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!
Related
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.
It is the first time I am using STL and I am confused about how should I deallocate the the memory used by these containers. For example:
class X {
private:
map<int, int> a;
public:
X();
//some functions
}
Now let us say I define the constructor as:
X::X() {
for(int i=0; i<10; ++i) {
map[i]=i;
}
}
Now my question is should I write the destructor for this class or the default C++ destructor will take care of deallocating the memory(completely)?
Now consider the modification to above class
class X {
private:
map<int, int*> a;
public:
X();
~X();
//some functions
}
Now let us say I define the constructor as:
X::X() {
for(int i=0; i<10; ++i) {
int *k= new int;
map[i]=k;
}
}
Now I understand that for such a class I need to write a destructor as the the memory allocated by new cannot be destructed by the default destructor of map container(as it calls destructor of objects which in this case is a pointer). So I attempt to write the following destructor:
X::~X {
for(int i=0; i<10; ++i) {
delete(map[i]);
}
//to delete the memory occupied by the map.
}
I do not know how to delete the memory occupied by the map. Although clear function is there but it claims to bring down the size of the container to 0 but not necessarily deallocate the memory underneath. Same is the case with vectors too(and I guess other containers in STL but I have not checked them).
Any help appreciated.
should I write the destructor for this class or the default C++ destructor will take care of deallocating the memory(completely)?
Yes it will. All the standard containers follow the principle of RAII, and manage their own dynamic resources. They will automatically free any memory they allocated when they are destroyed.
I do not know how to delete the memory occupied by the map.
You don't. You must delete something if and only if you created it with new. Most objects have their memory allocated and freed automatically.
The map itself is embedded in the X object being destroyed, so it will be destroyed automatically, and its memory will be freed along with the object's, once the destructor has finished.
Any memory allocated by the map is the responsibility of the map; it will deallocate it in its destructor, which is called automatically.
You are only responsible for deleting the dynamically allocated int objects. Since it is difficult to ensure you delete these correctly, you should always use RAII types (such as smart pointers, or the map itself) to manage memory for you. (For example, you have a memory leak in your constructor if a use of new throws an exception; that's easily fixed by storing objects or smart pointers rather than raw pointers.)
When a STL collection is destroyed, the corresponding destructor of the contained object is called.
This means that if you have
class YourObject {
YourObject() { }
~YourObject() { }
}
map<int, YourObject> data;
Then the destructor of YourObject is called.
On the other hand, if you are storing pointers to object like in
map<int, YourObject*> data
Then the destruct of the pointer is called, which releases the pointer itself but without calling the pointed constructor.
The solution is to use something that can hold your object, like a shared_ptr, that is a special object that will care about calling the holded item object when there are no more references to it.
Example:
map<int, shared_ptr<YourObject>>
If you ignore the type of container you're dealing with an just think of it as a container, you'll notice that anything you put in the container is owned by whomever owns the container. This also means that it's up to the owner to delete that memory. Your approach is sufficient to deallocate the memory that you allocated. Because the map object itself is a stack-allocated object, it's destructor will be called automatically.
Alternatively, a best practice for this type of situation is to use shared_ptr or unique_ptr, rather than a raw pointer. These wrapper classes will deallocate the memory for you, automatically.
map<int shared_ptr<int>> a;
See http://en.cppreference.com/w/cpp/memory
The short answer is that the container will normally take care of deleting its contents when the container itself is destroyed.
It does that by destroying the objects in the container. As such, if you wanted to badly enough, you could create a type that mismanaged its memory by allocating memory (e.g., in its ctor) but doesn't release it properly. That should obviously be fixed by correcting the design of those objects though (e.g., adding a dtor that releases the memory they own). Alternatively, you could get the same effect by just storing a raw pointer.
Likewise, you could create an Allocator that didn't work correctly -- that allocated memory but did nothing when asked to release memory.
In every one of these cases, the real answer is "just don't do that".
If you have to write a destructor (or cctor or op=) it indicates you likely do something wrong. If you do it to deallocate a resource more likely so.
The exception is the RAII handler for resources, that does nothing else.
In regular classes you use proper members and base classes, so your dtor has no work of its own.
STL classes all handle themselves, so having a map you have no obligations. Unless you filled it with dumb pointers to allocated memory or something like that -- where the first observation kicks in.
You second X::X() sample is broken in many ways, if exception is thrown on the 5th new you leak the first 4. And if you want to handle that situatuion by hand you end up with mess of a code.
That is all avoided if you use a proper smart thing, like unique_ptr or shared_ptr instead of int*.
Assume you have an object of class Fool.
class Fool
{
int a,b,c;
double* array ;
//...
~Fool()
{
// destroys the array..
delete[] array ;
}
};
Fool *fool = new Fool() ;
Now, I know you shouldn't, but some fool calls the destructor on fool anyway. fool->~Fool();.
Does that mean fool's memory is freed, (ie a,b,c are invalid) or does that mean only whatever deallocations in ~Fool() function occur (ie the array is deleted only?)
So I guess my question is, is a destructor just another function that gets called when delete is called on an object, or does it do more?
If you write
fool->~Fool();
You end the object's lifetime, which invokes the destructor and reclaims the inner array array. The memory holding the object is not freed, however, which means that if you want to bring the object back to life using placement new:
new (fool) Fool;
you can do so.
According to the spec, reading or writing the values of the fields of fool after you explicitly invoke the destructor results in undefined behavior because the object's lifetime has ended, but the memory holding the object should still be allocated and you will need to free it by invoking operator delete:
fool->~Fool();
operator delete(fool);
The reason to use operator delete instead of just writing
delete fool;
is that the latter has undefined behavior, because fool's lifetime has already ended. Using the raw deallocation routine operator delete ensures that the memory is reclaimed without trying to do anything to end the object's lifetime.
Of course, if the memory for the object didn't come from new (perhaps it's stack-allocated, or perhaps you're using a custom allocator), then you shouldn't use operator delete to free it. If you did, you'd end up with undefined behavior (again!). This seems to be a recurring theme in this question. :-)
Hope this helps!
The destructor call does just that, it calls the destructor. Nothing more and nothing less. Allocation is separate from construction, and deallocation from destruction.
The typical sequence is this:
1. Allocate memory
2. Construct object
3. Destroy object (assuming no exception during construction)
4. Deallocate memory
In fact, if you run this manually, you will have to call the destructor yourself:
void * addr = ::operator new(sizeof(Fool));
Fool * fp = new (addr) Fool;
fp->~Fool();
::operator delete(addr);
The automatic way of writing this is of course Fool * fp = new Fool; delete fp;. The new expression invokes allocation and construction for you, and the delete expression calls the destructor and deallocates the memory.
Does that mean fool's memory is freed, (ie a,b,c are invalid) or does
that mean only whatever deallocations in ~Fool() function occur (ie
the array is deleted only?)
Fool::~Fool() has zero knowledge about whether the Fool instance is stored in dynamic storage (via new) or whether it is stored in automatic storage (i.e. stack objects). Since the object ceases to exist after the destructor is run, you can't assume that a, b, c and array will stay valid after the destructor exits.
However, because Fool::~Fool() knows nothing about how the Fool was allocated, calling the destructor directly on a new-allocated Fool will not free the underlying memory that backs the object.
You should not access a, b, and c after the destructor is called, even if it's an explicit destructor call. You never know what your compiler puts in your destructor that might make those values invalid.
However, the memory is not actually freed in the case of an explicit destructor call. This is by design; it allows an object constructed using placement new to be cleaned up.
Example:
char buf[sizeof (Fool)];
Fool* fool = new (buf) Fool; // fool points to buf
// ...
fool->~Fool();
The easiest place to see that the destructor is distinct from deallocation via delete is when the allocation is automatic in the first place:
{
Fool fool;
// ~Fool called on exit from block; nary a sign of new or delete
}
Note also the STL containers make full use of the explicit destructor call. For example, a std::vector<> treats storage and contained objects lifetime quite separately.
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.
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();
}