Question

我想使用单个分隔符将wstring拆分为vector<wstring>。此字符在头文件中定义为单个char。为了保持代码的清洁和可读性，我真的想在一行上做到这一点:) 我找不到要使用的谓词，所以我决定使用C ++ 11 lambda。

#include    <boost/algorithm/string/split.hpp>
#include    <vector>
#include    <string>

constexpr char separator = '.';     // This is how it's declared in some header file

int main()
{
    std::wstring text( L"This.is.a.test" );

    std::vector<std::wstring> result;
    // can't use is_any_of() unless i convert it to a wstring first.
    boost::algorithm::split( result, text, [](wchar_t ch) -> bool { return ch == (wchar_t) separator; });

    return 0;
}

不幸的是，这会导致编译错误（clang 3.3）：

clang++ -c -pipe -fPIC -g -std=c++11 -Wextra -Wall -fPIE -DQT_QML_DEBUG -DQT_DECLARATIVE_DEBUG -I/usr/include -I/usr/lib64/qt5/mkspecs/linux-clang -I../splittest -I. -o debug/main.o ../splittest/main.cpp
In file included from ../splittest/main.cpp:1:
In file included from /usr/include/boost/algorithm/string/split.hpp:16:
/usr/include/boost/algorithm/string/iter_find.hpp:148:13: error: non-type template argument refers to function 'failed' that does not have linkage
            BOOST_CONCEPT_ASSERT((
            ^~~~~~~~~~~~~~~~~~~~~~
/usr/include/boost/concept/assert.hpp:44:5: note: expanded from macro 'BOOST_CONCEPT_ASSERT'
    BOOST_CONCEPT_ASSERT_FN(void(*)ModelInParens)
    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/include/boost/concept/detail/general.hpp:70:6: note: expanded from macro 'BOOST_CONCEPT_ASSERT_FN'
    &::boost::concepts::requirement_<ModelFnPtr>::failed>    \
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/include/boost/algorithm/string/split.hpp:146:40: note: in instantiation of function template specialization 'boost::algorithm::iter_split<std::vector<std::basic_string<wchar_t>, std::allocator<std::basic_string<wchar_t> > >, std::basic_string<wchar_t>, boost::algorithm::detail::token_finderF<<lambda at ../splittest/main.cpp:13:44> > >' requested here
            return ::boost::algorithm::iter_split(
                                       ^
../splittest/main.cpp:13:23: note: in instantiation of function template specialization 'boost::algorithm::split<std::vector<std::basic_string<wchar_t>, std::allocator<std::basic_string<wchar_t> > >, std::basic_string<wchar_t>, <lambda at ../splittest/main.cpp:13:44> >' requested here
    boost::algorithm::split( result, text, [](wchar_t ch) -> bool { return ch == (wchar_t) separator; });
                      ^
/usr/include/boost/concept/detail/general.hpp:46:17: note: non-type template argument refers to function here
    static void failed() { ((Model*)0)->constraints(); }
                ^
1 error generated.

我做错了什么或者C ++ 11-lambdas不支持（完全？）在boost中？

还有另一种可读的单行解决方案吗？

我目前正在使用我在某个基础库中定义的自己的谓词is_char()，但我宁愿摆脱它。

我知道boost lambdas（尚未使用它们） - 但是它们真的应该用在C ++ 11代码中吗？

谢谢！

Answer 1

在肢体上，尝试在包含第一个boost头之前定义它（提防预编译头文件）：

#define BOOST_RESULT_OF_USE_DECLTYPE

或 - 反过来

#define BOOST_RESULT_OF_USE_TR1

我相信默认值已经改变。最近。对于特定的编译器，所以这很好地解释了它。

如何使用C ++ 11 lambda作为boost谓词？

1 个答案: