In C++, does initializing a global variable with itself have undefined behaviour?

Question

int i = i;

int main() { 
 int a = a;
 return 0;
} 

int a = a surely has undefined behaviour (UB), and more details on it is in Is reading an uninitialized value always an undefined behaviour? Or are there exceptions to it?.

But what about int i = i? In C++ we are allowed to assign nonconstant values to globals. i is declared and zero initialized (since it has file scope) before the declaration is encountered. In which case we are assigning 0 to it later in the definition. Is it safe to say this does not have UB?

Answer 1

Surprisingly, this is not undefined behavior.

Static initialization [basic.start.static]

Constant initialization is performed if a variable or temporary object with static or thread storage duration is constant-initialized. If constant initialization is not performed, a variable with static storage duration or thread storage duration is zero-initialized. Together, zero-initialization and constant initialization are called static initialization; all other initialization is dynamic initialization. All static initialization strongly happens before any dynamic initialization.

Important parts bold-faced. "Static initialization" includes global variable initialization, "static storage duration" includes global variables, and the above clause is applicable here:

int i = i;

This is not constant-initialization. Therefore, zero-initialization is done according to the above clause (for basic integer types zero-initialization means, unsurprising, that it's set to 0). The above clause also specifies that zero initialization must take place before dynamic initialization.

So, what happens here:

1. i is initialized to 0.
2. i is then dynamically initialized, from itself, so it still remains 0.