Re: Is this reordering valid?

From:
"James Kanze" <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
29 Oct 2006 11:03:18 -0500
Message-ID:
<1162123582.008920.91040@e3g2000cwe.googlegroups.com>
Martin Bonner wrote:

pongba@gmail.com wrote:

In "C++ and the Perils of Double-Checked Locking", Scott Meyers and
Andrei Alexandrescu brought up a good point on how it is impossible to
avoid the reorder problem.
But I think there might be something that I would disagree. Look at the
code below:

C* pc = new C();

It involves three steps:

1. allocation of raw memory
2. construction of C
3. assignment

The question is, is it really allowed by the standard that 2 and 3
could be reordered?


I can't see any reason why not. Why do you think it isn't?


I'm not sure myself, but consider something like the following:

    struct Data {} ;
    struct C ;
    C* defaultInstance = new C ;

    struct C
    {
        C() {
            if ( defaultInstance == NULL ) {
                value = new Data ;
                defaultInstance = this ;
            } else {
                value = defaultInstance->value ;
            }
        }
        D* value ;
    } ;

I suspect that most programmers would expect this to work.
Although like you, I'm not really sure that the standard
guarantees it, and there is a simple work-around, guaranteed to
work: wrap the new in a function which returns the pointer.
(This introduces a sequence point between the new expression and
the assignment.)

Note that even wrapping the new in a function doesn't make
double checked locking work, because the guarantees given by the
standard only affect the observable behavior in a single thread.
Another thread is not guaranteed to see the writes in the same
order they occur; in fact, the standard doesn't guarantee any
order of the writes, as long as changing the order doesn't
affect observable behavior.

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