在大型项目上，复制分配操作员内存泄漏

Question

I have been working on a big project for the past few months. Now I'm finally done with that project. But in my copy assignment operator I have a memory leak. Valgrind shows that it's the original value of data_ that leaks.

Here is my copy assignment operator code.

Value& Value::operator=(const Value& r)
{
  data_ = copy(r.data_);
  delete data_; 
  return *this;
}

Anyone that can help me with this problem? I would really appreciate it.

Answer 1

I believe you wanted to write this:

delete data_;     //here it makes sense: delete the current value
data_ = copy(r.data_);  //copy now

not this one:

data_ = copy(r.data_); //LEAK, as data_ is pointing to current value
delete data_;         //here you're deleting the copied one

Make sure that data_ always points to a valid memory — else delete it conditionally.

Answer 2

This doesn't make sense:

data_ = copy(r.data_);
delete data_; 

because if data_ was pointing to allocated memory and you overwrite it with

data_ = copy(r.data_);

and then delete the newly copied area, you have a memory leak (you can no longer refer to the allocated memory you originally had).

Big bonus for deleting the just-copied memory: if you ever were to actually use the _data pointer you'll get undefined behavior.

You probably meant to write

template <typename T>
Value<T>& Value<T>::operator=(const Value<T>& r)
{
  delete data_; // Free this object's memory
  data_ = copy(r.data_); // And now get a copy (hopefully a deep one) of the new memory
  return *this;
}

A small caveat: the above code even if fixed has no strong exception guarantee : if memory copy fails for whatever reason, you might end up with an inconsistent object (since data_ has already been deleted).

Answer 3

The problem is that data_ is deleted right after it is copied!

  data_ = copy(r.data_);
  delete data_;          <<< PROBLEM

The best solution may be to employ copy-and-swap idiom ( What is the copy-and-swap idiom? ).

template <typename T>
Value<T>& Value<T>::operator=(const Value<T> rhs)  // NOTE: pass by value
{
  swap(data_, rhs.data_);  // either std::swap or a custom swap,
                           // hard to say without knowing the type of data_
  return *this;
}

Another alternative, a direct enhancement to OP's code, may be the following.

template <typename T>
Value<T>& Value<T>::operator=(const Value<T>& r)
{
  // 1) Allocate new data. If, for some reason, the allocation throws,
  // the original data_ stays intact. This offers better
  // exception safety.
  ... new_data = ...;

  // 2) Copy r.data to new_data (note: deep copy desired)
  new_data = copy(r.data_); 

  // 3) Destroy original data_
  delete data_;

  // 4) Point data_ to new_data
  data_ = new_data;

  return *this;
}