Re: Thread and C++

From:
peter koch <peter.koch.larsen@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 30 Mar 2009 13:58:42 -0700 (PDT)
Message-ID:
<6b4ec77e-6cff-43c4-872e-4bdfbcc9a84d@j39g2000yqn.googlegroups.com>
On 30 Mar., 22:16, Ioannis Vranos <ivra...@freemail.gr> wrote:

peter koch wrote:


[snip]

The book that mentions this is:

"C++ GUI Programming with Qt 4 (2nd Edition) - The official C++/Qt book":

http://www.qtsoftware.com/developer/books/cpp-gui-programming-with-qt...


[snip]

Regarding volatiles, in multithreading chapter (Chapter14) the books ment=

ions:

"class Thread: public QThread
{
    Q_OBJECT

    public:
        Thread();

        void setMessage(const QString &message);
        void stop();

    protected:
        void run();

    private:
        QString messageStr;
        volatile bool stopped;

};

The Thread class is derived from QThread and reimplements the run() funct=

ion. It provides two additional

functions: setMessage() and stop().

The stopped variable is declared volatile because it is accessed from dif=

ferent threads and we want to be sure

that it is freshly read every time it is needed. If we omitted the volati=

le keyword, the compiler might

optimize access to the variable, possibly leading to incorrect results".


That is clearly wrong: if the boolean "stopped" is modified using a
mutex only, there is no need for the volatile. What might happen is
that stopped is not modified under the control of a mutex: if the only
change that might take place is a change from false to true, it is my
understanding that such a change will be safe. And even if there might
be perverted situations where this might not be the case, I am
confident that no implementation ever would behave like that.
If we assume that stopped is accessed without using a mutex, it is
possible that the value written by one thread will never propagate to
another. The problem just is that this will not change with volatile.

/Peter

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