0x

When using placement new operator, destruction isn’t as transparent as usual..

Will the destructor be invoked?

#include <iostream>
using std::cout;
using std::endl;

struct A {
    A () { cout << "constructed." << endl; }
    ~A () { cout << "destructed." << endl; }
    int x; // so the class takes more than (the default) 1byte
};

int main () {
    // using char isn't as portable as we'd like
    void *buff = new char[sizeof(A)];
    new (buff) A; // placement new
    delete[] buff;
}

Hint: having used placement new, you are also responsible for (explicitly) destructing that object.

§ One Response to Destructors and placement new

• Sumant says:

  19/08/2009 at 23:06

  In such cases, the destructor of the class should be invoked explicitly. This is one of the rare cases where it is necessary to invoke a destructor manually. Here is how it can be done.

  void *buff = new char[sizeof(A)];
  A *a = new (buff) A;
  a->~A();
  delete[] buff;

  Reply