C: Array initialization requires a brace-enclosed initializer list

Question

I'm just a beginner and have encountered a problem with an array of pointers. Could you show me where the mistake is?

int ini()
{
    int *tab[N];
    int i, j, a, b;
    for (i = 0; i < N; i++)
    {
        tab[i] = (int*)malloc(M*sizeof(int));
    }
    if (tab == NULL)
        return -1;
    scanf_s("%d  %d", &a, &b);
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < M; j++)
        {
            *(*(tab+i)+j) = rand() % (b - a + 1) + a;
        }
    }
    return tab;
}

int main()
{
    int i, j, *tablica[N] = ini();
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < M; j++)
        {
            printf("%d  ", *(*(tablica+i) + j));
        }
        printf("\n");
    }
    system("PAUSE");
    return 0;
}

The task itsef is simple and I can do it in the other way, but I just wanted to use this:

*(*(tab+i)+j)

instead of that:

*(tab + N*i + j)

since the second option wouldn't always work.

I'll be glad if you could give me a hand. :)

Welcome to Stack Overflow! [Please see this discussion on why not to cast the return value of `malloc()` and family in `C`.](http://stackoverflow.com/q/605845/2173917). — Sourav Ghosh, Dec 28 '15 at 10:56
To begin with, what does the `ini` function return? I mean, what is it's return type? To continue you can only initialize an array by using "a brace-enclosed initializer list", no other way, and you can't *assign* an array later. To further continue, what the `init` function tries to do is return a pointer to a local variable, which will lead to *undefined behavior*. To end, the expression `*(tablica+i)` is equivalent to `tablica[i]`, a syntax that is easier to read and understand. — Some programmer dude, Dec 28 '15 at 11:00
`tab == NULL` : `tab` never become `NULL`. also `tab` is local variable. — BLUEPIXY, Dec 28 '15 at 11:05
You're trying to dynamically initialize `tablica` in main by calling `ini`. To do that, you need to pass it to `ini` as a parameter. — Tom Karzes, Dec 28 '15 at 11:14

score 1 · Accepted Answer · answered Dec 28 '15 at 11:06

A couple of obvious things:

you return a locally declared array (int *tab[N];) outside of its scope, therefore it's going to be garbage there. This must be your most visible problem.
you malloc the elements of that array inside your ini() method but never release them, therefore you are getting a memory leak.
checking tab for NULL does not make much sense, because tab would never be NULL; tab[i], on the other hand, can be NULL and could be checked for that after malloc.
you don't check that your tablica is not -1.

score 0 · Answer 2 · answered Dec 28 '15 at 11:47

I wanted to upgrade that program (and to make it work in the same way):

int ini()
{
int *tab;
int i, j, a, b;
for (i = 0; i < N; i++)
tab = (int*)malloc(N*M*sizeof(int));
if (tab == NULL)
    return -1;
scanf_s("%d  %d", &a, &b);
for (i = 0; i < N; i++)
{
    for (j = 0; j < M; j++)
    {
        *(tab + N*i +j ) = rand() % (b - a + 1) + a;
    }
}
return tab;
}


int main()
{
int i, j, *tablica = ini();
for (i = 0; i < N; i++)
{
    for (j = 0; j < M; j++)
    {
        printf("%d  ", *(tablica + N*i + j));
    }
    printf("\n");
}
system("PAUSE");
return 0;
}

C: Array initialization requires a brace-enclosed initializer list

2 Answers2