Initialize array in function

Question

When I want to initialize a pointer to an array through the function, I am doing the following:

Initialize and destroy array through functions:

int initArr(int **b)
{
    int *arr = (int *) malloc(sizeof(int)*2);

    if(arr == NULL)
        return 0;

    *b = arr;
    arr = NULL;
    return 1;
}

void destroyArr(int *b)
{
    free(b); 
    b = NULL;
}

Initialize pointer to array:

int *pArr;
int initStatus = initArr(&pArr);

if(initStatus == 0)
{
    printf("%s", "error");
    return 0;
}

Working with pointer to array:

*pArr = 1;
*(pArr + 1) = 2;

printf("0 = %i\n", *pArr);
printf("1 = %i\n", *(pArr + 1));

Destroy pointer to array:

destroyArr(pArr);
pArr = NULL;

Is this correct and safe?

Most people find pArr[1] = 2; easier to read than your *(pArr + 1) = 2;. They are equivalent. — wildplasser
– wildplasser, Commented Jun 3, 2012 at 14:03
In C, usually success is indicated by 0, and non-zero is a kind of failure. The call to malloc will return NULL or a valid address, but standard fare is the function that does that would normally return 0 for success. — octopusgrabbus
– octopusgrabbus, Commented Jun 3, 2012 at 21:58

Kevin · Accepted Answer · 2012-06-03 13:57:31Z

4

I haven't tested it, but it appears correct. A minor comment, though: you don't need to set arr or b to NULL, they're at the very end of their scope and can't be (safely) accessed after that anyway.

answered Jun 3, 2012 at 13:57

Kevin

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wildplasser · Accepted Answer · 2012-06-03 14:22:12Z

3

The initArr function can be reduced to:

int initArr(int **b)
{
    *b = malloc(2 * sizeof **b);

    return *b ? 1 : 0;
}

answered Jun 3, 2012 at 14:22

wildplasser

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Comments

Loupi · Accepted Answer · 2012-06-03 14:13:13Z

2

In fact, initArr and destroyArr do not add any value at all. They wrap the standard C functions to return and do the exact same thing.

Also, you could use array indexers [] to access individual members of the allocated array.

Here is a working equivalent:

int* pArr = (int*) malloc(2 * sizeof(int));
if (pArr) {
    pArr[0] = 1;
    pArr[1] = 2;
    printf("0 = %d\n", pArr[0]);
    printf("1 = %d\n", pArr[1]);
    free(pArr);
}

answered Jun 3, 2012 at 14:13

Loupi

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Kevin · Accepted Answer · 2012-06-03 13:53:52Z

-1

Looks good but the free in destroyArr() does not give you anything since you are not passing a double pointer to it.

edited Jun 3, 2012 at 13:53

Kevin

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answered Jun 3, 2012 at 13:51

ola1olsson

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1 Comment

Kevin Over a year ago

Actually it does. Setting it to NULL doesn't affect the variable in the calling function but b does point to the memory intended to be freed and so frees it properly.

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