尝试换行函数返回值时,"<class name>不提供调用运算符"错误

Question

我正在尝试编写一个函数,它将一个仿函数作为参数,调用仿函数,然后返回包含在一个函数中的返回值boost::shared_ptr.

以下拒绝编译,我完全没有想法.我得到"std :: vector <std :: string>不提供调用操作符"(粗略地).我在Mac OS X上使用Clang 3.1.

template< typename T >
boost::shared_ptr< T > ReturnValueAsShared(
    boost::function< T() > func )
{
  return boost::make_shared< T >( func() );
}

这是我尝试使用它的上下文:

make_shared< packaged_task< boost::shared_ptr< std::vector< std::string > > > >(
   bind( ReturnValueAsShared< std::vector< std::string > >,
      bind( [a function that returns a std::vector< std::string >] ) ) );

编辑:这是一个完整的独立测试用例.这段代码无法使用相同的错误进行编译,对于我的生活,我看不出有什么问题:

#include <boost/make_shared.hpp>
#include <boost/shared_ptr.hpp>
#include <boost/function.hpp>
#include <boost/bind.hpp>

#include <string>
#include <vector>

std::vector< std::string > foo( std::string a )
{
  std::vector< std::string > vec;
  vec.push_back( a );
  return vec;
}

template< typename T >
boost::shared_ptr< T > ReturnValueAsShared(
    boost::function< T() > func )
{
  return boost::make_shared< T >( func() );
}

int main()
{
  auto f = boost::bind( ReturnValueAsShared< std::vector< std::string > >,
                        boost::bind( foo, std::string("a") ) );
  f();

} // main

这是错误输出:

In file included from testcase.cpp:3:
In file included from /usr/local/include/boost/function.hpp:64:
In file included from /usr/local/include/boost/preprocessor/iteration/detail/iter/forward1.hpp:47:
In file included from /usr/local/include/boost/function/detail/function_iterate.hpp:14:
In file included from /usr/local/include/boost/function/detail/maybe_include.hpp:13:
/usr/local/include/boost/function/function_template.hpp:132:18: error: type 'std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > >' does not provide a call operator
          return (*f)(BOOST_FUNCTION_ARGS);
                 ^~~~
/usr/local/include/boost/function/function_template.hpp:907:53: note: in instantiation of member function 'boost::detail::function::function_obj_invoker0<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > >, std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >::invoke' requested here
        { { &manager_type::manage }, &invoker_type::invoke };
                                                    ^
/usr/local/include/boost/function/function_template.hpp:722:13: note: in instantiation of function template specialization 'boost::function0<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >::assign_to<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >' requested here
      this->assign_to(f);
            ^
/usr/local/include/boost/function/function_template.hpp:1042:5: note: in instantiation of function template specialization 'boost::function0<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >::function0<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >' requested here
    base_type(f)
    ^
/usr/local/include/boost/bind/bind.hpp:243:43: note: in instantiation of function template specialization 'boost::function<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > ()>::function<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >' requested here
        return unwrapper<F>::unwrap(f, 0)(a[base_type::a1_]);
                                          ^
/usr/local/include/boost/bind/bind_template.hpp:20:27: note: in instantiation of function template specialization 'boost::_bi::list1<boost::_bi::bind_t<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > >, std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > (*)(std::basic_string<char>), boost::_bi::list1<boost::_bi::value<std::basic_string<char> > > > >::operator()<boost::shared_ptr<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >, boost::shared_ptr<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > > (*)(boost::function<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > ()>), boost::_bi::list0>' requested here
        BOOST_BIND_RETURN l_(type<result_type>(), f_, a, 0);
                          ^
testcase.cpp:27:4: note: in instantiation of member function 'boost::_bi::bind_t<boost::shared_ptr<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > >, boost::shared_ptr<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > > (*)(boost::function<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > ()>), boost::_bi::list1<boost::_bi::bind_t<std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > >, std::vector<std::basic_string<char>, std::allocator<std::basic_string<char> > > (*)(std::basic_string<char>), boost::_bi::list1<boost::_bi::value<std::basic_string<char> > > > > >::operator()' requested here
  f();
   ^
1 error generated.

这里有一些更多的线索.以下代码编译得很好,但这对我没有帮助,因为这不是我想要的代码:)

#include <boost/make_shared.hpp>
#include <boost/shared_ptr.hpp>
#include <boost/function.hpp>
#include <boost/bind.hpp>

#include <string>
#include <vector>

std::vector< std::string > foo()
{
  std::vector< std::string > vec;
  return vec;
}

template< typename T >
boost::shared_ptr< T > ReturnValueAsShared(
    boost::function< T() > func )
{
  return boost::make_shared< T >( func() );
}

int main()
{
  auto f = boost::bind( ReturnValueAsShared< std::vector< std::string > >,
                        foo );
  f();

} // main

Answer 1

某些结构（例如bind）返回中间“表达式”类型，您实际上并不希望在鼻子上捕获这些类型。在这种情况下，您不能通过 捕获类型auto，并且可能需要指定显式转换，否则就不存在唯一的、单个用户定义的转换链。在您的情况下，将绑定表达式的显式转换添加到function：

typedef std::vector<std::string> G;
auto f = boost::bind(ReturnValueAsShared<G>,
             static_cast<boost::function<G()>(boost::bind(foo, std::string("a")))
                    );

（这本身实际上对我不起作用，但如果您使用相应的std结构，它确实有效。）