c++ - 使用 boost::named_condition 的错误

Question

我无法编译这个简单的示例（在 VS 2008 上）：

#include <boost/interprocess/sync/named_condition.hpp>
#include <boost/thread/mutex.hpp>
#include <string>

int main(int argc, const char* argv)
{
    boost::interprocess::named_condition cond(boost::interprocess::create_only, "somd.notify");
    boost::mutex the_mutex;
    while (true)
    {
        const boost::system_time timeout = boost::get_system_time() + boost::posix_time::milliseconds(500);
        boost::mutex::scoped_lock lock(the_mutex);
        if (cond.timed_wait(lock, timeout))
        {
            // Do something useful
        }
    }

    return 0;
}

我得到的编译错误是（boost v1.38.0）：

1>....\boost/interprocess/sync/named_condition.hpp(326) : error C2273: 'function-style cast' : illegal as right side of '->' operator
1>        ..\src\main.cpp(13) : see reference to function template instantiation 'bool boost::interprocess::named_condition::timed_wait<boost::mutex::scoped_lock>(L &,const boost::posix_time::ptime &)' being compiled
1>        with
1>        [
1>            L=boost::mutex::scoped_lock
1>        ]

想法？

Answer 1

我猜你必须使用不同的互斥锁类型：例如 boost::interprocess::interprocess_mutex 。

最好的后卫

托尔斯滕