Question

In C++ is a return value guaranteed to be created before automatic variables in the function are destroyed? Notice Basket::get:

class Basket
{
public:
  // Gift is a struct containing safely copyable things like int or string
  Gift gift;
  // Used to protect access and changes to gift
  Mutex mutex;

  // Copy gift into present, while locked to be thread safe
  void put (const Gift & gift)
  {
    Lock lock(mutex);   // Constructor locks, destructor unlocks mutex
    this->gift = gift;  // Gift assignment operator
  }

  // Return a memberwise-copy of gift, tries to be thread safe (but is it?)
  Gift get ()
  {
    Lock lock(mutex);  // Constructor locks, destructor unlocks mutex
    return gift;       // Gift copy constructor
  }
};

I need Basket::get to perform its Gift copy constructor (of the temp object returned) prior to destruction of the lock object. Otherwise the gift object being returned can be corrupted by a simultaneous call to put.

My tests show the gift copy is indeed created before lock destruction, however, is it guaranteed? If not, I’ll need to create a second temporary inside the function, such as:

  Gift get ()
  {
    Gift result;
    {
      Lock lock(mutex);
      result = gift;
    }
    return result;
  }

Answer 1

Yes, the auto variable will remain in scope until after the return is finished. This is especially true if you are using a compiler that optimizes the return, eg:

Gift get() 
{ 
    Lock lock(mutex);
    return gift;
} 

Gift g = basket.get();

Which would be equivilent to this sequence:

Gift g;
Lock lock(mutex);
g = Gift(gift);
lock.~Lock();

May be optimized to act more like this:

void get(Gift &ret) 
{ 
    Lock lock(mutex);
    ret = gift;
} 

Gift g;
basket.get(g);

Which would be equivilent to this sequence:

Gift g;
Lock lock(mutex);
g = gift;
lock.~Lock();

In other words, a temporary can be removed during the return.