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Agree totally.
by Dan Dimerman 03 Jun '03

03 Jun '03

Can anyone who successfuly wrote some code with BGL post it somewhere ?

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RE: [Boost-Users] int64_t problem
by Jim.Hyslop 02 Jun '03

02 Jun '03

Sigis wrote: > I am using OpenInventor and boost 1.29 librariess. Both of > them have defined > int64_t: > > typedef __int64 int64_t; [...] > The only solution I found is to add #undef just before define at > boost/cstdint.hpp: > #undef int64_t > typedef __int64 int64_t; > > But it looks as dirty solution. Maybe it is possilbe to solve > this problme in > more elegant way? First step is to send a complaint (or enhancement request, if you prefer :=) to OpenInventor - any new types they introduce should be wrapped in their own namespace. Also, if #undef really *does* work, then that means either your compiler is buggy (#undef should only work on macros, not typedefs), or OpenInventor are using a #define, not a typedef - double penalty points for that! (well, I suppose if it's a C library, that _might_ be acceptable. Barely.) If it is a #define, you could try wrapping the OpenInventor headers in another header: // Wrapper file for using OpenInventor with Boost [include guard #ifdefs] #include "OpenInventorHeader.h" #undef int64_t #endif // #include guard In any source file that uses both Boost and OpenInventor, include the wrapper header instead of the OpenInventor header. You'll want to check for any include order problems, i.e. you may have to ensure all Boost headers are included after all OpenInventor headers. Subtle issue, though - could this violate the ODR (One Definition Rule)? I think we'd have to see exactly how OpenInventor defines/typedefs int64_t to be sure. Other possible solutions: - Decouple Boost and OpenInventor, by wrapping the OpenInventor objects you are using in proxy objects. The proxies will use an opaque handle to the OpenInventor objects. Use these proxy objects wherever you currently use OpenInventor. This, by the way, has the advantage of decoupling your application from OpenInventor, allowing you the option to move to a different library if you ever decide that should be necessary. - If OpenInventor's int64_t is a typedef, put it inside an OpenInventor namespace (this probably won't work if you do not have the source code to OI in order to rebuild the binaries). If it's not a typedef, then make it a typedef in an OpenInventor namespace (same caveats). - If OpenInventor really does use a typedef, and if the typedef is inside an OI namespace, then hunt down and destroy any using directives (i.e. "using namespace OI;" or "using OI::whatever;") in any headers, and move any using directives inside your source files *after* all #include directives. If you find a "using namespace" directive in any of the OpenInclude directives, then that's triple penalty points against them, and definite cause for a complaint (not an enhancement request :=). - [radical solution, probably unrealistic] change library vendors, and be sure to tell OI why you are dropping them (because they don't play well with everyone else). - <sigh> change library vendors, and be sure to tell Boost why you are dropping them :=( Boost plays nicely - int64_t is a typedef, and it is declared inside namespace boost. -- Jim

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No more than 10 types in boost::tuple?
by Tarjei Knapstad 02 Jun '03

02 Jun '03

I tried constructing a tuple with 12 types today without much luck. I've read through the headers to see if there was any obvious way of getting around the 10 limit but couldn't find one (I half expected to find something along the way of vectorxx<>, in MPL). Am I being stupid, or is there really no way of constructing a tuple with more than 10 elements? Also, are there any plans of integrating and/or making boost::tuple dependant on the MPL? I realize that tuple predates the MPL, but there's no use in duplicating efforts. boost::tuple has it's own compile time IF<> and also provides it's own typelist class which are both readily available in the MPL. Cheers, -- Tarjei

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How to compile jam.exe with Visual Studio .NET (Win2000) where to download it?
by rasasenkuviene 01 Jun '03

01 Jun '03

I was trying to build boost-jam and boost-build executable with build.bat in Visual Studio .NET (Windows 2000) from downloaded sources. However, ended with single non-functional jam0.exe and truncated to zero length my own compiler cl.exe from Visual Studio .NET. Nmake coul'nt be used, because makefiles were not included. What is going on? How to compile or obtain needed boost-jam or boost-build executables under Windows2000? Or this BOOST is intended for UNIX only?

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Newbee Questions on Directory Structure of Boost Libs.
by Adalbert Perbandt 01 Jun '03

01 Jun '03

Hi there, I'm new to Boost and have some problems understanding the way Boost libs. are built and installed. I downloaded boost_1_30_0. unpacked the archive and built the test library (I had installed bjam previously). I did the build step on two different platforms and with different compilers: - On Linux (RedHat 7.3) with gcc/3.2.2 and - on Sun SPARC Solaris 2.7 with Sun's Forte 6.2 compiler. I created an installation directory /opt/boost_1.30.0. Into this directory I copied the directory tree below boost/test and libs/test: /opt/boost_1.30.0/boost/test/... /opt/boost_1.30.0/libs/test/... I think so far everything is correct, isn't it? I do not completely understand the directory structure generated during the build process: ../boost_1_30_0/libs/test/build/bin/libboost_prg_exec_monitor.a /libboost_prg_exec_monitor.so /libboost_test_exec_monitor.a /libboost_test_exec_monitor.so /libboost_unit_test_framework.a /libboost_unit_test_framework.so Below the directories given above a specific subdirectory tree was created. Example: .../libboost_prg_exec_monitor.a/ /sunpro/ /debug/ /runtime-link-dynamic/ /runtime-link-static/ /threading-multi/libboost_prg_exec_monitor.a /threading-single/libboost_prg_exec_monitor.a >From this I conclude: 1. The builds are separated from each other in compiler specific subdirectory trees, e.g. the "sunpro" subtree contains the build results from $ bjam "-sTOOLS=sunpro" i.e. the sun forte 6.2 compiler and the "gcc" subtree contains the build results from $ bjam "-sTOOLS=gcc" i.e. the gcc 3.2.2 compiler on Linux. Obviously there is no separation by OS platform (Solaris or Linux). What would I have to do to build the lib. with gcc on Solaris? Where would the products be installed? Furthermore there is no separation by compiler version (e.g. gcc 2.95.2 is not separated from gcc 3.2.2). Is this considered unnecessary? 2. The libs were built with debug symbols (-g option) only), correct? I'm not familiar with jam and therefore don't know what to do to get them built with optimization active. It seems to me as if this should be rather simple because in ../boost_1_30_0/tools/build/sunpro-tools.jam obviously two variants of the CFLAGS variable for optimization are defined: flags sunpro CFLAGS <optimization>speed : -fast -O4 ; flags sunpro CFLAGS <optimization>space : -O2 -xspace ; Any hints are welcome. Thanks, Adalbert Perbandt.

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RE: [Boost-Users] Regex: matching either of 2 fields, in any order
by john.wismar＠alldata.com 31 May '03

31 May '03

That's it exactly. And your explanation was a lot more succinct than mine, too! Thanks. ------------------------------ John Wismar BSA - AllData Offline Products john.wismar(a)alldata.com > -----Original Message----- > From: Jim.Hyslop [mailto:jim.hyslop@leitch.com] > Sent: Friday, May 30, 2003 6:10 AM > To: 'Boost-Users(a)yahoogroups.com' > Subject: RE: [Boost-Users] Regex: matching either of 2 > fields, in any orde r > > > John Maddock wrote: > [unknown] wrote (please try to attribute quotes) > > > I can see how I could do this in a loop to replace the different > > > subfields. Is there another step I would need to take to > > > guarantee that > > > there is at most one of each type of subfield? > > > > Well if you search for "C(\\d+)|O(\\w+)" it can only match > > one or the other > > can't it? Or am I missing something? > I could be wrong, but I interpreted the OP's question to mean: is it > possible to have the regex generate an error if more than one > sub-field of a > specific type is found, i.e. the following would be invalid: > > ~C123~C345 > > The original poster can correct me if I've misinterpreted anything. > > -- > Jim > > ------------------------ Yahoo! Groups Sponsor > ---------------------~--> > Get A Free Psychic Reading! Your Online Answer To Life's > Important Questions. > http://us.click.yahoo.com/Lj3uPC/Me7FAA/ySSFAA/EbFolB/TM > -------------------------------------------------------------- > -------~-> > > Info: <http://www.boost.org> > Wiki: <http://www.crystalclearsoftware.com/cgi-bin/boost_wiki/wiki.pl> > Unsubscribe: <mailto:boost-users-unsubscribe@yahoogroups.com> > > > Your use of Yahoo! Groups is subject to > http://docs.yahoo.com/info/terms/ > > >

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Using Bind with Signals (version 2)
by Darren Vincent Hart 30 May '03

30 May '03

Similar to my last post, this is a slightly different implementation where the handler has a signal of its own. Now I can't seem to bind to the handler functor, getting a compiler message like: /usr/include/boost/bind.hpp:60: error: `bool (handler::*)(char)' is not a class, struct, or union type The goal is that when the keypress signal is emitted, it will execute the handler which will receive the char (key), test it, and emit its signal (sig) if appropriate. Any help would again be much appreciated, the short code follow, with the problematic bind statement commented: Thanks in advance, Darren Hart compile: g++ test.cpp -o test -lboost_signals // test.cpp ---------------------------------------------------- #include <iostream> #include <boost/signal.hpp> #include <boost/function.hpp> #include <boost/bind.hpp> #include <boost/bind/apply.hpp> using namespace std; struct widget { boost::signal<bool ()> sig; boost::signal<bool (char)> keypress; bool test() { cout << "widget::test() - sig empty: " << (sig.empty() ? "true" : "false") << endl; sig(); return true; } }; struct handler { char key_; boost::signal<bool ()> sig; handler(char key) : key_(key) { } bool operator()(char key) { cout << "handler::operator() - received a " << key << endl; if (!sig.empty() && key == key_) { cout << "\texecuting slot\n\t"; sig(); } cout << endl; return true; } }; struct no_op { bool operator()() { cout << "no_op::operator() - doing nothing" << endl; return true; } bool test() { cout << "no_op::test()" << endl; return true; } }; int main() { // instantiate two widgets widget widget_a; widget widget_b; // instanstiate the functions object (and test them) no_op no_op_; handler handler_a('A'); handler_a.sig.connect(no_op_); no_op_(); handler_a('A'); // connect the function objects to the widgets widget_a.sig.connect(no_op_); // THE PROBLEM IS HERE - HOW DO I BIND TO handler_a.test(char) ? widget_b.keypress.connect( boost::bind(&handler::operator(), &handler_a) ); // END PROBLEM cout << "signalling widget_b.keypress('B')" << endl; widget_b.keypress('B'); // call the handler 3 times, but never execute its slot cout << "\nsignalling widget_b.keypress('A')" << endl; widget_b.keypress('A'); // call the handler 3 times, executing the slot return 0; }

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Re: Using Bind on Signals
by Peter Dimov 30 May '03

30 May '03

darrenvhart wrote: > --- In Boost-Users(a)yahoogroups.com, "Peter Dimov" <pdimov@m...> wrote: >> >> You can use apply<> from <boost/bind/apply.hpp> instead of 'call': >> >> bind(apply<bool>(), bind(&widget::sig, &widget_a)) (); // should call >> widget_a->sig > > Thank you for the reply. > > I have changed the line to use apply<bool>() and included the header > file, but it won't compile complaining about noncopyable members of > the signal of the widget. Any ideas what that is all about ? Please change every 'F f' to 'F & f' in boost/bind/apply.hpp; this will take care of the problem. I've already updated the CVS version. The problem is that a signal is noncopyable and cannot be passed by value.

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Using Bind on Signals
by Darren Vincent Hart 30 May '03

30 May '03

I am building a widget set using the signals and slots library. I have the need to be able create a functor (FA) that receives another (FB) as an argument to its constructor. I will connect FA to the signal of a widget, when FA is run it will perform a test and execute FB if appropriate. I want to be able to use the signal member of another widget as FB, so I am trying to use bind to get that function object for the signal. I am experiencing two problems: 1) When the member functor (FB) is bound to the signal of another widget, the slots don't get called. (No error results, but the connected slot doesn't print anything so I am assuming it isn't being executed). 2) I can't get it to compile when the signal has a <bool ()> signature, I can only get it to compile with a <void ()> signature. I am trying to assign the result of bind to a boost::function<bool ()> but I get an error saying it can't make the conversion, what type should I be using if not boost::function (note that it works if the signal is void())? The following is a commented test_case demonstrating the problem. Any assistance in binding to signals and compiling those with a bool() signature would be greatly appreciated. Thanks in advance, Darren Hart compile with: g++ test.cpp -o test -lboost_signals // test.cpp ---------------------------------------------------------- #include <iostream> #include <boost/signal.hpp> #include <boost/function.hpp> #include <boost/bind.hpp> using namespace std; struct widget { boost::signal<bool ()> sig; boost::signal<bool (char)> keypress; bool test() { cout << "widget::test() - sig empty: " << (sig.empty() ? "true" : "false") << endl; sig(); return true; } }; template <typename T> struct handler { boost::function<T ()> slot_; handler(boost::function<T ()> slot) : slot_(slot) { } bool operator()(char letter) { cout << "handler::operator() - received a " << letter << endl; if (letter == 'A') { cout << "\texecuting slot\n\t"; slot_(); cout << endl; } return true; } }; struct no_op { bool operator()() { cout << "no_op::operator() - doing nothing" << endl; return true; } bool test() { cout << "no_op::test()" << endl; return true; } }; int main() { // instantiate two widgets widget widget_a; widget widget_b; /* The goal is to connect the hanlder to widget_b which will then call * widget_a's sig signal. The problem is that even though widget_a's * test says that sig() isn't empty, it doesn't execute when bound * directly in handler_b. It does however work when called as part * of the test() member function as in handler_c. We must get * handler_b to work in order to do this in libstk! We can't compile * at all if the signal returns a bool. */ // instanstiate the function objects no_op no_op_; // bind to a no_op member, successul handler<bool> handler_a( boost::bind(&no_op::test, &no_op_) ); // bind to a widget signal handler, BROKEN //handler<bool> handler_b( boost::bind<bool>(&widget::sig, &widget_a) ); // <-- Broken // bind to a widget member, successul handler<bool> handler_c( boost::bind(&widget::test, &widget_a) ); // connect the function objects to the widgets widget_a.sig.connect(no_op_); widget_b.keypress.connect(0, handler_a); //widget_b.keypress.connect(1, handler_b); // fix handler_b above, then uncomment this widget_b.keypress.connect(2, handler_c); cout << "attempting to bind to widget_a's sig signal" << endl; // store the function object returned by binding a widget signal, BROKEN //boost::function<bool ()> fp = boost::bind(&widget::sig, &widget_a); // <-- BROKEN // executre the functor returned by binding a widget signal, BROKEN // (builds, but doesn't really call widget_a.sig()) boost::bind(&widget::sig, &widget_a)(); // <-- Test if binding works at all, it doesn't // signal the keypress signals, they will test the character and execute their // slots if it equals 'A' cout << "signalling widget_b.keypress('B')" << endl; widget_b.keypress('B'); // call the handler 3 times, but never execute its slot cout << "***\nThis correctly executed the three handlers which, also correctly," << " do not execute their slots\n***" << endl; cout << "signalling widget_b.keypress('A')" << endl; widget_b.keypress('A'); // call the handler 3 times, executing the slot cout << "***\nThis executes the three handlers, but handler_b calls slot_() " << "and widget_a's signal does not execute the slot it has, which the " << "test() in handler_c successfully executes.\n***" << endl; return 0; } // --------------------------------------------------------------------

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RE: [Boost-Users] Regex: matching either of 2 fields, in any order
by john.wismar＠alldata.com 30 May '03

30 May '03

The detail I'm concerned about -- and it may not be a problem because I may be overlooking something -- is that the string I'm matching against can have zero or one 'C' field and zero or one 'O' fields. It looks like the "C(\\d+)|O(\\w+)" expression will match either of them if at least one is present, but after that, I would need optionally to match the other one. Matching the same field again would be an error. In other words: ~C12345~OF <-- valid (1 of each) ~OF <-- valid (1 'O' field) ~C12345 <-- valid (1 'C' field) ~ZAQ~C12345 <-- valid (1 'C' field) ~OV~C67890 <-- valid (1 of each) ~A23~OV~DFG~C67890~ZAQ <-- valid (1 of each) ~A23~DFG <-- valid (neither) ~C12345~OF~C67890 <-- invalid (2 'C' fields) ~C12345~OF~OV <-- invalid (2 'O' fields) ~ASD~C12345~OF~OV <-- invalid (2 'O' fields) Thanks again for your help!! ------------------------------ John Wismar BSA - AllData Offline Products john.wismar(a)alldata.com > -----Original Message----- > From: John Maddock [mailto:john_maddock@compuserve.com] > Sent: Friday, May 30, 2003 3:43 AM > To: Boost-Users(a)yahoogroups.com > Subject: Re: [Boost-Users] Regex: matching either of 2 > fields, in any order > > > > I can see how I could do this in a loop to replace the different > > subfields. Is there another step I would need to take to > guarantee that > > there is at most one of each type of subfield? > > Well if you search for "C(\\d+)|O(\\w+)" it can only match > one or the other > can't it? Or am I missing something? > > John.

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