CODEWITHSUNDEEP
c
Johnny_Recursion
Johnny_Recursion

Reputation: 37

Warning: Return from incompatible pointer type

The code below is producing a compiler warning: return from incompatible pointer type. The type I'm returning seems to be the issue but I cant seem to fix this warning.

I have tried changing the type of hands to int *. Also have tried returning &hands. 

int * dealDeck(int numPlayers, int numCards, int cardDeck[])
{
    static int hands[MAX_PLAYERS][MAX_CARDS]={0};

    int start = 0;
    int end = numCards;
    int player, hand, j;
    int card;

    for(player = 0; player < numPlayers; player++)
    {
        for(hand = start, j=0; hand < end; hand++,j++)
        {
            card =  cardDeck[hand];
            hands[player][j] = card;
        }
        start = end;
        end += numCards;
    }

    return hands;
}

This function should return a pointer to the array "hands". This array is then passed to another function which will print out its elements.

Upvotes: 0

Views: 722

Answers (2)

veprolet
veprolet

Reputation: 391

First of all, you have declared return type of int *, which would mean, that you are trying to return an array, while you want to return a 2-dimensional array. The proper type for this would usually be int **, but that won't cut it here. You opted to go with static, fixed size array. That means, that you need to return pointer to some structures of size MAX_CARDS * sizeof(int) (and proper type, which is the real problem here). AFAIK, there is no way to specify that return type in C*.

There are many alternatives though. You could keep the static approach, if you specify only up to 1 size (static int *hands[MAX_PLAYERS] or static int **hands), but then you need to dynamically allocate the inner arrays.

The sane way to do it is usually "call by reference", where you define the array normally before calling the function and you pass it as a parameter to the function. The function then directly modifies the outside variables. While it will help massively, with the maintainability of your code, I was surprised to find out, that it doesn't get rid of the warning. That means, that the best solution is probably to dynamically allocate the array, before calling the function and then pass it as an argument to the function, so it can access it. This also solves the question of whether the array needs to be initialized, and whether = {0} is well readable way to do it (for multidimensional array) , since you'll have to initialize it "manually".

Example:

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

#define PLAYERS 10
#define DECKS   20

void foo(int **bar)
{
    bar[0][0] = 777;
    printf("%d", bar[0][0]);
    /*
     * no point in returning the array you were already given
     * but for the purposes of curiosity you could change the type from
     * void to int ** and "return bar;"
     */
}

int main()
{
    int **arr;

    arr = malloc(sizeof(int *) * PLAYERS);
    for (size_t d = 0; d < DECKS; d++) {
        /* calloc() here if you need the zero initialization */
        arr[d] = malloc(sizeof(int) * DECKS);
    }
    foo(arr);

    return 0;
}

*some compilers call such type like int (*)[20], but that isn't valid C syntax

Upvotes: 1

Kubessandra
Kubessandra

Reputation: 11

The hands variable is not an int * this is a int ** So you need to return a int **

This is a 2d array.

Upvotes: 1

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