How will this variable be initialized?

Question

I have the following

char mem_pool[1024*1024*64]; 

int main() {
    // ... 
}

I'm trying to get a thorough understanding how will mem_pool be initialized. After a lot of search my conclusions are :

• it's a static initialization (not as in static the keyword, but as in "run before the program does - during static initialization phase")
• it will run in 2 phases : zero initialization and default initialization (the second phase won't do anytning)
• it's an array of PODs so the default initialization for every element should apply, but due to the previous 2 points we won't have an array of indeterminable values (as we would with a char ar[N] in a function scope) but an array of zeroes.

Could someone help me dissambiguate what's guaranteed by the language and correct me if I'm wrong ?

I also thought of doing any of the following

char mem_pool[1024*1024*64] {}; 
char mem_pool[1024*1024*64] = ""; 

I suspect it's a better/recommended practice, but for now I need to understand my initial question.

Answer 1

Your understanding is correct.

The array's elements will all be zero-initialised, because the array has static storage duration:

[C++11: 3.6.2/2]: Variables with static storage duration (3.7.1) or thread storage duration (3.7.2) shall be zero-initialized (8.5) before any other initialization takes place. [..]

[C++11: 8.5/5]: To zero-initialize an object or reference of type T means:

• if T is a scalar type (3.9), the object is set to the value 0 (zero), taken as an integral constant expression, converted to T;
• if T is a (possibly cv-qualified) non-union class type, each non-static data member and each base-class subobject is zero-initialized and padding is initialized to zero bits;
• if T is a (possibly cv-qualified) union type, the object’s first non-static named data member is zero-initialized and padding is initialized to zero bits;
• if T is an array type, each element is zero-initialized;
• if T is a reference type, no initialization is performed.

If it did not have static storage duration, the elements would all have indeterminate values:

[C++11: 8.5/11]: If no initializer is specified for an object, the object is default-initialized; if no initialization is performed, an object with automatic or dynamic storage duration has indeterminate value. [..]

[C++11: 8.5/6]: To default-initialize an object of type T means:

• if T is a (possibly cv-qualified) class type (Clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
• if T is an array type, each element is default-initialized;
• otherwise, no initialization is performed.