Example:
Class *_obj1;
Class *_obj2;
void doThis(Class *obj) {}
void create() {
Class *obj1 = new Class();
Class obj2;
doThis(obj1);
doThis(&obj2);
_obj1 = obj1;
_obj2 = &obj2;
}
int main (int argc, const char * argv[]) {
create();
_obj1->doSomething();
_obj2->doSomething();
return 0;
}
This creates 2 objects, creates pointers to them, then main() calls a method on each. The Class object creates a char* and stores the C string "Hello!" in it; the ~Class() deallocator frees the memory. The doSomething() method prints out "buff: %s" using printf(). Simple enough. Now if we run it we get this:
Dealloc
Buff: Hello!
Buff: ¯ø_ˇ
Obviously the stack object does not work here - it's obvious that when the function exits the pointer _obj2 is pointing at a location in the stack. This is why I used heap objects in my previous question, which people told me was "stupid".
So, the first question is: if how can I convert the stack object (obj2) to a heap object so it's not deallocated after create() exits? I want a straight answer, not an arrogant "you're doing it wrong" as so many have done. Because in this case stack objects cannot work so heap objects seem to be the only way. EDIT: Also, converting back to a stack object would be useful as well.
The second question: the specific example of heap objects being "wrong" was creating a new vector<string>* using the new operator. If dynamically allocating STL objects is wrong, then what's the right way? Obviously if you create them as stack objects it fails because they're immediately deallocated, but I've been told (again, by a very high-ranking member) that dynamically allocating them can corrupt the heap. So what's the right way to do it?
So, the first question is: if how can I convert the stack object (obj2) to a heap object so it's not deallocated after create() exits? I want a straight answer,
The straight answer is: You can't "convert" an object between the stack and heap. You can create a copy of the object that lives in the other space, as others have pointed out, but that's it.
The second question: the specific example of heap objects being "wrong" was creating a new vector* using the new operator. If dynamically allocating STL objects is wrong, then what's the right way? Obviously if you create them as stack objects it fails because they're immediately deallocated, but I've been told (again, by a very high-ranking member) that dynamically allocating them can corrupt the heap.
Dynamically allocating STL objects will not on its own corrupt the heap. (No idea where you might have heard that.)
If you want to use a stack-allocated STL object outside of the function that you created it in, you can't, since the stack space in which the object resides is only valid inside the function that created it.
You can, however, return a copy of the object:
std::vector<char> SomeFunc()
{
std::vector<char> myvector;
// myvector.operations ...
return myvector;
}
As I said, though, this will return a copy of the object, not the original object itself -- that would be impossible, since the stack that contains the object is unwound after the function returns.
One other option is to have the caller pass in a reference / pointer to the object that your function manipulates, if this makes sense for your particular scenario:
void SomeFunc(std::vector<char>& destination)
{
// destination.operations ...
}
void AnotherFunc()
{
std::vector<char> myvector;
SomeFunc(myvector);
}
As you can see, you've still allocated everything on the stack, and you avoid the (sometimes consequential) overhead of relying on the copy-constructor to return a copy of the object.
So, the first question is: if how can I convert the stack object (obj2) to a heap object so it's not deallocated after create() exits?
This line:
_obj2 = &obj2;
Change to:
_obj2 = new Class(obj2); // Create an object on the heap invoking the copy constructor.
I want a straight answer, not an arrogant "you're doing it wrong" as so many have done.
Thats as straight an answer as you can get. Obviously you are new to C++, So I am sure this will nto work as intended because you have probably made a couple of mistakes in the defintion of the class "Class" (by the way terrible name).
Also, converting back to a stack object would be useful as well.
class obj3(*_obj2); // dereference the heap object pass it to the copy constructor.
The second question: the specific example of heap objects being "wrong" was creating a new vector<string>* using the new operator. If dynamically allocating STL objects is wrong, then what's the right way?
Why do you dynamically allocate the vector. Just create it locally.
std::vector<std::string> funct()
{
std::vector<std::string> vecString;
// fill your vector here.
return vecString; // Notice no dynamic allocation with new,
}
Using new/delete is using C++ like C. What you need to read up on is smart pointers. These are obejcts that control the lifespan of the object and automatically delete the object when they go out of scope.
std::auto_ptr<Class> x(new Class);
Here x is a smart pointer (of type auto_ptr) when it goes out of scope the object will be deleted. But you can return an auto_ptr to the calling function and it will be safely transfered out of the function. Its actually a lot more complicated than that and you need a book.
Obviously if you create them as stack objects it fails because they're immediately deallocated,
Its de'allocated when it goes out of scope.
but I've been told (again, by a very high-ranking member) that dynamically allocating them can corrupt the heap.
If you do it incorrectly. Which given your knowledge is very likely. But hard to verify since you have not provided the definition of Class.
So what's the right way to do it?
Learn why you should use stack objects
Learn what smart pointers are.
Learn how to use smart pointers to control lifespans of objects.
Learn the different types of smart pointers.
Look up what the separation of concerns is (you are not following this basic principle).
You have to either copy-construct a new heap object (Class * foo = new Class(obj2)) or assign the stack object to a heap object (*obj1 = obj2).
the only way is to copy object.
Change declaration to:
Class _obj2;
and assign:
_obj2 = obj2;
Taking the address of a stack variable won't magically transfer it into heap. You need to write a proper copy-constructor for your class and use _obj2 = new Class(obj2);.
As for STL containers, they allocate their data on the heap anyway, why would you want to allocate container itself on the heap? Put them in a scope that will keep them alive as long as you need them.
Your stack object is created inside the create function and is deallocated as soon you get out of scope of the function. The pointer is invalid.
You could change Class* obj2 to Class obj2 and the assign (which means copy) the object by obj2 = obj2;
I think you're really trying to ask "How can I return an object created inside my function?" There are several valid ways:
Allocate on the heap and return a pointer
Use an automatic variable and return its value, not a pointer (the compiler will copy it)
Let the caller provide storage, either by pointer or reference parameter, and build your object there.
Related
My following question is on memory management. I have for example an int variable not allocated dynamically in a class, let's say invar1. And I'm passing the memory address of this int to another classes constructor. That class does this:
class ex1{
ex1(int* p_intvar1)
{
ptoint = p_intvar1;
}
int* ptoint;
};
Should I delete ptoint? Because it has the address of an undynamically allocated int, I thought I don't need to delete it.
And again I declare an object to a class with new operator:
objtoclass = new ex1();
And I pass this to another class:
class ex2{
ex2(ex1* p_obj)
{
obj = p_obj;
}
ex1* obj;
};
Should I delete obj when I'm already deleting objtoclass?
Thanks!
Because it has the address of an undynamically allocated int I thought I don't need to delete it.
Correct.
Should I delete obj when I'm already deleting objtoclass?
No.
Recall that you're not actually deleting pointers; you're using pointers to delete the thing they point to. As such, if you wrote both delete obj and delete objtoclass, because both pointers point to the same object, you'd be deleting that object twice.
I would caution you that this is a very easy mistake to make with your ex2 class, in which the ownership semantics of that pointed-to object are not entirely clear. You might consider using a smart pointer implementation to remove risk.
just an appendix to the other answers
You can get rid of raw pointers and forget about memory management with the help of smart pointers (shared_ptr, unique_ptr).
The smart pointer is responsible for releasing the memory when it goes out of scope.
Here is an example:
#include <iostream>
#include <memory>
class ex1{
public:
ex1(std::shared_ptr<int> p_intvar1)
{
ptoint = p_intvar1;
std::cout << __func__ << std::endl;
}
~ex1()
{
std::cout << __func__ << std::endl;
}
private:
std::shared_ptr<int> ptoint;
};
int main()
{
std::shared_ptr<int> pi(new int(42));
std::shared_ptr<ex1> objtoclass(new ex1(pi));
/*
* when the main function returns, these smart pointers will go
* go out of scope and delete the dynamically allocated memory
*/
return 0;
}
Output:
ex1
~ex1
Should I delete obj when I'm already deleting objtoclass?
Well you could but mind that deleting the same object twice is undefined behaviour and should be avoided. This can happen for example if you have two pointers for example pointing at same object, and you delete the original object using one pointer - then you should not delete that memory using another pointer also. In your situation you might as well end up with two pointers pointing to the same object.
In general, to build a class which manages memory internally (like you do seemingly), isn't trivial and you have to account for things like rule of three, etc.
Regarding that one should delete dynamically allocated memory you are right. You should not delete memory if it was not allocated dynamically.
PS. In order to avoid complications like above you can use smart pointers.
You don't currently delete this int, or show where it's allocated. If neither object is supposed to own its parameter, I'd write
struct ex1 {
ex1(int &i_) : i(i_) {}
int &i; // reference implies no ownership
};
struct ex2 {
ex2(ex1 &e_) : e(e_) {}
ex1 &e; // reference implies no ownership
};
int i = 42;
ex1 a(i);
ex2 b(a);
If either argument is supposed to be owned by the new object, pass it as a unique_ptr. If either argument is supposed to be shared, use shared_ptr. I'd generally prefer any of these (reference or smart pointer) to raw pointers, because they give more information about your intentions.
In general, to make these decisions,
Should I delete ptoint?
is the wrong question. First consider things at a slightly higher level:
what does this int represent in your program?
who, if anyone, owns it?
how long is it supposed to live, compared to these classes that use it?
and then see how the answer falls out naturally for these examples:
this int is an I/O mapped control register.
In this case it wasn't created with new (it exists outside your whole program), and therefore you certainly shouldn't delete it. It should probably also be marked volatile, but that doesn't affect lifetime.
Maybe something outside your class mapped the address and should also unmap it, which is loosely analogous to (de)allocating it, or maybe it's simply a well-known address.
this int is a global logging level.
In this case it presumably has either static lifetime, in which case no-one owns it, it was not explicitly allocated and therefore should not be explicitly de-allocated
or, it's owned by a logger object/singleton/mock/whatever, and that object is responsible for deallocating it if necessary
this int is being explicitly given to your object to own
In this case, it's good practice to make that obvious, eg.
ex1::ex1(std::unique_ptr<int> &&p) : m_p(std::move(p)) {}
Note that making your local data member a unique_ptr or similar, also takes care of the lifetime automatically with no effort on your part.
this int is being given to your object to use, but other objects may also be using it, and it isn't obvious which order they will finish in.
Use a shared_ptr<int> instead of unique_ptr to describe this relationship. Again, the smart pointer will manage the lifetime for you.
In general, if you can encode the ownership and lifetime information in the type, you don't need to remember where to manually allocate and deallocate things. This is much clearer and safer.
If you can't encode that information in the type, you can at least be clear about your intentions: the fact that you ask about deallocation without mentioning lifetime or ownership, suggests you're working at the wrong level of abstraction.
Because it has the address of an undynamically allocated int, I
thought I don't need to delete it.
That is correct. Simply do not delete it.
The second part of your question was about dynamically allocated memory. Here you have to think a little more and make some decisions.
Lets say that your class called ex1 receives a raw pointer in its constructor for a memory that was dynamically allocated outside the class.
You, as the designer of the class, have to decide if this constructor "takes the ownership" of this pointer or not. If it does, then ex1 is responsible for deleting its memory and you should do it probably on the class destructor:
class ex1 {
public:
/**
* Warning: This constructor takes the ownership of p_intvar1,
* which means you must not delete it somewhere else.
*/
ex1(int* p_intvar1)
{
ptoint = p_intvar1;
}
~ex1()
{
delete ptoint;
}
int* ptoint;
};
However, this is generally a bad design decision. You have to root for the user of this class read the commentary on the constructor and remember to not delete the memory allocated somewhere outside class ex1.
A method (or a constructor) that receives a pointer and takes its ownership is called "sink".
Someone would use this class like:
int* myInteger = new int(1);
ex1 obj(myInteger); // sink: obj takes the ownership of myInteger
// never delete myInteger outside ex1
Another approach is to say your class ex1 does not take the ownership, and whoever allocates memory for that pointer is the responsible for deleting it. Class ex1 must not delete anything on its destructor, and it should be used like this:
int* myInteger = new int(1);
ex1 obj(myInteger);
// use obj here
delete myInteger; // remeber to delete myInteger
Again, the user of your class must read some documentation in order to know that he is the responsible for deleting the stuff.
You have to choose between these two design decisions if you do not use modern C++.
In modern C++ (C++ 11 and 14) you can make things explicit in the code (i.e., do not have to rely only on code documentation).
First, in modern C++ you avoid using raw pointers. You have to choose between two kinds of "smart pointers": unique_ptr or shared_ptr. The difference between them is about ownership.
As their names say, an unique pointer is owned by only one guy, while a shared pointer can be owned by one or more (the ownership is shared).
An unique pointer (std::unique_ptr) cannot be copied, only "moved" from one place to another. If a class has an unique pointer as attribute, it is explicit that this class has the ownership of that pointer. If a method receives an unique pointer as copy, it is explicit that it is a "sink" method (takes the ownership of the pointer).
Your class ex1 could be written like this:
class ex1 {
public:
ex1(std::unique_ptr<int> p_intvar1)
{
ptoint = std::move(p_intvar1);
}
std::unique_ptr<int> ptoint;
};
The user of this class should use it like:
auto myInteger = std::make_unique<int>(1);
ex1 obj(std::move(myInteger)); // sink
// here, myInteger is nullptr (it was moved to ex1 constructor)
If you forget to do "std::move" in the code above, the compiler will generate an error telling you that unique_ptr is not copyable.
Also note that you never have to delete memory explicitly. Smart pointers handle that for you.
This question already has answers here:
Does std::list::remove method call destructor of each removed element?
(6 answers)
Closed 9 years ago.
I have a class that contains pointers, the class inherits nothing
class MyClass
{
public:
MyClass();
~MyClass();
private:
//i have pointers here
};
MyClass::~MyClass()
{
print("destroyed..");
}
Now i have to use this class as a pointer in vector like this:
vector<MyClass*> classes;
Push some classes in here but when i remove an element:
classes.remove(index);
The destructor doesn't get called,and i think that I have a memory leak.
So how do i make it call the destructor
A vector of pointers does nothing to delete the pointers when they get removed or cleared from it. The vector cannot know if the pointers are dynamically allocated or not. It is not it's job to call delete.
It is up to you to call delete on the pointers, if and when it is necessary. There are not enough details in your question to determine whether it is necessary at all (you haven't shown how the objects pointed to are allocated). But since you claim there is a memory leak, this could indicate that they are dynamically allocated. The immediate solution is to call delete:
delete *it;
classes.erase(it); // vector has no remove member function
A safer solution is to store unique ownership smart pointers, such as std::unique_ptr<MyClass>. The standard library also provides smart pointers for shared and weak ownership. See Smart Pointers.
All the above is assuming that you do actually need to store a pointer. In general, it is safer and clearer to store values:
std::vector<MyClass> classes; // but don't call it "classes". A vector stores objects.
That's one of the reasons why you should avoid using std::vector<MyClass*> at first place. There's an ugly memory management connected with it and it won't stay as easy as classes.remove(index);
Basically, for every new a delete must be called and for every new[] a delete[] must be called, no matter whether you use this pointer as a local variable or you put it into the vector:
vector<MyClass*> vec;
vec.push_back(new MyClass()); // <-- object has been created
...
delete classes[index]; // <-- object shall be destructed
// the delete call will automatically invoke the destructor if needed
...
// now you can remove the dangling pointer from the vector
Just note that once the object has been destructed, any (old) reference to this object is invalid and trying to access this object using such reference (dangling pointer) will yield undefined behavior.
Firstly, std::vector has no remove, you probably mean erase.
Secondly, you need to manually call delete on whatever you're removing:
vector<MyClass*> classes;
auto iter = <iterator to index to remove>;
delete *iter;;
classes.erase(iter);
Or, to avoid all this pain, use a std::unique_ptr<MyClass>.
It is unclear who is responsible for managing the lifetime of the objects pointed by the pointers inside classes. Have you pushed newed pointers into it, or have you pushed the addresses of automatic storage objects?
If you have done the former, then you must manually delete the pointer before removing it. Else, if you have done the latter, then you could just leave it as is, just leaving the pointed-to objects destroy themselves as they leave their respective scopes. If you have mixed newed and non-newed pointers, whose possibility isn't that remote as you would think, then you're definitely damned, undefined behavior making demons fly out of your nose.
These kinds of situations involving pointers are very ambiguous, and it is generally recommended not to use pointers at all, and make the std::vector store plain objects, which makes your object lifetime management much simpler and the making the declaration just speak for itself.
vector<MyClass> classes; // Do this instead
You have to manually delete your pointers before your application exit or after your class object is removed from vector.
// Delete all
vector<MyClass*>::iterator it = classes.begin();
while (it != classes.end()) {
delete *it;
it = classes.erase(it);
}
Tip: Never add stack constructed pointers like following:
MyClass m;
classes.push_back(&m);
Edit: As suggested by other member the better solution is:
MyClass m(/* ... */);
vector<MyClass> classes;
classes.push_back(m);
However please note, you have to properly implement the copy constructor especially if your class has pointer data members that were created with new.
Make a temp pointer to hole MyClass* pointer before you remove it from your vector.
vector<MyClass*> classes;
//push some classes in here but
//when i remove an element
MyClass* temp = classes[index];
classes.remove(index);
// call delete temp; if you want to call the destructor thus avoid memory leak.
delete temp;
To avoid memory leak, remember never to loose control of heap object, always keep a a pointer or reference to it before object release.
It seems that you want your vector to be manager of your items.
Take a look at boost::ptr_vector class
its basically a wrapper around std::vector class.
You declare that this vector is the "holder" of these pointers, and if you remove them from this containers you want them to be deleted.
#include <boost/ptr_container/ptr_vector.hpp>
...
boost::ptr_vector<MyClass> myClassContainer;
myClassContainer.push_back(new MyClass());
myClassContainer.clear(); // will call delete on every stored object!
So I have a class called List which stores a vector of pointers to classes of type Object. List has a function called add which initialises an Object class and adds it's pointer to the vector. I thought that once the add function ended that the Object class would be destroyed and accessing the pointer would cause an exception. When I wrote a test program to test this it turned out that the Object class was never destroyed.
Are classes initialised inside a function ever destroyed once the function ends?
When are classes automatically destroyed?
Depends how you're creating the object. If you are doing it like this:
void add() {
Object obj;
vec.push_back(&obj);
}
Then you are creating obj with automatic storage duration. That means it will be destroyed when the add function ends. The pointer you have pushed into the vector will no longer point to a valid Object, so definitely don't do this.
You may, however, be doing this:
void add() {
Object* obj = new Object();
vec.push_back(obj);
}
If you are, you are creating the Object with dynamic storage duration and it will not be destroyed at the end of the function. The pointer you push into the vector will remain valid. However, if you do this, you need to remember to delete the object at a later time. If you don't, you'll have a leak.
The best option is to avoid using pointers at all, if you can. Just make the vector a std::vector<Object> and copy objects into it:
void add() {
vec.push_back(Object());
}
Or in C++11:
void add() {
vec.emplace_back();
}
If you really need pointers, prefer smart pointers.
In C++, if you don't use any memory manager, smart pointers or some kind of "managed" environment, classes created using keyword new. To avoid misunderstanding could you please post constructor of Object and method List.Add?
So, when you are creating new object like:
Object* o = new Object();
you reserving space in the heap and store pointer to that space in pointer 'o'. If you never calling
delete o;
explicitly your object will never deleted.
Basic Question: when does a program call a class' destructor method in C++? I have been told that it is called whenever an object goes out of scope or is subjected to a delete
More specific questions:
1) If the object is created via a pointer and that pointer is later deleted or given a new address to point to, does the object that it was pointing to call its destructor (assuming nothing else is pointing to it)?
2) Following up on question 1, what defines when an object goes out of scope (not regarding to when an object leaves a given {block}). So, in other words, when is a destructor called on an object in a linked list?
3) Would you ever want to call a destructor manually?
1) If the object is created via a pointer and that pointer is later deleted or given a new address to point to, does the object that it was pointing to call its destructor (assuming nothing else is pointing to it)?
It depends on the type of pointers. For example, smart pointers often delete their objects when they are deleted. Ordinary pointers do not. The same is true when a pointer is made to point to a different object. Some smart pointers will destroy the old object, or will destroy it if it has no more references. Ordinary pointers have no such smarts. They just hold an address and allow you to perform operations on the objects they point to by specifically doing so.
2) Following up on question 1, what defines when an object goes out of scope (not regarding to when an object leaves a given {block}). So, in other words, when is a destructor called on an object in a linked list?
That's up to the implementation of the linked list. Typical collections destroy all their contained objects when they are destroyed.
So, a linked list of pointers would typically destroy the pointers but not the objects they point to. (Which may be correct. They may be references by other pointers.) A linked list specifically designed to contain pointers, however, might delete the objects on its own destruction.
A linked list of smart pointers could automatically delete the objects when the pointers are deleted, or do so if they had no more references. It's all up to you to pick the pieces that do what you want.
3) Would you ever want to call a destructor manually?
Sure. One example would be if you want to replace an object with another object of the same type but don't want to free memory just to allocate it again. You can destroy the old object in place and construct a new one in place. (However, generally this is a bad idea.)
// pointer is destroyed because it goes out of scope,
// but not the object it pointed to. memory leak
if (1) {
Foo *myfoo = new Foo("foo");
}
// pointer is destroyed because it goes out of scope,
// object it points to is deleted. no memory leak
if(1) {
Foo *myfoo = new Foo("foo");
delete myfoo;
}
// no memory leak, object goes out of scope
if(1) {
Foo myfoo("foo");
}
Others have already addressed the other issues, so I'll just look at one point: do you ever want to manually delete an object.
The answer is yes. #DavidSchwartz gave one example, but it's a fairly unusual one. I'll give an example that's under the hood of what a lot of C++ programmers use all the time: std::vector (and std::deque, though it's not used quite as much).
As most people know, std::vector will allocate a larger block of memory when/if you add more items than its current allocation can hold. When it does this, however, it has a block of memory that's capable of holding more objects than are currently in the vector.
To manage that, what vector does under the covers is allocate raw memory via the Allocator object (which, unless you specify otherwise, means it uses ::operator new). Then, when you use (for example) push_back to add an item to the vector, internally the vector uses a placement new to create an item in the (previously) unused part of its memory space.
Now, what happens when/if you erase an item from the vector? It can't just use delete -- that would release its entire block of memory; it needs to destroy one object in that memory without destroying any others, or releasing any of the block of memory it controls (for example, if you erase 5 items from a vector, then immediately push_back 5 more items, it's guaranteed that the vector will not reallocate memory when you do so.
To do that, the vector directly destroys the objects in the memory by explicitly calling the destructor, not by using delete.
If, perchance, somebody else were to write a container using contiguous storage roughly like a vector does (or some variant of that, like std::deque really does), you'd almost certainly want to use the same technique.
Just for example, let's consider how you might write code for a circular ring-buffer.
#ifndef CBUFFER_H_INC
#define CBUFFER_H_INC
template <class T>
class circular_buffer {
T *data;
unsigned read_pos;
unsigned write_pos;
unsigned in_use;
const unsigned capacity;
public:
circular_buffer(unsigned size) :
data((T *)operator new(size * sizeof(T))),
read_pos(0),
write_pos(0),
in_use(0),
capacity(size)
{}
void push(T const &t) {
// ensure there's room in buffer:
if (in_use == capacity)
pop();
// construct copy of object in-place into buffer
new(&data[write_pos++]) T(t);
// keep pointer in bounds.
write_pos %= capacity;
++in_use;
}
// return oldest object in queue:
T front() {
return data[read_pos];
}
// remove oldest object from queue:
void pop() {
// destroy the object:
data[read_pos++].~T();
// keep pointer in bounds.
read_pos %= capacity;
--in_use;
}
~circular_buffer() {
// first destroy any content
while (in_use != 0)
pop();
// then release the buffer.
operator delete(data);
}
};
#endif
Unlike the standard containers, this uses operator new and operator delete directly. For real use, you probably do want to use an allocator class, but for the moment it would do more to distract than contribute (IMO, anyway).
When you create an object with new, you are responsible for calling delete. When you create an object with make_shared, the resulting shared_ptr is responsible for keeping count and calling delete when the use count goes to zero.
Going out of scope does mean leaving a block. This is when the destructor is called, assuming that the object was not allocated with new (i.e. it is a stack object).
About the only time when you need to call a destructor explicitly is when you allocate the object with a placement new.
1) Objects are not created 'via pointers'. There is a pointer that is assigned to any object you 'new'. Assuming this is what you mean, if you call 'delete' on the pointer, it will actually delete (and call the destructor on) the object the pointer dereferences. If you assign the pointer to another object there will be a memory leak; nothing in C++ will collect your garbage for you.
2) These are two separate questions. A variable goes out of scope when the stack frame it's declared in is popped off the stack. Usually this is when you leave a block. Objects in a heap never go out of scope, though their pointers on the stack may. Nothing in particular guarantees that a destructor of an object in a linked list will be called.
3) Not really. There may be Deep Magic that would suggest otherwise, but typically you want to match up your 'new' keywords with your 'delete' keywords, and put everything in your destructor necessary to make sure it properly cleans itself up. If you don't do this, be sure to comment the destructor with specific instructions to anyone using the class on how they should clean up that object's resources manually.
Pointers -- Regular pointers don't support RAII. Without an explicit delete, there will be garbage. Fortunately C++ has auto pointers that handle this for you!
Scope -- Think of when a variable becomes invisible to your program. Usually this is at the end of {block}, as you point out.
Manual destruction -- Never attempt this. Just let scope and RAII do the magic for you.
To give a detailed answer to question 3: yes, there are (rare) occasions when you might call the destructor explicitly, in particular as the counterpart to a placement new, as dasblinkenlight observes.
To give a concrete example of this:
#include <iostream>
#include <new>
struct Foo
{
Foo(int i_) : i(i_) {}
int i;
};
int main()
{
// Allocate a chunk of memory large enough to hold 5 Foo objects.
int n = 5;
char *chunk = static_cast<char*>(::operator new(sizeof(Foo) * n));
// Use placement new to construct Foo instances at the right places in the chunk.
for(int i=0; i<n; ++i)
{
new (chunk + i*sizeof(Foo)) Foo(i);
}
// Output the contents of each Foo instance and use an explicit destructor call to destroy it.
for(int i=0; i<n; ++i)
{
Foo *foo = reinterpret_cast<Foo*>(chunk + i*sizeof(Foo));
std::cout << foo->i << '\n';
foo->~Foo();
}
// Deallocate the original chunk of memory.
::operator delete(chunk);
return 0;
}
The purpose of this kind of thing is to decouple memory allocation from object construction.
Remember that Constructor of an object is called immediately after the memory is allocated for that object and whereas the destructor is called just before deallocating the memory of that object.
Whenever you use "new", that is, attach an address to a pointer, or to say, you claim space on the heap, you need to "delete" it.
1.yes, when you delete something, the destructor is called.
2.When the destructor of the linked list is called, it's objects' destructor is called. But if they are pointers, you need to delete them manually.
3.when the space is claimed by "new".
Yes, a destructor (a.k.a. dtor) is called when an object goes out of scope if it is on the stack or when you call delete on a pointer to an object.
If the pointer is deleted via delete then the dtor will be called. If you reassign the pointer without calling delete first, you will get a memory leak because the object still exists in memory somewhere. In the latter instance, the dtor is not called.
A good linked list implementation will call the dtor of all objects in the list when the list is being destroyed (because you either called some method to destory it or it went out of scope itself). This is implementation dependent.
I doubt it, but I wouldn't be surprised if there is some odd circumstance out there.
If the object is created not via a pointer(for example,A a1 = A();),the destructor is called when the object is destructed, always when the function where the object lies is finished.for example:
void func()
{
...
A a1 = A();
...
}//finish
the destructor is called when code is execused to line "finish".
If the object is created via a pointer(for example,A * a2 = new A();),the destructor is called when the pointer is deleted(delete a2;).If the point is not deleted by user explictly or given a new address before deleting it, the memory leak is occured. That is a bug.
In a linked list, if we use std::list<>, we needn't care about the desctructor or memory leak because std::list<> has finished all of these for us. In a linked list written by ourselves, we should write the desctructor and delete the pointer explictly.Otherwise, it will cause memory leak.
We rarely call a destructor manually. It is a function providing for the system.
Sorry for my poor English!
Using C++:
I currently have a method in which if an event occurs an object is created, and a pointer to that object is stored in a vector of pointers to objects of that class. However, since objects are destroyed once the local scope ends, does this mean that the pointer I stored to the object in the vector is now null or undefined? If so, are there any general ways to get around this - I'm assuming the best way would be to allocate on the heap.
I ask this because when I try to access the vector and do operations on the contents I am getting odd behavior, and I'm not sure if this could be the cause or if it's something totally unrelated.
It depends on how you allocate the object. If you allocate the object as an auto variable, (i.e. on the stack), then any pointer to that object will become invalid once the object goes out of scope, and so dereferencing the pointer will lead to undefined behavior.
For example:
Object* pointer;
{
Object myobject;
pointer = &myobject;
}
pointer->doSomething(); // <--- INVALID! myobject is now out of scope
If, however, you allocate the object on the Heap, using the new operator, then the object will remain valid even after you exit the local scope. However, remember that there is no automatic garbage collection in C++, and so you must remember to delete the object or you will have a memory leak.
So if I understand correctly you have described the following scenario:
class MyClass
{
public:
int a;
SomeOtherClass b;
};
void Test()
{
std::vector<MyClass*> v;
for (int i=0; i < 10; ++i)
{
MyClass b;
v.push_back(&b);
}
// now v holds 10 items pointers to strange and scary places.
}
This is definitely bad.
There are two primary alternatives:
allocate the objects on the heap using new.
make the vector hold instances of MyClass (i.e. std::vector<MyClass>)
I generally prefer the second option when possible. This is because I don't have to worry about manually deallocating memory, the vector does it for me. It is also often more efficient. The only problem, is that I would have to be sure to create a copy constructor for MyClass. That means a constructor of the form MyClass(const MyClass& other) { ... }.
If you store a pointer to an object, and that object is destroyed (e.g. goes out of scope), that pointer will not be null, but if you try to use it you will get undefined behavior. So if one of the pointers in your vector points to a stack-allocated object, and that object goes out of scope, that pointer will become impossible to use safely. In particular, there's no way to tell whether a pointer points to a valid object or not; you just have to write your program in such a way that pointers never ever ever point to destroyed objects.
To get around this, you can use new to allocate space for your object on the heap. Then it won't be destroyed until you delete it. However, this takes a little care to get right as you have to make sure that your object isn't destroyed too early (leaving another 'dangling pointer' problem like the one you have now) or too late (creating a memory leak).
To get around that, the common approach in C++ is to use what's called (with varying degrees of accuracy) a smart pointer. If you're new to C++ you probably shouldn't worry about these yet, but if you're feeling ambitious (or frustrated with memory corruption bugs), check out shared_ptr from the Boost library.
If you have a local variable, such as an int counter, then it will be out of scope when you exit the function, but, unless you have a C++ with a garbage collector, then your pointer will be in scope, as you have some global vector that points to your object, as long as you did a new for the pointer.
I haven't seen a situation where I have done new and my memory was freed without me doing anything.
To check (in no particular order):
Did you hit an exception during construction of member objects whose pointers you store?
Do you have a null-pointer in the container that you dereference?
Are you using the vector object after it goes out of scope? (Looks unlikely, but I still have to ask.)
Are you cleaning up properly?
Here's a sample to help you along:
void SomeClass::Erase(std::vector<YourType*> &a)
{
for( size_t i = 0; i < a.size(); i++ ) delete a[i];
a.clear();
}