<分区>
在 C++14 中,lambda 表达式可以通过使用捕获初始值设定项从变量中 move 来捕获变量。但是,这使得生成的闭包对象不可复制。如果我有一个接受 std::function
参数(我无法更改)的现有函数,我无法传递闭包对象,因为 std::function
的构造函数要求给定的仿函数是 CopyConstructible
。
#include <iostream>
#include <memory>
void doit(std::function<void()> f) {
f();
}
int main()
{
std::unique_ptr<int> p(new int(5));
doit([p = std::move(p)] () { std::cout << *p << std::endl; });
}
这会产生以下错误:
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/functional:1911:10: error:
call to implicitly-deleted copy constructor of '<lambda at test.cpp:10:7>'
new _Functor(*__source._M_access<_Functor*>());
^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/functional:1946:8: note: in
instantiation of member function 'std::_Function_base::_Base_manager<<lambda at test.cpp:10:7>
>::_M_clone' requested here
_M_clone(__dest, __source, _Local_storage());
^
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/functional:2457:33: note: in
instantiation of member function 'std::_Function_base::_Base_manager<<lambda at test.cpp:10:7>
>::_M_manager' requested here
_M_manager = &_My_handler::_M_manager;
^
test.cpp:10:7: note: in instantiation of function template specialization 'std::function<void
()>::function<<lambda at test.cpp:10:7>, void>' requested here
doit([p = std::move(p)] () { std::cout << *p << std::endl; });
^
test.cpp:10:8: note: copy constructor of '' is implicitly deleted because field '' has a deleted
copy constructor
doit([p = std::move(p)] () { std::cout << *p << std::endl; });
^
/usr/bin/../lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/bits/unique_ptr.h:273:7: note:
'unique_ptr' has been explicitly marked deleted here
unique_ptr(const unique_ptr&) = delete;
^
是否有合理的解决方法?
使用 Ubuntu clang 版本 3.5-1~exp1 (trunk) 进行测试