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c++ - 如何使用 C++11 lambda 作为 boost 谓词?

转载 作者:太空宇宙 更新时间:2023-11-04 13:54:57 25 4
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我想拆分一个wstring进入 vector<wstring>使用单个分隔符。此字符在头文件中定义为单个 char .为了保持代码的整洁和可读性,我真的很想在一行中完成此操作 :)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.

是我做错了什么还是 C++11-lambda 不(完全?)在 boost 中得到支持?

是否有另一种可读的单行解决方案?

我目前正在使用自己的谓词 is_char()我在一些基础库中定义了它,但我宁愿摆脱它​​。

我知道 boost lambda(还没有用过)——但它们真的应该用在 C++11 代码中吗?

谢谢!

最佳答案

冒险尝试在包含第一个 boost header 之前定义它(注意预编译 header ):

#define BOOST_RESULT_OF_USE_DECLTYPE

或者 - 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,所以这可以很好地解释它。

关于c++ - 如何使用 C++11 lambda 作为 boost 谓词?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21937236/

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