Thread: >> Destructor Question

Originally Posted by G4B3

I was playing around a little, and found out that this not only compiles, but the last call produces the result "5". I though a destructor destroys the class, so the program shouldn't be able to access the variable. Or is this accessing something out of bounds?

I uh, have never considered implementing a miscarriage instead of a constructor. Your program has undefined behaviour for two reasons:

• The explicit destructor call followed by the construction of the object on the stack means that the destructor will be called on a destroyed object (i.e., it will be called a second time, if that even makes sense).
• The member function printVal() is called on a destroyed object.

There is only one case where calling the destructor explicitly is valid: when the instance was created with placement new.

Code:

#include <iostream>
using namespace std;

class Test {
    int m_Val;

    public:

    void printVal() { cout << m_Val << endl; }

    ~Test() { cout << "~Test\n"; }
    Test(int val) { m_Val = val; }
};

int main()
{
    char buffer[sizeof(Test)];

    //put a Test object into the buffer of bytes
    Test* p_test = new (buffer) Test(42);
    p_test->printVal();

    //now call the destructor, which wouldn't be called otherwise
    p_test->~Test();
    return 0;
}

Originally Posted by G4B3

Anon: I'm not sure I understand what's going on there...I've never seen new used in such a way. Do you think you could give a more detailed explanation?

You might want to search the Web for "placement new".

In particular, the article from C++ FAQ Lite would be a good starting point. The Wikipedia article also looks trustworthy. Danny Kalev's devx article is too short, but if you search for array placement new instead, his relevant devx article would be a good read.