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();
}
Related
I come from a C background. I used to allocate memory/array, for example, sending it to somewhere else, and the pointer stayed there, even after, the scope where it was allocated was destroyed.
Now, if I do the same with vectors : initialize, pass it by reference, and then keep that reference saved somewhere. After the initial method, that actually created the vector, goes out of scope, would it be destroyed ? Or as the pointer to it is still saved, it will be kept.
std::vector frees the memory it holds when it is destroyed. Holding a reference to an object that gets destroyed is very bad. Accessing it is UB. General rule: Do not store references, only use them where you can be sure that the object exists for the entire scope, e.g. as parameter of a function.
If you want to keep a copy of the data, simply copy the std::vector. No reference needed.
If you want to be able to access the data from different locations, and have it live as long as at least one location still has a reference/pointer to it, don't use std::vector, use std::shared_ptr.
If you want to combine the benefits of std::vector with the benefits of shared memory that lives until the last place lets go of it, combine them: std::shared_ptr<std::vector<...>>.
If you want to have the std::vector live in one place for a bit, and then live in another place but not in the first place anymore, use move semantics:
std::vector<int> foo = {1, 2, 3};
std::vector<int> bar = std::move(foo); // bar now holds the data that foo held, foo is empty, no copy was performed
A pointer to an array on the stack will not keep the array alive, I suppose thats the same in C. Now a vector is not an array. It is a wrapper around a heap allocated array which frees the memory of the array when it goes out of scope.
... because a pointer to it is still saved it will still be kept?
No! A pointer or reference to a std::vector will not keep the vector alive. The canonical wrong example is:
std::vector<T>& foo() {
std::vector<T> x;
return x;
} // baaam !
The reference returned from the function is dangling. YOu cannot do anything with it. The correct way would be to return by value and rely on return value optimization:
std::vector<T> foo() {
std::vector<T> x;
return x;
}
If you do:
auto y = foo();
No copying is involved, thanks to NRVO.
PS: Compilers should warn you about returning a reference to a local variable.
No if I do the same with vectors, I initialize a vector, pass it by reference, then keep the reference saved somewhere. After the initial method that actually created the vector goes out of scope, would it be destoryed?
Yes.
or because a pointer to it is still saved it will still be kept?
No.
You can have dangling pointers in C++ just like you can in C. It's exactly the same.
This is also often the case for references (though in some cases the lifetime of the object is extended for a little bit).
It is true that the vector's data is internally managed by the vector, but that's by-the-by since you're asking about the vector itself. Forget the vector and ask the same question about an int, then you'll realise the answer is just as you'd expect it to be in C.
The existing answers are good, will add here:
block A
{
vector<int> my_vec(10, 0);
//...something...
}
block B
The my_vec vector will go out of scope and be destroyed at the closing bracket. Now that's stl (Standard Template Library) vectors.
You can also use C-style arrays (but with different syntax). For very large arrays, I have found the allocation time with a dynamically-allocated array to be faster than STL vectors.
//static, goes out of scope and memory handled at the end of code block
int arr0[10];
//dynamic, not destroyed unless delete called.
int* arr1 = new int[10];
//...work with arr1...
delete [] arr1;
Just like in C, you need to take care of de-allocating any memory you create using new.
I was thinking about a this situation not for a real implementation but to understand better how pointers works.
class foo(){
foo();
~foo();
void doComplexThings(const std::vector<int*>& v){
int* copy;
for(int i = 0; i < v.size(); i++){
copy = v[i];
// do some stuffs
}
}
}
main(){
std::vector<int*> myVector; // suppose we have 100 elements
doComplexThings(myVector);
for(int i = 0; i < myVector.size(); i++){
delete myVector[i];
}
myVector.clear();
}
Ok, I know that have no sense to copy v[i] inside an other pointer, but I was thinking: copy do a memory leak?
After the execution of doComplexThings(), copy will continue to exist and will occupy space in the heap?
After deleting all elements it will continue to exist and point to a deallocated memory?
So logically if I do this things with complex objects I'll keep occupy the memory with unreference object? Or copy is saved in the stack because I don't use new? And at the end of doComplexThings it will be deleted?
I'm a bit confused, thanks!
There is some confusion on the topic of pointers in the C++ community. While it is true that smart pointers have been added to the library to alleviate problems with dynamic memory allocation, raw pointers are not obsolete. In fact, whenever you want to inspect another object without owning it, you should use a reference or raw pointer, depending on which suits your needs. If the concept of ownership is unclear to you, think of an object as being owned by another object if the latter is responsible for cleaning up afterwards (deleting the former).
For example most uses of new and delete should be replaces with the following (omitting std for brevity):
{
auto ptr_to_T = make_unique<T>(//constructor params);
do_stuff_with_smart_ptr(ptr_to_T);
do_stuff_with_T(*ptr_to_T);
do_stuff_with_raw_ptr(ptr_to_T.get());
} // automatic release of memory allocated with make_unique()
Notice how a function that takes a T* doesn't need a smart pointer if it doesn't keep a copy of the T* it is given, because it doesn't affect the lifetime of the object. The object is guaranteed to be alive past the return point of do_stuff_with_T() and its function signature signals that it doesn't own the object by taking a raw pointer.
On the other hand, if you need to pass the pointer to an object that is allowed to keep the pointer and reference it later, it is unclear when the object will need to be destroyed and most importantly by whom. This is solved via a shared pointer.
ClassThatNeedsSharedOwnership shared_owner;
{
auto ptr_to_T = make_shared<T>(//constructor params);
shared_owner.set_T(ptr_to_T);
// do a lot of stuff
}
// At this point ptr_to_T is destroyed, but shared_owner might keep the object alive
So how does the above factor in to your code. First of all, if the vector is supposed to own (keep alive) the ints it points to, it needs to hold unique_ptr<int> or shared_ptr<int>. If it is just pointing to ints held by something else, and they are guaranteed to be alive until after the vector is destroyed, you are fine with int*. In this case, it should be evident that a delete is never necessary, because by definition your vector and the function working on the vector are not responsible for cleaning-up!
Finally, you can make your code more readable by changing the loop to this (C++11 which you've tagged in the post):
for (auto copy : v){
// equivalent to your i-indexed loop with copy = v[i];
// as long as you don't need the value of i
do_stuff_to_int_ptr(copy);
// no delete, we don't own the pointee
}
Again this is only true if some other object holds the ints and releases them, or they are on the stack but guaranteed to be alive for the whole lifetime of vector<int*> that points to them.
No additional memory is allocated on the heap when you do this:
copy = v[i];
variable copy points to the same address as v[i], but no additional array is allocated, so there would be no memory leak.
A better way of dealing with the situation is to avoid raw pointers in favor of C++ smart pointers or containers:
std::vector<std::vector<int>> myVector;
Now you can remove the deletion loop, which is an incorrect way of doing it for arrays allocated with new int[length] - it should use delete[] instead:
delete[] myVector[i];
Basically you're illustrating the problem with C pointers which lead to the introduction of C++ unique and shared pointers. If you pass a vector of allocated pointers to an opaque member function, you've no way of knowing whether that function hangs onto them or not, so you don't know whether to delete the pointer. In fact in your example you don't seem to, "copy" goes out of scope.
The real answer is that you should only seldom use allocated pointers in C++ at all. The stl vector will serve as a safer, easier to use version of malloc / new. Then you should pass them about as const & to prevent functions from changing them. If you do need an allocated pointer, make one unique_ptr() and then you know that the unique_ptr() is the "owner" of the memory.
always delete pointer even if it is just in function call stack?
Isn't it disappeared when function stack released?
// just Simple class
class CSimple{
int a;
}
// just simple function having pointer.
void simpleFunc(){
CSimple* cSimple = new CSimple();
cSimple->a = 10;
//.. do sth
delete cSimple; // <<< Always, do I have to delete 'cSimple' to prevent the leak of memory?
}
void main(){
for( int =0 ; i< 10 ; i++){
simpleFunc();
}
}
when function stack released?
It is true that "CSimple *csimple" goes away when the function returns.
However, there's a big difference between the pointer, and what it's pointed to.
When a pointer object gets destroyed, nothing happens to whatever the pointer is pointing to. There isn't just one, but two objects here:
The pointer.
What it's pointing to.
In this case, the pointer is pointing to an object in dynamic scope that was created with new.
Nothing is going to happen to this object, otherwise, so you will leak memory.
Therefore, this object needs to be deleted.
After you understand, and fully wrap your brain around this concept, your next step will be to open your C++ book to the chapter that talks about the std::unique_ptr and std::shared_ptr classes, which will take care of these pesky details, for you. You should learn how to use them. Modern C++ code rarely needs to delete something; rather these smart pointers do all the work.
Yes.
On scope exit (ex. when function exists or block { ... } finishes), all objects created on stack will be destroyed and memory will be freed.
This applies to your case, ie. the pointer will be destroyed and memory occupied by the pointer will be freed. The object pointed by the pointer will not be cleared.
This is a common problem and a nightmare when you deal with multiple flow paths (if-else ladders, many return statements) and exceptions.
To solve this problem, we employ 2 main strategies:
RAII
Smart pointers (std::unique_ptr, boost::scoped_ptr, legacy std::auto_ptr, etc).
RAII - without academic consideration - is just creating object on stack, like this:
{
std::string s;
fancy_object obj;
}
When we exit he scope, obj and s destructors will be called duing stack unwinding. Compiler ensures this for all flow paths and will keep proper order of deallocations for us.
If you need to allocate memory on heap, using new, use a smart pointer.
int foo()
{
std::unique_ptr<Object> o(new Object);
... some more code ...
if( something ) { return -1 }
... some more code ...
if( something_else ) { return -2 }
else { not yet }
return 0;
}
As you can see, we can leave the scope using 3 "exists". Normally, you'd need to clear your memory in all cases, which is prone to human errors.
Instead of clearing the object manually in all 3 palces, we rely on automatic destructor call for objects created on stack. Compiler will figure it out. When we leave the scope, std::unique_ptr destructor will be called, calling delete on Object.
Don't be affraid of smart poiners. They are not "slow", "bloat" or other nonsense. Smart poiners are designed to have no overhead on access, adding extra security.
Very similar technique is used for locks. Check out std::lock_guard class.
Yes, you must delete the data that is being pointed to.
The pointer itself is on the stack and does not need to be deleten.
You can, however, store cSimple on the stack, then you don't have to delete it:
void simpleFunc(){
CSimple cSimple; // no new
cSimple.a = 10; // "." instead of "->"
//.. do sth
// no deletion
}
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!
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.