Question

I was reading about SIOF from a book and it gave an example :

//file1.cpp
extern int y;
int x=y+1;

//file2.cpp
extern int x;
int y=x+1;  

Now My question is :
In above code, will following things happen ?

1. while compiling file1.cpp, compiler leaves y as it is i.e doesn’t allocate storage for it.
2. compiler allocates storage for x, but doesn’t initialize it.
3. While compiling file2.cpp, compiler leaves x as it is i.e doesn’t allocate storage for it.
4. compiler allocates storage for y, but doesn’t initialize it.
5. While linking file1.o and file2.o, now let file2.o is initialized first, so now:
   Does x gets initial value of 0? or doesn’t get initialized?

Answer 1

The initialization steps are given in 3.6.2 “Initialization of non-local objects” of the C++ standard:

Step 1: x and y are zero-initialized before any other initialization takes place.

Step 2: x or y is dynamically initialized – which one is unspecified by the standard. That variable will get the value 1 since the other variable will have been zero-initialized.

Step 3: the other variable will be dynamically initialized, getting the value 2.