Initialization of Objects with Static Storage Duration in C vs C++ [duplicate]

Answer 1

Since a compiler takes statements and emits low-level code for the CPU to use, you have to split apart what is really happening here. It would go something like this:

1. Create a memory slot for "i".
2. Initialize the memory to zero (normal default behavior).
3. Read the value of "i" (which is zero).
4. Add 1.
5. Store it in "i".

Answer 2

In C++ it is syntactically correct. In C you can initialize a global variable only with a constant. So your code would not compile in C.

In C this is legal BTW

int main()
{
   int i = i+1;
}

3.3.1/1 Point of declaration

The point of declaration for a name is immediately after its complete declarator and before its initializer (if any).

The behaviour is well defined as per §3.6.2/1 which says:

"Objects with static storage duration (3.7.1) shall be zero-initialized (8.5) before any other initialization takes place."

Answer 3

(notice: I'm referring to the current C++ standard)

I'm not really sure about this, but, if my interpretation of the standard is correct, the code should be fine and not UB.

The first initialization of that variable is the zero-initialization of objects with static storage duration that happens before any other initialization takes place (§3.6.2 ¶1).

So, first of all i is set to zero.

Then, dynamic initialization (i.e. non-zero and non-constant initialization) takes place, so it uses the current value of i (0) to actually initialize it again. At the end it should evaluate to 1.

This seems confirmed by §8.5 ¶6, that explicitly says:

The memory occupied by any object of static storage duration shall be zero-initialized at program startup before any other initialization takes place. [Note: in some cases, additional initialization is done later. ]

(If you find some flaw in the analysis please just tell me in the comments and I'll be glad to correct/delete the answer, it's slippery floor and I'm conscious of it :) )

Answer 4

You cannot assign a value to variable using another variable outside any function. The statement i + 1; is evaluated during runtime, while int i = i + 1; is outside any function so it needs to be evaluated in compile time.

Answer 5

Whether its really syntactically illegal I'm not sure (it would definitly be valid inside the method). However, as you suggest, the is a semantic problem and the compiler should issue a warning as i was used without initialization. IMO the C/C++ compiler does not generally warn of such thing however (Java for instance would give an error), though you might turn on such warning by adding -Wall parameter to gcc.

Answer 6

Your code is not legal C.

If your compiler compiles it without a diagnostic,
your compiler is not a C compiler

You must use constants to initialize a variable.

In your code, the initializer expression ( i + 1 ) is not a constant.

This violates 6.7.8/4:

All the expressions in an initializer [...] shall be constant expressions or string literals.