Explicit call to the destructor

Question

Can i still access an object after making an explicit call to its destructor?

for example,

class A{
        public:
            A(){
                cout<<"Constructor\n";
                x=5;
            }
            ~A(){
                cout<<"Destructor\n";
            }
        int x;
    };


int main() {
    ios_base::sync_with_stdio(false);

    A obj;
    obj.~A();
    obj.x=4;
    cout<<obj.x<<endl;

    return 0;
}

gives output

Constructor Destructor 4 Destructor

How am I able to access obj.x even after calling the destructor? If the explicit call does not destroy the object, then what does it do?

What code do you think was exeuted when you called the dtor? What effect did it have? — Peter - Reinstate Monica, Apr 19 '15 at 20:44
Also define "destroy", with particular emphasis on data integrity and memory management ;-). — Peter - Reinstate Monica, Apr 19 '15 at 20:47
duplicate of [this question](http://stackoverflow.com/questions/7396059/explicit-call-to-destructor-is-not-destroying-my-object-why?rq=1) — Titus, Apr 19 '15 at 20:58

Arkanosis · Accepted Answer · 2015-04-19T20:51:00.720

0

It calls the destructor, but it doesn't deallocate the memory. Memory will be deallocated at the end of the function.

BTW, if you don't understand what happens, you probably shouldn't do that: the destructor will be called a second time at the end of the scope, which may cause issues if it can't be called twice (which is often the case).

edited Apr 19 '15 at 20:51

answered Apr 19 '15 at 20:42

Arkanosis

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Even "deallocation" will (most often) not prevent later access to that memory, as little as a dtor call "destroys" the object. – Peter - Reinstate Monica Apr 19 '15 at 20:44
You shouldn't operate a light switch? – Peter - Reinstate Monica Apr 19 '15 at 20:45
if the object still exists ,then what does the destructor do ? – grv95 Apr 19 '15 at 20:52
It runs the code in the “destructor” method. Destroying an object is a two-step process: call its destructor, then free the memory. Well, in general, because if you don't allocate memory to begin with (using placement new, for example), then there's no memory to free at the end. – Arkanosis Apr 19 '15 at 20:55
In your case, the object is stack-allocated (you haven't used `new`), so it will be destructed (ie. the destructor will be called) at the end of the scope (the next closing bracket) and the memory will be freed at the end of the function. – Arkanosis Apr 19 '15 at 20:56
@grv95 Define "exists". – Peter - Reinstate Monica Apr 19 '15 at 21:45

Explicit call to the destructor

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