CODEWITHSUNDEEP
carrayspointersdouble-pointer
Tsuhirito
Tsuhirito

Reputation: 13

I have doubts about double pointer as parameters for a swap function

I hope you can help me! I added this function to my array.c to shuffle the array elements and it works with static Item *a; static int dim; in my array.c.

static Item *a; 
static int dim = 3;

a=malloc( dim * sizeof(Item);

void random_array() {
                                  int i , j;
                                  Item t;
                                  for (i = dim - 1;i > 0; i--)
                                  {
                                      j = rand() % (i + 1);
                                      t = a[i];
                                      a[i] = a[j];
                                      a[j] = t;
                                     //swap(a[i], a[j]);
                                  }
                           }

item.h

typedef void *Item;

item-int.c

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

Item inputItem(){
    int *p;
    p=malloc(sizeof(int));
    scanf("%d",p);
    return p;
}

void outputItem(Item item){
    int *p;
    p=item;
    printf("%d ",*p);
}

int cmpItem(Item item1,Item item2){
    int *p1,*p2;
    p1 = item1;
    p2 = item2;
    return *p1 - *p2;

}

So I tried with a function swap (a[i], a[j]) from the file utils.c but it doesn't work.

void swap (Item pa, Item pb) {
    Item t;
    t = pa;
    pa = pb;
    pb = t;
}

It works with these instructions:

void swap (Item *pa, Item *pb) 
{
    Item t;
    t = *pa;
    *pa = *pb;
    *pb = t;
}

I know that with "typedef void *Item' Item t is like void *t and Item *a is like void **a, right? So with double pointers I have a dynamic array with many pointers thanks to a=malloc(numberelements*(sizeof(Item)); and with a[i]=malloc(sizeof(Item) every pointer points to a memory location with many bytes?

Here a image.

So for t I don't need malloc because with t = a[i] t points to memory location pointed from a[i], right? If yes, why do I have to use in my swap function t = *pa etc. instead of t = pa ? I have several doubts about double pointers as function parameters. I hope you can resolve that. Thanks in advance!

Upvotes: 1

Views: 163

Answers (1)

Michael Dorgan
Michael Dorgan

Reputation: 12515

Ok, let's start from here:

typedef int Item;  // Always get this backwards, need to check...

void swap (Item *pa, Item *pb) 
{
    Item t;
    t = *pa;
    *pa = *pb;
    *pb = t;
}

void foo()
{
    Item i= (Item), j=(Item)2;  // These casts are not needed, but help later explanations

    // Must take address so swap can change original values of i and j.
    swap(&i, &j);

    Item *ip = &i;
    Item *jp = &j;

    // This also works because we do the dereference above.
    swap(ip, jj);
}

Either are fine. What about?

typedef int******* Item;

void swap (Item *pa, Item *pb) 
{
    Item t;
    t = *pa;
    *pa = *pb;
    *pb = t;
}

void foo()
{
    Item i= (Item)1, j=(Item)2;

    // Must take address so swap can change original values of i and j.
    swap(&i, &j);

    int *ip = &i;
    int *jp = &j;

    // This also works because we do the dereference above.
    swap(ip, jp);
}

Again, the type doesn't matter. It's how it's used. If you were to forget the &, the compiler would warn you about type mismatch.

Upvotes: 0

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