Why is the dtor being called (using annoymous/lambda func)

问题

I am trying to mimic a finally like effect. So i thought i should run a quick dirty test.

The idea was to use Most Important const to stop destruction and to put the finally block in a lambda. However apparently i did something wrong and its being called at the end of MyFinally(). How do i solve this problem?

#include <cassert>
template<typename T>
class D{
    T fn;
public:
    D(T v):fn(v){}
    ~D(){fn();}
};

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

int d;
class A{
    int a;
public:
    void start(){
        int a=1;
        auto v = MyFinally([&]{a=2;});
        try{
            assert(a==1);
            //do stuff
        }
        catch(int){
            //do stuff
        }
    }
};
int main() {
    A a;
    a.start();
}

My Solution code (Note: You can not have two finally in the same block. as expect. But still kind of dirty)

#include <cassert>
template<typename T>
class D{
    T fn; bool exec;
public:
    D(T v):fn(v),exec(true){}
    //D(D const&)=delete //VS doesnt support this yet and i didnt feel like writing virtual=0
    D(D &&d):fn(move(d.fn)), exec(d.exec) {
      d.exec = false;
    }

    ~D(){if(exec) fn();}
};
template<typename T>
D<T> MyFinally(T t) { return D<T>(t); }


#define FINALLY(v) auto OnlyOneFinallyPlz = MyFinally(v)

int d;
class A{
public:
    int a;
    void start(){
        a=1;
        //auto v = MyFinally([&]{a=2;});
        FINALLY([&]{a=2;});
        try{
            assert(a==1);
            //do stuff
        }
        catch(int){
            FINALLY([&]{a=3;}); //ok, inside another scope
            try{
                assert(a==1);
                //do other stuff
            }
            catch(int){
                //do other stuff
            }
        }
    }
};
void main() {
    A a;
    a.start();
    assert(a.a==2);
}

Funny enough, if you remove the & in MyFinally in the original code it works -_-.

回答1:

// WRONG! returning a reference to a temporary that will be
// destroyed at the end of the function!
template<typename T>
const D<T>& MyFinally(T t) { return D<T>(t); }

You can fix it my introducing a move constructor

template<typename T>
class D{
    T fn;
    bool exec;

public:
    D(T v):fn(move(v)),exec(true){}

    D(D &&d):fn(move(d.fn)), exec(d.exec) {
      d.exec = false;
    }

    ~D(){if(exec) fn();}
};

And then you can rewrite your toy

template<typename T>
D<T> MyFinally(T t) { return D<T>(move(t)); }

Hope it helps. No "const reference" trick is needed when you work with auto. See here for how to do it in C++03 with const references.

回答2:

Your code and Sutter's are not equivalent. His function returns a value, yours returns a reference to an object that will be destroyed when the function exits. The const reference in the calling code does not maintain the lifetime of that object.

回答3:

The problem stems from the use of a function maker, as demonstrated by Johannes.

I would argue that you could avoid the issue by using another C++0x facility, namely std::function.

class Defer
{
public:
  typedef std::function<void()> Executor;

  Defer(): _executor(DoNothing) {}

  Defer(Executor e): _executor(e) {}
  ~Defer() { _executor(); }

  Defer(Defer&& rhs): _executor(rhs._executor) {
    rhs._executor = DoNothing;
  }

  Defer& operator=(Defer rhs) {
    std::swap(_executor, rhs._executor);
    return *this;
  }

  Defer(Defer const&) = delete;

private:
  static void DoNothing() {}
  Executor _executor;
};

Then, you can use it as simply:

void A::start() {
  a = 1;
  Defer const defer([&]() { a = 2; });

  try { assert(a == 1); /**/ } catch(...) { /**/ }
}

回答4:

Well the problem is explained by others, so I will suggest a fix, exactly in the same way Herb Sutter has written his code (your code is not same as his, by the way):

First, don't return by const reference:

template<typename T>
D<T> MyFinally(T t) 
{
   D<T> local(t); //create a local variable
   return local; 
}

Then write this at call site:

const auto & v = MyFinally([&]{a=2;}); //store by const reference

This became exactly like Herb Sutter's code.

Demo : http://www.ideone.com/uSkhP

Now the destructor is called just before exiting the start() function.

---

A different implementation which doesn't use auto keyword anymore:

struct base { virtual ~base(){} };

template<typename TLambda>
struct exec : base 
{
   TLambda lambda;
   exec(TLambda l) : lambda(l){}
   ~exec() { lambda(); }
};

class lambda{
    base *pbase;
public:
    template<typename TLambda>
    lambda(TLambda l): pbase(new exec<TLambda>(l)){}
    ~lambda() { delete pbase; }
};

And use it as:

lambda finally = [&]{a=2;  std::cout << "finally executed" << std::endl; }; 

Looks interesting?

Complete demo : http://www.ideone.com/DYqrh

回答5:

You could return a shared_ptr:

template<typename T>
std::shared_ptr<D<T>> MyFinally(T t) {
    return std::shared_ptr<D<T>>(new D<T>(t));
}

来源：https://stackoverflow.com/questions/6167017/why-is-the-dtor-being-called-using-annoymous-lambda-func