The initialization of static variables in C

Question

I have a question about the initialization of static variables in C. I know if we declare a global static variable that by default the value is 0. For example:

static int a; //although we do not initialize it, the value of a is 0

but what about the following data structure:

typedef struct
{
    int a;
    int b;
    int c;
} Hello;

static Hello hello[3];

are all of the members in each struct of hello[0], hello[1], hello[2] initialized as 0?

question from:https://stackoverflow.com/questions/13251083/the-initialization-of-static-variables-in-c

Answer 1

Yes, all members are initialized for objects with static storage. See 6.7.8/10 in the C99 Standard (PDF document)

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, then:
— if it has pointer type, it is initialized to a null pointer;
— if it has arithmetic type, it is initialized to (positive or unsigned) zero;
— if it is an aggregate, every member is initialized (recursively) according to these rules;
— if it is a union, the first named member is initialized (recursively) according to these rules.

To initialize everything in an object, whether it's static or not, to 0, I like to use the universal zero initializer

sometype identifier0 = {0};
someothertype identifier1[SOMESIZE] = {0};
anytype identifier2[SIZE1][SIZE2][SIZE3] = {0};

---

There is no partial initialization in C. An object either is fully initialized (to 0 of the right kind in the absence of a different value) or not initialized at all.
If you want partial initialization, you can't initialize to begin with.

int a[2]; // uninitialized
int b[2] = {42}; // b[0] == 42; b[1] == 0;
a[0] = -1; // reading a[1] invokes UB