如何捕获一个unique_ptr到lambda表达式？ [英] How to capture a unique_ptr into a lambda expression?

问题描述

我尝试过以下操作：

  std :: function< void（）> getAction（std :: unique_ptr< MyClass>&& psomething）{
 //调用者赋予psomething的所有权
 return [psomething]（）{
 psomething-> do_some_thing ; 
 // psomething预计将在此点之后释放
}; 
} 
  

但它不编译。任何想法？

更新：

根据建议，需要一些新的语法来明确指定我们需要转移对lambda的所有权，我现在考虑以下语法：

  std :: function< void（）> getAction（std :: unique_ptr< MyClass> psomething）{
 //调用者赋予psomething的所有权
 return [auto psomething = move（psomething）]（）{
 psomething-> do_some_thing （）; 
 // psomething预计将在此点之后释放
}; 
} 
  

这将是一个很好的候选人吗？

UPDATE 1：

我将展示我的实现 move 和 copy 如下：

  template< typename T> 
 T copy（const T& t）{
 return t; 
} 
 
 // process lvalue references 
 template< typename T> 
 T move（T& t）{
 return std :: move（t）; 
} 
 
 class A {/*...*/}; 
 
 void test（A& a）; 
 
 int main（int，char **）{
 A a; 
 test（copy（a））; // OK，copied 
 test（move（a））; // OK，moved 
 test（A（））; // OK，临时对象
 test（copy（A（）））; // OK，正在复制临时对象
 //您可以通过让副本接受T& 
 // test（move（A（）））;你不应该移动一个临时对象
 //这是不好的有一个右值版本的移动。 
 // test（a）;禁止，你必须说天气你要复制或移动
 //从一个左值引用。 
} 
  

解决方案

code> unique_ptr 中。事实上，如果你想永久捕获lambda中的任何东西，它必须可复制;

这可能被认为是C ++ 11中的一个缺陷，但是你需要一些语法来明确地说你想移动 unique_ptr 值插入lambda。 C ++ 11规范是非常谨慎的措词，以防止对命名变量的隐式移动;这就是为什么 std :: move 存在，这是一个好的东西。

要做你想要的，需要使用 std :: bind （这将是半卷积的，需要一个短序列绑定）或只返回一个常规的旧对象。

此外，从不 unique_ptr c>&&&& ，除非你真正写它的移动构造函数。只是把它的价值;用户可以通过值提供它的唯一方法是使用 std :: move 。事实上，除非你正在编写移动构造函数/赋值运算符（或实现转发函数），否则不要采取任何&&& / p>

I have tried the following:

std::function<void ()> getAction(std::unique_ptr<MyClass> &&psomething){
    //The caller given ownership of psomething
    return [psomething](){ 
        psomething->do_some_thing();
        //psomething is expected to be released after this point
    };
}

But it does not compile. Any ideas?

UPDATE:

AS suggested, some new syntax is required to explicitly specify we need to transfer the ownership to the lambda, I am now thinking about the following syntax:

std::function<void ()> getAction(std::unique_ptr<MyClass> psomething){
    //The caller given ownership of psomething
    return [auto psomething=move(psomething)](){ 
        psomething->do_some_thing();
        //psomething is expected to be released after this point
    };
}

Would it be a good candidate?

UPDATE 1:

I will show my implementation of move and copy as following:

template<typename T>
T copy(const T &t) {
    return t;
}

//process lvalue references
template<typename T>
T move(T &t) {
    return std::move(t);
}

class A{/*...*/};

void test(A &&a);

int main(int, char **){
    A a;
    test(copy(a));    //OK, copied
    test(move(a));    //OK, moved
    test(A());        //OK, temporary object
    test(copy(A()));  //OK, copying temporary object
    //You can disable this behavior by letting copy accepts T &  
    //test(move(A())); You should never move a temporary object
    //It is not good to have a rvalue version of move.
    //test(a); forbidden, you have to say weather you want to copy or move
    //from a lvalue reference.
}

解决方案

You cannot permanently capture a unique_ptr in a lambda. Indeed, if you want to permanently capture anything in a lambda, it must be copyable; merely movable is insufficient.

This could be considered a defect in C++11, but you would need some syntax to explicitly say that you wanted to move the unique_ptr value into the lambda. The C++11 specification is very carefully worded to prevent implicit moves on named variables; that's why std::move exists, and this is a good thing.

To do what you want will require either using std::bind (which would be semi-convoluted, requiring a short sequence of binds) or just returning a regular old object.

Also, never take unique_ptr by &&, unless you are actually writing its move constructor. Just take it by value; the only way a user can provide it by value is with a std::move. Indeed, it's generally a good idea to never take anything by &&, unless you're writing the move constructor/assignment operator (or implementing a forwarding function).

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