Question

我正在以这种方式捕获lambda表达式中的unique_ptr：

auto str = make_unique<string>("my string");
auto lambda = [ capturedStr = std::move(str) ] {
   cout << *capturedStr.get() << endl;
};
lambda();

在我尝试将capturedStr移动到另一个unique_ptr之前，它很有用。例如，以下内容无效：

auto str = make_unique<string>("my string");
auto lambda = [ capturedStr = std::move(str) ] {
    cout << *capturedStr.get() << endl;
    auto str2 = std::move(capturedStr); // <--- Not working, why?
};
lambda();

以下是编译器的输出：

.../test/main.cpp:11:14: error: call to implicitly-deleted copy
constructor of 'std::__1::unique_ptr<std::__1::basic_string<char>,
std::__1::default_delete<std::__1::basic_string<char> > >'
        auto str2 = std::move(capturedStr);
             ^      ~~~~~~~~~~~~~~~~~~~~~~ ../include/c++/v1/memory:2510:31: note: copy constructor is implicitly
deleted because 'unique_ptr<std::__1::basic_string<char>,
std::__1::default_delete<std::__1::basic_string<char> > >' has a
user-declared move constructor
    _LIBCPP_INLINE_VISIBILITY unique_ptr(unique_ptr&& __u) _NOEXCEPT
                              ^ 1 error generated.

为什么不能移动capturedStr？

Answer 1

默认情况下，lambda的operator ()为const，您无法从const对象移动。

如果要修改捕获的变量，请将其声明为mutable。

auto lambda = [ capturedStr = std::move(str) ] () mutable {
//                                             ^^^^^^^^^^
    cout << *capturedStr.get() << endl;
    auto str2 = std::move(capturedStr);
};

Answer 2

auto lambda = [ capturedStr = std::move(str) ] {
   cout << *capturedStr.get() << endl;
   auto str2 = std::move(capturedStr); // <--- Not working, why?
};

为了提供更多细节，编译器正在有效地进行这种转换：

class NameUpToCompiler
{
    unique_ptr<string> capturedStr;  // initialized from move assignment in lambda capture expression

    void operator()() const
    {
        cout << *capturedStr.get() << endl;
        auto str2 = std::move(capturedStr);  // move will alter member 'captureStr' but can't because of const member function.
    }
}

在lambda上使用mutable将从operator（）成员函数中删除const，从而允许更改成员。

Answer 3

更明确地提出建议：添加mutable：http://coliru.stacked-crooked.com/a/a19897451b82cbbb

#include <memory>

int main()
{
    std::unique_ptr<int> pi(new int(42));

    auto ll = [ capturedInt = std::move(pi) ] () mutable { };
}

捕获并移动c ++ 14 lambda表达式中的unique_ptr

3 个答案: