使用boost :: future .then（）的编译错误

Question

我正在尝试使用boost :: future .then（）功能。 该片段摘自Boost 1.54.0线程同步文档

#include <string>  
#include <boost/thread/future.hpp>
int main() {
  boost::future<int> f1 = boost::async([]() { return 123; });
  boost::future<std::string> f2 = f1.then([](boost::future<int> f)->std::string {
                                            int x = f.get();
                                            return ("Done" + std::to_string(x));
                                            });
}

设定 ：
Ubuntu 13.04
g ++版本g ++（Ubuntu 4.8.1-2ubuntu1〜13.04）4.8.1
Boost版本1.54.0

命令行 ：

g++  then_test.cc -std=c++0x -DBOOST_THREAD_VERSION=4 -I        /home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost -L /home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/stage/lib -static -lboost_thread-mt -lboost_date_time-mt  -lboost_system-mt -lpthread

错误：

g++  then_test.cc -std=c++0x -DBOOST_THREAD_VERSION=4 -I /home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost -L /home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/stage/lib -static -lboost_thread-mt -lboost_date_time-mt  -lboost_system-mt -lpthread
then_test.cc: In function ‘int main()’:
then_test.cc:10:44: error: no matching function for call to ‘boost::future<int>::then(main()::__lambda1)’
                                           });
                                            ^
then_test.cc:10:44: note: candidates are:
In file included from then_test.cc:2:0:
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp:1598:9: note: template<class F> boost::future<typename boost::result_of<F(boost::future<R>&)>::type> boost::future<R>::then(F&&) [with F = F; R = int]
         then(BOOST_THREAD_FWD_REF(F) func);
         ^
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp:1598:9: note:   template argument deduction/substitution failed:
In file included from then_test.cc:2:0:
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp: In substitution of ‘template<class F> boost::future<typename boost::result_of<F(boost::future<R>&)>::type> boost::future<R>::then(F&&) [with F = F; R = int] [with F = main()::__lambda1]’:
then_test.cc:10:44:   required from here
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp:62:29: error: no type named ‘type’ in ‘struct boost::result_of<main()::__lambda1(boost::future<int>&)>’
 #define BOOST_THREAD_FUTURE future
                             ^
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp:3840:3: note: in expansion of macro ‘BOOST_THREAD_FUTURE’
   BOOST_THREAD_FUTURE<R>::then(BOOST_THREAD_FWD_REF(F) func)
   ^
In file included from then_test.cc:2:0:
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp:1601:9: note: template<class F> boost::future<typename boost::result_of<F(boost::future<R>&)>::type> boost::future<R>::then(boost::launch, F&&) [with F = F; R = int]
         then(launch policy, BOOST_THREAD_FWD_REF(F) func);
         ^
/home/prakash/maidsafe/MaidSafe/build/boost_1_54_0/src/boost/boost/thread/future.hpp:1601:9: note:   template argument deduction/substitution failed:
then_test.cc:10:44: note:   cannot convert ‘<lambda closure object>main()::__lambda1{}’ (type ‘main()::__lambda1’) to type ‘boost::launch’
                                           });

如果我在这里缺少什么，请告诉我。

Answer 1

通过引用.then（）传递future可以修复gcc 4.8和clang上的编译问题。

对于Windows和gcc 4.7，我们还需要定义BOOST_RESULT_OF_USE_DECLTYPE 。 （根据Xeo的评论 ）。 对于gcc 4.8和clang，它似乎已经可用。

boost::future<std::string> f2 = f1.then([](boost::future<int>& f)->std::string {
                                                             ^