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.. — Sourav Ghosh
– Sourav Ghosh, Commented Dec 28, 2015 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
– Some programmer dude, Commented Dec 28, 2015 at 11:00
tab == NULL : tab never become NULL. also tab is local variable. — BLUEPIXY
– BLUEPIXY, Commented Dec 28, 2015 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
– Tom Karzes, Commented Dec 28, 2015 at 11:14

Ashalynd · Accepted Answer · 2015-12-28 11:06:48Z

1

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.

answered Dec 28, 2015 at 11:06

Ashalynd

12.6k2 gold badges36 silver badges38 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

wis.niowy · Accepted Answer · 2015-12-28 11:47:02Z

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;
}

Collectives™ on Stack Overflow

C: Array initialization requires a brace-enclosed initializer list

2 Answers 2

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Linked

Related