在链接静态库时初始化静态对象

Question

What are the rules for initialisation of static object declared in another shared library? For instance, consider the following:

file X.hpp:

struct X {
   X ();
   static X const s_x;
};

struct Y {
   Y (X const &) {}
};

file X.cpp:

#include "X.hpp"
#include <iostream>

X::X ()
{
   std::cout << "side effect";
}

X const X::s_x;

I compiled X.cpp in a static library libX.a , and I tried to link the following executable against it (file main.cpp):

#include "X.hpp"

int main ()
{
     (void)X::s_x;  // (1)
     X x = s_x;     // (2)
     Y y = s_x;     // (3)
 }

with only (1) or (2), nothing happens. But if I add (3), the static object is initialised (ie "side effect" is printed). (I use gcc 4.6.1).

Is there any way to predict what will happen here?

I don't understand how the instruction (2) does not force the X::s_x object to be default-constructed, whereas (3) does.

EDIT: build commands:

g++ -c X.cpp
g++ -c main.cpp
ar rcs libX.a X.o
g++ -o test main.o -L. -lX

Answer 1

By default on many platforms, if your program doesn't reference any symbols from a given object file in a static library, the whole object file (including static initializers) will be dropped. So the linker is ignoring Xo in libX.a because it looks like it is unused.

There are a few solutions here:

1. Don't depend on the side-effects of static initializers. This is the most portable/simple solution.
2. Introduce some fake dependency on each file by referencing a dummy symbol in a way the compiler will not see through (like storing the address into a externally-visible global).
3. Use some platform-specific trick to retain the objects in question. For example, on Linux you can use -Wl,-whole-archive ao ba -Wl,-no-whole-archive .