Re: About initializing non-local object

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 19 May 2008 01:58:22 -0700 (PDT)
Message-ID:
<6e9ec038-d399-47e0-a0c0-0d9e174ac545@b64g2000hsa.googlegroups.com>
On May 17, 8:47 am, Jidilo <jid...@gmail.com> wrote:

On 16 mai, 19:10, Jidilo <jid...@gmail.com> wrote:

Like:
//file1.cpp
int g_var1 = 3;

//file2.cpp
extern int g_var1;
int g_var2 = g_var1;

No matter what the initialization order is, when g_var2 is
initialized, g_var1 will be in an initialized state.


In this case, the initialization order is defined. Static
initialization (g_var1) always takes place before dynamic
intialization (g_var2). To extend my previous example:


OK but what if:

extern int x1;
extern int x2;
int x1 = x2;
int x2 = x1;

?


That's all dynamic initialization, so first zero initialization
will set both to 0, then dynamic initialization will occur in
the declaration order (within any single file)


Still having question about this. I have try this:

class A {
public:
    A(const A& a, const std::string& name) { cout << name << endl; }
};

file 1:
-------
extern A a2;
A a1(a2, "a1");

file 2:
-------
extern A a1;
A a2(a1, "a2");

And got no compilation error. In this case, what is the
behaviour ? is it well defined ?


It's defined behavior, but unspecified: both variables will
first be zero initialized, then one of the variable's
constructors will be called, then the other one. The first
constructor will see a zero initialized A const&.

You might try it, adding something to A so that you can see the
initialization, say:

    class A
    {
    public:
        A( A const& other, std::string const& name )
            : myFirst( 1 )
            , mySecond( other.myFirst )
        {
            std::cout << name << std::endl ;
        }
        int myFirst ;
        int mySecond ;
    } ;

Use this in the two files above, and try linking them in
different orders. After initialization, both objects will have
myFirst == 1, but one of them will have mySecond == 1, the other
== 0. Which one is not specified, but with most compilers,
you'll find that it will depend on the order the object files
are incorporated into the build. (In an unspecified way; you
can't reliably use this for controlling order of initialization,
even unportably.)

--
James Kanze (GABI Software) email:james.kanze@gmail.com
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