Can the compiler deal with the initialization order of static variables correctly if there is dependency?

Question

Can the compiler deal with the initialization order of static variables correctly if there is dependency? For example, I have

a.h:

struct A { static double a; };

a.cpp:

#include "a.h"
double A::a = 1;

b.h:

struct B { static double b; };

b.cpp:

#include "b.h"
#include "a.h"
double B::b = A::a;

Wouldn't `A.cpp` fail to compile due to duplicate definition? — Mark B, Sep 16 '13 at 18:12

score 6 · Accepted Answer · edited Sep 16 '13 at 18:34

6

Within translation units the order of such initialization is specified. Across translation units the order is not specified.

So in your case, since statics will get zero initialized by default, B::b will definitely be 0 or 1.

edited Sep 16 '13 at 18:34

SirGuy

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answered Sep 16 '13 at 18:24

Mark B

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Wow. So the short answer is unfortunately `no`. Thanks. I have a new related question here: http://stackoverflow.com/questions/18834855/can-the-compiler-deal-with-the-initialization-order-of-static-variables-correctl – user1899020 Sep 16 '13 at 18:26

score 1 · Answer 2 · answered Sep 16 '13 at 18:26

Nope. Static initialization order is undefined by the C++ standard. You can HOPE that your compiler is smart, but there's no guarantee.

Think about what would happen if you also have a "c" variable which is assigned with "b", and the "a" variable is assigned with "c". Then you've have a circular dependency, and it would compile, but you would end up with garbage values.

Can the compiler deal with the initialization order of static variables correctly if there is dependency?

2 Answers2

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