When is a non-pointer member of a class destructed? Example:
class foo {
private:
int a;
public:
foo(int sa):a(sa){}
~foo(){}//does anything need to be done here?
};
{
foo(10);
}//the destructor is called
Should anything be done inside the destructor? Thanks!
No, not a thing. a will be destroyed after any code in your destructor completes. In a case like this, you don't even need to declare a destructor; the compiler will do the right thing on its own.
An object's contents are destroyed in inverse order of their appearance in the class definition after the execution of the object's destructor.
A non-pointer member of an object is destructed after the containing object's destructor has completed.
Related
I'm trying to understand what is allowed to do in destructors.
The standard says: "For an object with a non-trivial destructor, referring to any non-static member or base class of the object after the destructor finishes execution results in undefined behavior".
cppreference describes destruction sequence this way: "For both user-defined or implicitly-defined destructors, after the body of the destructor is executed, the compiler calls the destructors for all non-static non-variant members of the class".
Does this mean, that in the following code calling method from its member's destructor is UB? Or by "referring" standard means something particular?
struct Foo {
Foo(Callback cb) : cb_(cb) {}
~Foo() {
// body of Bar destructor finished at this moment;
// cb_() calls Bar::call_me()
cb_();
}
Callback cb_;
};
struct Bar {
// pass callback with captured this
Bar() : foo_([this]() { call_me(); }) {
}
void call_me() {
}
// foo is a member, its destructor will be called after Bar destructor
Foo foo_;
};
Also, what does the phrase "after the destructor finishes" from the standard mean exactly? After the body of a destructor finishes? Or after all members and base classes destroyed?
I think the answer to the last question is the key to understanding what is allowed and what is not.
The destructor of Bar has not finished, and therefore referring to a member of Bar, and indeed calling a member function of Bar within its destructor is OK.
Calling member functions of the super object can be a bit precarious though, since member functions may access sub objects, and some sub objects may have already been destroyed by the time the member function is called, and in that case accessing the destroyed objects would result in undefined behaviour. This is not the case in your example.
Or by "referring" standard means something particular?
I think it means to form a pointer or a reference to a sub object. As is done in the example below the rule.
Also, what does the phrase "after the destructor finishes" from the standard mean exactly? After the body of a destructor finishes? Or after all members and base classes destroyed?
The latter.
The body is executed first, then the destructor calls the sub object destructors, and then the destructor has finished.
For a simple class like this:
class X {
public:
//...
private:
int *ptr;
};
X::~X() {
delete ptr;
}
I have written a destructor to free the memory pointed to by ptr. Now, I am wondering, if my destructor stays like this, when is ptr actually destroyed?
Thanks
I think this article might answer most of your questions
"Destructors are implicitly called when an automatic object (a local
object that has been declared auto or register, or not declared as
static or extern) or temporary object passes out of scope. They are
implicitly called at program termination for constructed external and
static objects. Destructors are invoked when you use the delete
operator for objects created with the new operator."
More specifically:
"The destructors of base classes and members are called in the reverse
order of the completion of their constructor:
The destructor for a class object is called before destructors for
members and bases are called.
Destructors for nonstatic members are called before destructors for
base classes are called.
Destructors for nonvirtual base classes are called before destructors
for virtual base classes are called."
delete invokes the destructor of the object that is being deleted and then frees the memory it occupied.
The destructor will calls in the end of scope where the specific instance are.
The local version of ptr for a given instance of X will be destroyed when the object goes out of scope. For example:
int main()
{
X obj;
for (int i=0; i<10; i++) {
X tmp;
// do work with tmp
...
// tmp goes out of scope here
// the destructor tmp::~X will be called here, immediately after each iteration ends
}
// obj goes out of scope here
// the destructor obj::~X called when the program ends
}
Having a simple class:
class A {
public:
A() {}
void set(int value) { value_ = value; }
private:
int value_;
};
and its global instance:
A a;
Is it OK to call method set on a not yet constructed object a? That can happen when for example a.set(123) is called from a constructor of another global object in another translation unit.
Will the value in the object a set by calling a.set(123) remain when the non-parametric and empty constructor of A is later called for object a?
Is it ok to call method set on a not yet constructed object a?
No. You may not call member functions for an object that has not yet begun construction.
(Since the answer is no, your second question requires no answer.)
If you may need to access this global instance from multiple translation units during dynamic initialization, you can use the Meyers singleton technique:
A& global_a()
{
static A a;
return a;
}
a will be initialized when global_a() is first called. Note that in a multithreaded program you may need to concern yourself with synchronization of the initialization.
When you write
A a;
a is a constructed object now. In case A is a then A default constructor was already been called
If in 1) you mean it's ok to call set in the constructor, then yes, that's fine because it isn't a virtual method. You cannot call a virtual method in the constructor.
As for 2), what you're asking isn't really clear. The constructor is only called once (although there are ways around that sort of thing, but don't do them) and that is when the object is first created. You can't call the constructor on a a second time so the question doesn't really make sense.
Does the default destructor in C++ classes automatically delete members that are not explicitly allocated in code? For example:
class C {
public:
C() {}
int arr[100];
};
int main(void) {
C* myC = new C();
delete myC;
return 0;
}
Does delete myC deallocate myC's arr automatically? Or do I need to write C's destructor to do this explicitly?
The constructor (in the absence of any ctor-initializer-list) calls the default constructor for each subobject.
Since you have no base classes and your member variables are primitive types, it will do nothing at all.
Same with the destructor. Yours is implicitly compiler-generated since you haven't declared one, and it will call the destructor for each subobject. Again that's trivial because your only subobject is an aggregate of primitives.
Now, all memory of the class will be freed when you delete it. Since the array is embedded inside the class, it's part of the same memory region and will be freed at the same time.
The implicitly defined (default) destructor will call the destructor for each member. In the case of a member array, it will call the destructor for each element of the array.
Note that pointers don't have destructors; you need to manually delete them. You don't have this problem in the provided example, but it's something to be aware of.
If your class/struct contains a pointer, and you explicitly allocate something for that pointer to refer to, then you normally need to write a matching delete in the dtor. Members that are directly embedded into the class/struct will be created and destroyed automatically.
class X {
int x;
int *y;
public:
X() : y(new int) {}
~X() : { delete y; }
};
Here X::x will be created/destroyed automatically. X::y (or, to be technically correct, what X::y points at) will not -- we allocate it in the ctor and destroy it in the dtor.
Anything that you call new for must have a corresponding delete. If you didn't call new to create an instance of something then you don't have to call delete.
You don't have to write a destructor. C++ class has the default destructor to delete the object after 'return 0' to recycle the memory.
#include<iostream>
using namespace std;
class A
{
public:
int i;
A() {cout<<"A()"<<endl;}
~A() {cout<<"~A()"<<endl;}
};
class B:public A
{
public:
int j;
B(): j(10)
{
this->i=20;
this->~A();
}
};
int main()
{
B abc;
cout<<"i="<<abc.i<<" j="<<abc.j<<endl;
}//main
Two questions:
How come A's destructor gets called like an ordinary function instead of destroying the object? (or is it some kind of rule that the base class will be destroyed only if the child class's destructor calls the base class's destructor?) I was trying out this sample code to find out how the destructor works. So if simply calling the destructor function does not destruct the object, then there is obviously some other kind of call that calls the destructor and only then the object is destructed. What's so special in that kind of call and what call is it?
Is there a way to have an initialization list for A in B's constructor? Something like this:
class B:public A
{
B(): j(10), A():i(20) {}
};
Destructor is like any other normal function which you can call (but you should never do it unless you use a placement new). When you call delete on a object two things happen: Destructor is called for cleanup and then operator delete is called to release the memory allocated for the object. Here the second step is not happening.
No, you can not call it like that. What you can do is some thing like this:
class A
{
public:
A(int n) : i(n){}
};
class B : public A
{
public:
B() : A(20), j(10){}
};
The base class's destructor should be virtual. Here, as it's created on the stack, it's not problem, but anyway..
No, but you can call the class A() constructor in the initialize list of B's constructor, like this:
B(): A( .. ), ...
A* a = new B();
//..
delete a;
will not call B's destructor unless class A destructor is virtual. That's why STL containers should not be derived - theirs destructors are not virtual.
#Nav: no, your understanding of "destroyed" is just wrong. When an object's destructor is called, the object is destroyed. You seem to believe that the memory it resided in evaporates entirely, but that never happens. The object no longer exists, but some garbage data is typically left over by the object, and if you're willing to break the rules of C++ and invoke undefined behavior, then you can read those leftover bytes, and they'll look like the object, and because there are no runtime checks on whether you're accessing a valid object, you can often treat them as an object. Which you do.
It's illegal, it's undefined behavior, but in practice it often works.
Once again, a destructor does not physically vaporize the memory. Your RAM still has the same capacity after a destructor has executed. Conceptually, the object no longer exists once the destructor has run. But the data it contained is still there in memory.
For point:
This is an undefined behaviour but only ~A() is called though an instance of class B because ~A() is not declared virtual. See more on Wikipedia.
No. For derived classes, first call your parent class, then assign parameters.
For point 1) on Wikipedia:
having no virtual destructor, while
deleting an instance of class B will
correctly call destructors for both B
and A if the object is deleted as an
instance of B, an instance of B
deleted via a pointer to its base
class A will produce undefined
behaviour.
Example (for point 2):
B(): A(), j(10) {}
or
B(): A() {j = 10;}
1) Destructor calling order in C++ is reverse order of the constructor calling order. So first derived class object get destroy and then base class object.
2) No.
In the code that you are giving, you are indeed destroying the base class and as such i. Calling a destructor and then using the dead object is undefined behavior - it may work or it may crash.
Should i was something that is more complex that an int (for example a vector), trying to do anything with that would probably result in a crash.
If you call ~SomeClass() yourself, explicitly, the destructor function will be called. Which leaves the object (in this case, the base class part of the object) in a destroyed state.
Since the destructor is not virtual, the destructor of derived classes will not be called, but base classes of SomeClass will be destroyed too.
Trying to find out if A is really destroyed by just using the i member, is not a good test. In fact, you can't test this, since using the object results in undefined behavour. It may work, or it may not (in your case, it probably will print "i=20 j=10", but i is already destroyed).