create an array of pointers as commented by WhozCraig, just use array_of_pointers[i], where i = the index to the sorted array of pointers. — rcgldr
– rcgldr, Commented Sep 9, 2016 at 8:51
Thumbs up for drawing arrows and boxes. Everybody trying to solve issues with lists in C should do that. — Frerich Raabe
– Frerich Raabe, Commented Sep 9, 2016 at 8:58

2501 · Accepted Answer · 2016-09-09 09:00:56Z

3

The image you've shown is not correct. After the initializations:

int *nizptr = niz;
int **niz2ptr = &nizptr;

the state looks like this:

The pointer niz2ptr is not pointing to an array, but to a pointer. Thus the indexing you make here:

temp = *(niz2ptr + i);

which is identical to:

temp = niz2ptr[i];

will read out of bounds of when i is greater than one. This happens because it tries to read at the address &nizptr+i instead of niz+i.

To fix the code, the type of variable temp must be changed to int, not a pointer to int.

Since nizptr is pointing to an array, you can index it:

temp = nizptr[i];

To achieve the same with niz2ptr, first dereference it to get the value of nizptr, and the index it:

temp = (*niz2ptr)[i];

This should be done whenever niz2ptr is used.

edited Sep 9, 2016 at 9:00

answered Sep 9, 2016 at 8:43

2501

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2 Comments

honeyPot Over a year ago

That picture solved my problem i understood after that

rcgldr Over a year ago

There's still just one pointer here, you need an array of pointers in order to sort an array of pointers, as answered by WhozCraig.

WhozCraig · Accepted Answer · 2016-09-09 08:53:21Z

For some reason you're bent on using a pointer to pointer for this task. None is needed. Your picture shows a value array and a pointer array, where the goal is only the pointer array is changed; the value array stays the same. So do just that:

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

int main()
{
    int niz[7] = { 4, 2, 5, 7, 6, 1, 3};
    int *nizp[7];
    int i;

    // load pointers, show original order with addresses
    for (i=0; i<7; ++i)
    {
        nizp[i] = niz+i;
        printf("%p: %d\n", (const void*)(nizp[i]), *(nizp[i]));
    }
    fputc('\n', stdout);

    // sort the pointer array; not the value array
    int swapped = 1;
    int n = 7;
    while (swapped && n-- > 0)
    {
        swapped = 0;
        for (i=0; i<n; ++i)
        {
            if (*(nizp[i+1]) < *(nizp[i]))
            {
                void *tmp = nizp[i+1];
                nizp[i+1] = nizp[i];
                nizp[i] = tmp;
                swapped = 1;
            }
        }
    }

    // show the sorted pointer array, dereferenced
    for(i = 0; i < 7; i++)
        printf("%p: %d\n", (const void*)(nizp[i]), *(nizp[i]));
    fputc('\n', stdout);

    // prove the original sequence is still unsorted
    for(i = 0; i < 7; i++)
        printf("%p: %d\n", (const void*)(niz+i), niz[i]);
    fputc('\n', stdout);

    return 0;
}

Output (addresses are system dependent)

0x7fff5fbff980: 4
0x7fff5fbff984: 2
0x7fff5fbff988: 5
0x7fff5fbff98c: 7
0x7fff5fbff990: 6
0x7fff5fbff994: 1
0x7fff5fbff998: 3

0x7fff5fbff994: 1
0x7fff5fbff984: 2
0x7fff5fbff998: 3
0x7fff5fbff980: 4
0x7fff5fbff988: 5
0x7fff5fbff990: 6
0x7fff5fbff98c: 7

0x7fff5fbff980: 4
0x7fff5fbff984: 2
0x7fff5fbff988: 5
0x7fff5fbff98c: 7
0x7fff5fbff990: 6
0x7fff5fbff994: 1
0x7fff5fbff998: 3

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