在C ++ 14 Lambda表达式中捕获并移动unique_ptr [英] Capture and move a unique_ptr in a c++14 lambda expression

问题描述

我正在通过lambda表达式捕获这种方式：

I am capturing a unique_ptr in a lambda expression this way:

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

效果很好，直到我尝试移动 capturedStr 到另一个unique_ptr。例如，以下操作不起作用：

It works great until I try to move capturedStr to another unique_ptr. For instance, the following is not working:

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 ？

推荐答案

lambda的运算符（）为 const 默认状态，并且您不能从 const 对象移动。

The operator () of a lambda is const by default, and you can't move from a const object.

将其声明为可变。

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

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