I was reading this question Does calling a destructor explicitly destroy an object completely? where this situation comes up in code.
Object* aWidget = new Widget(); //allocate and construct
aWidget->~Object(); //destroy and DON'T deallocate
From the answers, I undrestand that the memory region is in fact not deallocated in this situation. My question is (more out of curiosity than anything):
How can I delete the memory pointed to by aWidget after the two lines of code above have executed? I would assume calling delete aWidget; would fail because it would try to run the destructor on an already-destructed object. Could you call free(aWidget) or something like that instead to just target the memory?
free would, factually speaking, be my best guess. However I don't think you can do anything without invoking UB. How did you arrive at a requirement to invoke the destructor like that?
Calling free on an object allocated with new is undefined behavior.
I suggest you keep it simple, and call delete.
However, if you want to do this, you can, in some cases, call delete even if you previously called the destructor explicitly. If you call it explicitly, you can view it as a function. The memory isn't freed, so I'm guessing setting member pointers to NULL after you destroy them would be enough to prevent you from running into any trouble. (because calling delete on a NULL pointer is a no-op).
For example, the following should be ok:
class A
{
public:
int * x;
A()
{
x = new int[10];
}
~A()
{
delete[] x;
x = NULL;
}
};
int main()
{
A* a = new A;
a->~A();
delete a;
return 0;
}
Related
I studding C++ concept. I'm confused in constructor and Destractor concept. Destructor will be invoked implicitly by the compiler upon exit from the program. But in my program, only constructor called.
#include <iostream>
using namespace std;
class A
{
int i;
public:
A()
{
cout<<"Constructor call"<<endl;
}
~A()
{
cout<<"Destructtor call"<<endl;
}
};
int main()
{
A *a = new A();
}
Output:
Constructor call
So, I have question : Why destructor not call implicit by the compiler upon exit program?
Why destructor not call implicit by the compiler upon exit program?
Because dynamically allocated objects are not destroyed automatically. That is how the language has been specified. Tracking the destruction of dynamic objects would require runtime / memory overhead.
Solution: Do not leak dynamically allocated objects.
There are 2 "types" of variable lifetime in C++. Dynamic and automatic.
Automatic is something like :
void foo()
{
int i; // automatic storage
} // destroys i
Here, i will be destructed the moment it leaves scope (after foo returns). You can check this for yourself by making your object have an automatic lifetime:
int main()
{
A a; // calls A's constructor and allocates memory.
} //calls A's destructor and frees the memory used.
You'll see that both the constructor and the destructor will get called.
The second form is dynamically allocated memory. You can dynamically allocate memory by using new (like you did) and free that memory by using delete. The compiler won't do this for you.
This means that, to destruct your object you have to call delete explicitly yourself:
int main()
{
A* a = new A();
delete a; // calls A's destructor and frees the memory used.
}
If you do not call delete (like your code) then the moment the program leaves main the pointer a is destroyed and now we have a piece of memory that nothing can reach and thus nobody can clean up (with delete) which means you're leaking memory.
However, modern operating systems reclaim all memory used by the program automatically the moment the program ends so at this point it won't matter too much. This does mean your destructor won't be called as you've just witnessed.
Dynamically allocated memory allows you to do some nifty tricks like controlling the lifetime of your objects to the point where you want to destroy them explicitly yourself with delete. The problem with using new and delete is that it's so easy to forget a single delete and your program will already leak memory, which is why it's advised to stay away from this way of allocation.
If for some reason you absolutely need dynamic lifetime then use a smart pointer like std::unique_ptr that will call new and delete for you the moment it goes out of scope.
You create the object dynamically with new. The destructor will only be called when the objected is deleted with delete.
To prevent memory leaks, you could/should use smart pointers like unique_ptr.
In any case, the memory itself will of course be released when the process ends.
The destructor is never called in your code because the object is never destroyed.
You dynamically allocate the A object and you never deallocate it. Add delete a; into main() and you'll see the destructor in action:
int main()
{
A *a = new A();
delete a;
return 0;
}
In your main function, you create a pointer variable that you must delete it by delete a at end of main function.
Because you have a memory leak. You dynamically created an object with new, promising that you would manage the lifetime of the object and clean it up later with delete. Then you broke that promise.
If you create an object in the usual fashion:
A a;
then it will be destroyed automatically for you.
#include <iostream>
struct ABC{
int A;
ABC(int i = 1) : A(i) {}
~ABC() {
std::cout << A << std::endl;
}
void destruct() {
delete this;
}
};
int main() {
ABC A1(2);
A1.destruct();
return 0;
}
Output:
2
2
I have this code in which I'm trying to manually delete structure variable. Doing so, I realized the destructor gets called twice here. Why is this happening? Why isn't it getting deleted when destruct() is called?
The delete this call causes undefined behaviour, which means that anything at all can happen.
delete may only be used for objects created by new.
For automatic objects with a non-trivial destructor (e.g. your A1), it is not possible to "destroy them early" unless you also create another ABC in the same place before the scope ends. In other words you can't "turn off" the destruction process that occurs when the scope ends.
Why isn't [my object] getting deleted when destruct() is called?
When an object gets destructed in C++, it does not mean that the object disappears. It means that the clean-up code from the destructor gets executed, and that any access to the object's members from that point on is invalid.
When you call destruct() you try to free object's memory by calling delete. This is undefined behavior in itself, because you have not allocated the object with new. This call causes the first printout.
However, since your object is in automatic memory, C++ is required to call its destructor when the object gets out of scope. This is the call that causes the second printout.
Note: You can fix your code by allocating A1 in dynamic memory:
int main() {
ABC *A1 = new ABC(2);
A1->destruct();
return 0;
}
Now you get a single printout (demo). However, the practice of hiding delete in a member function is questionable.
This is RAII working plus you committing suicide on an object. Calling the destructor on an object is almost always wrong! And calling the destructor twice is always wrong, as it invokes undefined behaviour.
You have to understand that C++ is handling memory for you, if you just let it:
struct Foo{};
int main() {
Foo f; // automatic storage, gets destroyed
// when object gets out of scope
Foo* g = new Foo(); // heap allocated
delete g; // only here you have to delete
}
Just remember: Do not delete anything that you did not create via new (thanks to Mike Vine for the comment). And do not use (naked) heap allocation unless you need to.
Two points to consider here :-
1) Destructor for stack objects will always be called when they go out of scope. So no need to worry for their deallocation.
2) You cannot & should not use delete on the object allocated on stack. In general, you should not use delete this as long as you are not sure that this will be executed only as a consequence of deleting heap objects and after that you are not referring to that object.
I have the following code and I get stack overflow error can anyone please explain me What's wrong here. from my understanding this pointer points to current object so why I cant delete it in a destructor;
class Object
{
private:
static int objCount;
public:
int getCount()
{
int i =10;
return i++;
}
Object()
{
cout<< "Obj Created = "<<++objCount<<endl;
cout <<endl<<this->getCount()<<endl;
}
~Object()
{
cout<<"Destructor Called\n"<<"Deleted Obj="<<objCount--<<endl;
delete this;
}
};
int Object::objCount = 0;
int _tmain(int argc, _TCHAR* argv[])
{
{
Object obj1;
}
{
Object *obj2 = new Object();
}
getchar();
return 0;
}
You're doing delete this; in your class's destructor.
Well, delete calls the class's destructor.
You're doing delete this; in your class's destructor.
...
<!<!<!Stack Overflow!>!>!>
(Sorry guys I feel obliged to include this... this might probably spoil it sorrrryyyy!
Moral of the boring story, don't do delete this; on your destructor (or don't do it at all!)
Do [1]
Object *obj = new Object();
delete obj;
or much better, just
Object obj;
[1]#kfsone's answer more accurately points out that the stack overflow was actually triggered by obj1's destructor.
'delete this' never makes sense. Either you're causing an infinite recursion, as here, or you're deleting an object while it is still in use by somebody else. Just remove it. The object is already being deleted: that's why you're in the destructor.
The crash you are having is because of the following statement:
{
Object obj1;
}
This allocates an instance of "Object" on the stack. The scope you created it in ends, so the object goes out of scope, so the destructor (Object::~Object) is invoked.
{
Object obj1;
// automatic
obj1.~Object();
}
This means that the next instruction the application will encounter is
delete this;
There are two problems right here:
delete calls the object's destructor, so your destructor indirectly calls your destructor which indirectly calls your destructor which ...
After the destructor call, delete attempts to return the object's memory to the place where new obtains it from.
By contrast
{
Object *obj2 = new Object();
}
This creates a stack variable, obj2 which is a pointer. It allocates memory on the heap to store an instance of Object, calls it's default constructor, and stores the address of the new instance in obj2.
Then obj2 goes out of scope and nothing happens. The Object is not released, nor is it's destructor called: C++ does not have automatic garbage collection and does not do anything special when a pointer goes out of scope - it certainly doesn't release the memory.
This is a memory leak.
Rule of thumb: delete calls should be matched with new calls, delete [] with new []. In particular, try to keep new and delete in matching zones of authority. The following is an example of mismatched ownership/authority/responsibility:
auto* x = xFactory();
delete x;
Likewise
auto* y = new Object;
y->makeItStop();
Instead you should prefer
// If you require a function call to allocate it, match a function to release it.
auto* x = xFactory();
xTerminate(x); // ok, I just chose the name for humor value, Dr Who fan.
// If you have to allocate it yourself, you should be responsible for releasing it.
auto* y = new Object;
delete y;
C++ has container classes that will manage object lifetime of pointers for you, see std::shared_ptr, std::unique_ptr.
There are two issues here:
You are using delete, which is generally a code smell
You are using delete this, which has several issues
Guideline: You should not use new and delete.
Rationale:
using delete explicitly instead of relying on smart pointers (and automatic cleanup in general) is brittle, not only is the ownership of a raw pointer unclear (are you sure you should be deleting it ?) but it is also unclear whether you actually call delete on every single codepath that needs it, especially in the presence of exceptions => do your sanity (and that of your fellows) a favor, don't use it.
using new is also error-prone. First of all, are you sure you need to allocate memory on the heap ? C++ allows to allocate on the stack and the C++ Standard Library has containers (vector, map, ...) so the actual instances where dynamic allocation is necessary are few and far between. Furthermore, as mentioned, if you ever reach for dynamic allocation you should be using smart pointers; in order to avoid subtle order of execution issues it is recommend you use factory functions: make_shared and make_unique (1) to build said smart pointers.
(1) make_unique is not available in C++11, only in C++14, it can trivially be implemented though (using new, of course :p)
Guideline: You shall not use delete this.
Rationale:
Using delete this means, quite literally, sawing off the branch you are sitting on.
The argument to delete should always be a dynamically allocated pointer; therefore should you inadvertently allocate an instance of the object on the stack you are most likely to crash the program.
The execution of the method continues past this statement, for example destructors of local objects will be executed. This is like walking on the ghost of the object, don't look down!
Should a method containing this statement throw an exception or report an error, it is difficult to appraise whether the object was successfully destroyed or not; and trying again is not an option.
I have seen several example of usage, but none that could not have used a traditional alternative instead.
I'm wonder to know is it possible to delete an object through destructor method?
Constructor and destructor of my class:
class cal
{
public:
cal()
{
days = 0;
day = 1;
month = 1;
year = 1300;
leap = true;
};
~cal()
{
delete this;
}
}*calendar = new cal;
How can I delete this pointer through class?
P.S
I forgot to write the following code
cal *calandar = new cal[];
I want to use it in heap not stack
I want to use this class (objects) frequently (many of that object) imagine how many time should I write delete and it make hard understanding, troubleshooting and tracing codes I want them to been destroyed automatically (in heap)
I used following code in my class when I exec "delete[] calendar" it reduce my rams occupied (amount of ram that is used) does it work properly (destroy all objects) by exiting program? Because I use GNU/Linus and it destructs all objects with or without that lines I'm concern about leaking in windows
void DisposeObject() { delete this; }
No. By the time the destructor is called, the object is already being destroyed, so delete this is not valid.
The behaviour is undefined, but most likely it will call the destructor recursively, leading to a stack overflow.
You can write the code like this:
class cal
{
public:
cal()
{
};
~cal()
{
}
void DisposeObject()
{
delete this;
}
}
And it will call your destructor.
You should not call delete this from your destructor since it will call the destructor recursively, possibly leading to a stack overflow. Anyway, the code you wrote suffers from undefined behavior.
Refer to this question for in-depth discussion: Is delete this allowed?
I`m wonder to know is it possible to delete an object through destructor method
Even supposing this was valid - The problem is that nothing would ever call the destructor, so this would never have an effect. The destructor is called when the calling code deletes the object, which in turn can delete any internal objects as needed.
In your case, it doesn't look like you need to do anything in your destructor, however, as you don't have any resources being allocated that need to be explicitly deleted (at least in what you're showing).
in the following code:
class x
{
private:
someRef& m_ref;
public:
x(someRef& someRef):m_ref(someRef)
{
}
do I need to do:
~x()
{
delete m_ref;
}
which by the way doesnt work without getting the pointer...
basically I'm asking: Do I need to call a destructor on a reference member?
No.
You only need to delete an object if you own it. If you were passed a reference, it means that someone else owns it, thus it's unnecessary and thankfully the language prevents it.
I don't think one actually strictly speaking ever deletes even pointers. What you delete are dynamically allocated objects (or arrays of objects) that the pointer is a handle for. If the object originates from a call to new and it is the responsibility of this class to clean up after this object, then you call delete.
It is technically possible that a reference might be referring to a dynamically allocated object:
int main()
{
//in principle a reference can also refer to a dynamically allocated object
x var(*new someRef);
}
//and if that is the intended usage:
x::~x()
{
delete &m_ref;
}
However, this would be incredibly bad style. By convention, the "owning" handle of a dynamically allocated object should not be a reference.
No. You can only delete pointers, not references, and even then you must only delete objects that you allocated using the new operator. And then you must be sure to delete them only once. Here is the case in which you would need to use delete in your destructor:
class x
{
private:
someObj* m_ptr;
public:
x():m_ptr(new someObj())
{
}
~x()
{
delete m_ptr;
}
But in general it's best to avoid even this and use smart pointers instead.
I want to clarify some misconceptions you seem to have that are beyond the intent of your question:
When a class's destructor is called all of it's members' destructors get called as well.
Calling delete is not the same as calling the destructor. delete explicitly calls the destructor and also calls operator delete at the objects location, it is a 2 part thing.
For a small bit of extra clarification I want to offer the following:
int *pi = new int;
//int& ir = pi; // can't do
// this a reference to the pointer but it is an error
// because or the type difference int& vs int* and
// static_cast won't help. reinterpret_cast would allow
// the assignment to take place but not help the 'delete ir'
int& ir = *pi; // this is OK - a reference to what the pointer points to.
// In other words, the the address of the int on the heap.
//delete ir; // can't do, it is a reference and you can't delete non-pointers.
delete &ir; // this works but it is still not "deleting a reference".
// The reference 'ir' is another name for the heap-based int.
// So, &ir is the address of that int, essentially a pointer.
// It is a pointer that is being used by delete, not a reference.