Declare and initialize after - c

Question

I'm working on an exercise and I can't declare and initialize on the same line, and with these instructions I don't know how I can do, I tried multiple tips but no one works.
Can you help me please ? I want to initialize my array in the next line.
This, is working fine:
int tab[n];
tab[0] = 5;
And, this, is not working
int tab[];
tab[n] = { 0 };

Answer 1

Issue #1. The compiler needs a size when declaring an array.
int tab[];
Will not compile.
C does allow for variable sized arrays.
int n = 5;
int tab[n];
Issue #2. You attempt to assign to the nth element in an array with size n, which is out of bounds, and will not work.
Issue #3. You cannot assign to an array directly, so this will not work:
int n = 5;
int tab[n];
tab = { 0 };
And the initializer syntax will not work in this context. You would have to use it inline with the declaration. Although, a variable-sized array may not be initialized this way anyway.
Legal:
#define N 5
int main() {
int tab[N] = { 0 };
return 0;
}
Illegal:
int main() {
int n = 5;
int tab[n] = { 0 };
return 0;
}

Answer 2

To assign an all zero byte pattern to int tab[n] after definition:
memset(tab, 0, sizeof tab);
If code needs a more complex assignment, code could then use individually for each array element as needed.
tab[0] = 5;
...

Answer 3

C has something which is called flexible array members.
struct flex
{
size_t size;
int x[];
};
struct flex a = {5, {1,2,3,4,5}};
int main(void)
{
struct flex *z;
z = malloc(sizeof(*z) + 3 * sizeof(z -> x[0]));
z -> size = 3;
}

Answer 4

You cannot specify an empty pair of brackets to declare a variable.
int tab[]; /* error */
And adding the assigment you add later, just gets things worse:
tab[5] = 3;
The first thing you need to learn is that an assignment is some kind of expression that allows you to make a calculation, while a declaration is a statement (not an expression) that allows you to introduce a new object (and allows you to initialize it)
An initializer allows an incomplete declaration to become complete, so in the case you do both things at the same time (declare and initialize) the code above, written in this form:
int tab[] = { [0] = 3, [1] = 5, [2] = 8 };
makes the array declaration to be complete and will initialize the elements as shown.
You can go further, and also specify an array length, and all the unspecified elements in the initializer will be initialized to 0 (in this case the object is complete from the declaration, but what is incomplete is the initializer, which fills the holes with zeros).
int tab[20] = { [0] = 3, [1] = 5, [19] = 6 };
(if the indices will go up from 0, you can avoid the [index] = , as in
int tab[20] = { 3, 5, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 6 };
)
Had you written your code as one declaration:
int tab[] = { [5] = 3 };
that would have been valid.
#include <stdio.h>
#define NELEM(_arr) (sizeof (_arr) / sizeof (_arr)[0])
int tab[] = { [5] = 3 };
int main()
{
for (int i = 0; i < NELEM(tab); i++)
printf("tab[%d] = %d\n", i, tab[i]);
return 0;
} /* main */
will show:
$ a.out
tab[0] = 0
tab[1] = 0
tab[2] = 0
tab[3] = 0
tab[4] = 0
tab[5] = 3
$ _