如何使用 C++11 λ 作为提升谓词

How to use C++11 lambda as boost predicate?

本文关键字:谓词 何使用 C++11      更新时间:2023-10-16

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。为了保持代码干净易读,我真的很想在一行:)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.

我做错了什么还是 C++11-lambda 在 boost 中不受(完全?

有没有另一种可读的单行解决方案?

我目前正在使用我在某些基本库中定义的自己的谓词is_char(),但我宁愿摆脱它。

我知道 boost lambdas(还没有使用过它们) - 但它们真的应该在 C++11 代码中使用吗?

谢谢!

在包含第一个 boost 标头之前,请尝试定义以下内容(当心预编译标头):

#define BOOST_RESULT_OF_USE_DECLTYPE

或者 - 相反

#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,这可以很好地解释它。