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利用std::shared_ptr 回收不同型別物件的記憶體

轉自:  http://stackoverflow.com/questions/5913396/why-do-stdshared-ptrvoid-work

利用std::function 進行deleter 函式的型別擦除。

===========================================ASK===========================================

I found some code using std::shared_ptr to perform arbitrary cleanup at shutdown. At first I thought this code could not possibly work, but then I tried the following:

#include <memory>
#include <iostream>
#include <vector>

class test {
public:
  test() {
    std::cout << "Test created" << std::endl;
  }
  ~test() {
    std::cout << "Test destroyed" << std::endl;
  }
};

int main() {
  std::cout << "At begin of main.\ncreating std::vector<std::shared_ptr<void>>" 
            << std::endl;
  std::vector<std::shared_ptr<void>> v;
  {
    std::cout << "Creating test" << std::endl;
    v.push_back( std::shared_ptr<test>( new test() ) );
    std::cout << "Leaving scope" << std::endl;
  }
  std::cout << "Leaving main" << std::endl;
  return 0;
}

This program gives the output:

At begin of main.
creating std::vector<std::shared_ptr<void>>
Creating test
Test created
Leaving scope
Leaving main
Test destroyed

I have some ideas on why this might work, that have to do with the internals of std::shared_ptrs as implemented for G++. Since these objects wrap the internal pointer together with the counter the cast from std::shared_ptr<test>

 to std::shared_ptr<void> is probably not hindering the call of the destructor. Is this assumption correct?

And of course the much more important question: Is this guaranteed to work by the standard, or might further changes to the internals of std::shared_ptr, other implementations actually break this code?

===========================================ANSWER=======================================

The trick is that std::shared_ptr performs type erasure. Basically, when a new shared_ptr is created it will store internally a deleter function (which can be given as argument to the constructor but if not present defaults to calling delete). When the shared_ptr is destroyed, it calls that stored function and that will call the deleter.

A simple sketch of the type erasure that is going on simplified with std::function, and avoiding all reference counting and other issues can be seen here:

template <typename T>
void delete_deleter( void * p ) {
   delete static_cast<T*>(p);
}

template <typename T>
class my_unique_ptr {
  std::function< void (void*) > deleter;
  T * p;
  template <typename U>
  my_unique_ptr( U * p, std::function< void(void*) > deleter = &delete_deleter<U> ) 
     : p(p), deleter(deleter) 
  {}
  ~my_unique_ptr() {
     deleter( p );   
  }
};

int main() {
   my_unique_ptr<void> p( new double ); // deleter == &delete_deleter<double>
}
// ~my_unique_ptr calls delete_deleter<double>(p)

When a shared_ptr is copied (or default constructed) from another the deleter is passed around, so that when you construct a shared_ptr<T> from a shared_ptr<U> the information on what destructor to call is also passed around in the deleter.