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static initialization order fiasco

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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?
like image 995
Happy Mittal Avatar asked Jun 14 '10 06:06

Happy Mittal

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1 Answers

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.

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Michael Burr Avatar answered Sep 29 '22 13:09

Michael Burr
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