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Utah C++ User Group started
by legalize+jeeves＠mail.xmission.com 22 Sep '14

22 Sep '14

[Please do not mail me a copy of your followup] I'm booting up a C++ User Group in Utah. We're bootstrapping out first meeting to decide a regular meeting time and location. Join the Utah C++ Programmers MeetUp group to hear about meetings: <http://www.meetup.com/Utah-Cpp-Programmers/> Follow related evens on the Utah C++ Users Group blog: <http://utahcpp.wordpress.com> -- "The Direct3D Graphics Pipeline" free book <http://tinyurl.com/d3d-pipeline> The Computer Graphics Museum <http://computergraphicsmuseum.org> The Terminals Wiki <http://terminals.classiccmp.org> Legalize Adulthood! (my blog) <http://legalizeadulthood.wordpress.com>

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[lockfree] stack needs undocumented trivial assignment
by Norbert Wenzel 21 Sep '14

21 Sep '14

I'm trying to replace some computation intensive loops in legacy code with parallel loops and thus tried to store the results in a lockfree stack. I've chosen stack over queue because of the requirements stated in the documentation, where stack only needs a copy ctor and queue additionally needs trivial assignment and trivial dtor. My result struct I want to store in the stack is *not* trivially assignable. When trying to compile the code containing my lockfree::stack the static assertion in line 75 of stack.hpp because it requires trivial assignment and trivial dtor. My question would be if this is simply an error in the documentation ie. if the requirements section should be copied over from lockfree::queue or if these static assertions should only be checked in some special case, eg. where the stack is of fixed size? thanks, Norbert

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[context] How to set stack pointer in a platform-independent way?
by TONGARI J 21 Sep '14

21 Sep '14

Hi, The problem had been discussed here: https://groups.google.com/forum/#!topic/boost-devel-archive/cl8B1oUPrVM But it's still not clear how to set the stack pointer portably, even after examining the code of Boost.Coroutine. For now, Boost.Context only supports downward stack, and Boost.Coroutine, which happened to be written by the same author, just assumes that the stack always grows downward, which is not really a portable implementation, if future Boost.Context supports upward stack, Boost.Coroutine has to be changed as well. Even than that, what does "beginning of the stack" really means? Suppose we have a stack of size 3, starting at [s]: [s-1][s][s+1][s+2][s+3] Why the beginning of a downward stack is [s+3], not [s+2]? If it's a upward stack, is beginning [s] or [s-1]? Thanks.

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from: giridhar
by giridhar 20 Sep '14

20 Sep '14

Salutations boost http://ariasking.com/morning.php/ttqhx3144vcfnpry/ giridhar

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[ASIO] [SSL] [SOCKS4] Layers
by Maciej Miszczak 19 Sep '14

19 Sep '14

Hello, I have been successfully using ASIO for a long time, but recently I try to use SSL over SOCKS 4 and I'm struggling with short read while asynchronous SSL handshaking. I have come across this: http://stackoverflow.com/questions/9943541/boost-asio-how-to-read-write-on-…] but it's only partially helpful. I haven't find any good explanation, what the layers really are (ex. I can't assign next_layer and lowest_layer to any of OSI layers). Could someone explain the meaning of next_layer and lowest_layer and how to use SSL stream with SOCKS 4 proxy. Best regards, Maciej Miszczak

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Re: [Boost-users] Boost XML Serialization
by Manasi Deshpande 19 Sep '14

19 Sep '14

Hello All, Currently I am working on DB Driven C++ API, where it also involves XML processing functionalities. That's where I came across Boost Serialization. We have a very peculiar story here, and it goes like this: We have 2 identical APIs. One in C# and another in C++. Both are DB driven similar APIs. There are instances, where any XML data serialized on C++ platform needs to be de serialized on C# and vice versa. In C# life is straight forward; one needs no bothering about serialization and de serialization. I think in boost you can de serialize the XML document only if it is built using Boost Serialization. I think it is the case in C# also. Is there any way out, where Boost Serialization can be customized? Please forgive me if I am not clear in stating my problem statement! -------------------------------------------------------------------------------------------------- This email and any attachments are confidential and are for the use of the addressee only. If you are not the addressee, you must not use or disclose the contents to any other person. Please immediately notify the sender and delete the email. Statements and opinions expressed here may not represent those of the company. Email correspondence is monitored by the company. This information may be subject to export control regulation. You are obliged to comply with such regulations. Renishaw plc (company number 1106260) and Wotton Travel Limited (company number 01973158) are companies registered in England and Wales with a registered office at New Mills, Wotton-under-Edge, Gloucestershire, GL12 8JR, United Kingdom, Telephone +44 1453 524524. --------------------------------------------------------------------------------------------------

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[logging+unittest]
by Witold E Wolski 19 Sep '14

19 Sep '14

Hi, I need to integrate some loggin into my unit to some extend bein integration tests (hardware). How best to call this function which sets up boost logging void init() { boost::log::add_file_log("sample.log"); boost::log::core::get()->set_filter ( boost::log::trivial::severity >= boost::log::trivial::info ); } if I have a test Runner looking like: #define BOOST_TEST_MODULE MasterTestSuite #include <boost/test/unit_test.hpp> Is there a way not have to add the suites manually Thanks. -- Witold Eryk Wolski

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[thread] memory synchronization multiple mutexes
by firespot 19 Sep '14

19 Sep '14

Hi, How does memory synchronization and visibility behave when multiple mutexes are involved? I have two groups of threads (boost::thread_group) and two mutexes (boost::mutex m1, boost::mutex m2) + associated condition variables, with each thread group being responsible to write to some variables (here Ptr, run) and consistently using "its" mutex for locking. However there is a piece of code that looks - much simplified - like this: void f() { { boost::unique_lock<boost::mutex> Lock(m1); Ptr = someAddress; // set the Ptr from 0 to some memory address } { boost::unique_lock<boost::mutex> Lock(m2); run = 1; // was 0, inform that execution can proceed condVar2.notify_all(); } ... // wait until Ptr has been reset to 0 { boost::unique_lock<boost::mutex> Lock(m1); while (Ptr != 0) condVar1.wait(Lock); } } and another function: void g() { { boost::unique_lock<boost::mutex> Lock(m2); while (run == 0) condVar2.wait(Lock); } // run the execution, e.g. std::copy(Ptr, Ptr + 100, Target); // finally all finished, signal back that Ptr is not needed any more { boost::unique_lock<boost::mutex> Lock(m1); Ptr = 0; condVar1.notify_all(); } } f() is only invoked from a thread of the first group, and g() only from a thread of the second group. Whenever Ptr might get changes the first mutex (m1) and its associated condition-variable is used, and whenver run might get changed the second mutex and its associated conditional variable is used. The control flow guarantees that run is not set to 1 prior to Ptr being set, and thus in g() the "running" itself (here the std::copy) cannot commence earlier. However, what puzzles me now: I used a different mutex for setting Ptr and setting/checking run. Does this have any impact whatsoever on memory visibility across the threads in the two thread-groups? Specifically, is it always guaranteed that when g() continues its execution (because run == 1) that it must see the correct value of Ptr (a non-NULL value)? Or is it possible that it still sees 0 and I'd have to make a separate locking involving m1 to guarantee a proper read for Ptr as anything written under a lock of m1 is also only guaranteed to be later properly read if the locking also used m1? thanks!

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[mpl] simple mpl::transform with placeholder?
by Andrew King 19 Sep '14

19 Sep '14

Hi, Lets say I have an mpl vector defined as : template <typename Traits> struct A { }; template <typename Traits> struct B { }; typedef mpl::vector<A<mpl::_>, B<mpl::_> > MyVec; struct MyTraits {}; How do I 'apply' or transform MyVec with MyTraits? ie. I want the mpl::vector ... boost::mpl::vector<A<MyTraits>, B<MyTraits>>; I realise I can do mpl::apply<A<mpl::_>, MyTraits> to get A<MyTraits> but I can't work out how to do the equivalent to the entire vector. mpl::transform seems like what I should be using. Transform seems to do apply internally, but I can't work out the syntax to do something like: using substituted = typename boost::mpl::transform<MyVec, boost::mpl::apply<boost::mpl<_>, MyTraits> >::type; (This isn't right!) It may be I am confused because inside the transform, boost::mpl::_1 is the item in the vector, not the 'inner' placeholder. Sorry if this is a basic question ... Regards, Andrew

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[Thread] Read-write locks
by lfrfly＠icqmail.com 18 Sep '14

18 Sep '14

I'm trying to create a standard read/write lock using Boost Threads, and I'm having a bit of difficulty mapping the Boost concepts onto the usual ones. A read/write lock should have several states: Locked for reading (multiple threads may lock at once), locked for writing (only one thread may lock, and no threads may be locked for reading at the same time), and unlocked. Ideally it would be nice if an RWL which is locked for reading may temporarily upgrade to locked for writing, then return to the locked for reading state. Boost supports several relevant notions of locking: shared locking, upgrade locking, and exclusive locking. This to me is a confusing distinction. In particular, a shared_mutex seems appropriate as the basis for a RWL, but is lock_upgrade() the equivalent of a write lock and lock_shared the equivalent of a read lock? This is unclear. If "upgrade locking" is not the same as "exclusive locking", then is an upgrade lock actually exclusive? Could someone please clarify this.

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