Re: [Boost-users] thread safety of the mutex ctor

From: Matthew Peltzer (goochrules_at_[hidden])
Date: 2005-08-17 17:03:56

On 8/17/05, Christopher Hunt <huntc_at_[hidden]> wrote:
> On 18/08/2005, at 1:19 AM, boost-users-request_at_[hidden] wrote:
> 
> > I envisage two threads accessing a function like this concurrently:
> >
> > template<typename T>
> > boost::shared_ptr<T>
> > my_func()
> > {
> >     static boost::mutex m;
> >     boost::mutex::scoped_lock l(m);
> >
> >     static boost::shared_ptr<T> p(new T);
> >     return p;
> > }
> >
> > and my worry is that both threads could attempt to do the static
> > initialisation of m concurrently.  Is there protection against this?
> > Is it even possible to protect against this?
>
> The scope of m is actually outside of your function given that it is
> static. Thus it will be initialised outside of your function generally
> before any other threads get to start up.

This does not sound right to me, so I whipped up a test case... on
HP-UX, given the code:

#include <iostream>
#include <string>

class C {
public:
        C(const std::string& str) {
                std::cout << "C::C(" << str << ")" << std::endl;
                str_ = str;
        }

        ~C() {
                std::cout << "C::~C(" << str_ << ")" << std::endl;
        }

private:
        std::string str_;
};

C c("global");

void f() {
        static C c("static");
}

int main(int argc, char* argv[]) {
        C c("local");
        f();

        return 0;
}

the output will be:

> g++ -o static static.cpp
> ./static
C::C(global)
C::C(local)
C::C(static)
C::~C(local)
C::~C(static)
C::~C(global)

Given my experience, this is what I expected.  The static instance is
not created until the function is called, and is destroyed just before
global instances are destroyed.

So, I would say that there is a race condition in the original code
with mutex m, if there could be two threads entering my_func() for the
first time.

-- Matthew Peltzer
-- goochrules_at_[hidden]