When we call
delete[] array; // array is a pointer to an array
or
delete[] vector; // vector is a pointer to a vector
Assuming these are array/vector of linked list head pointers: Do these 2 statements call the destructors for every item in them? Or should we loop through them, delete each item and then delete[] the array or vector itself?
Let me be more specific,
If the array holds a head pointer with its destructor is called it deletes its next node or that it points to.
Your second code snippet is incorrect. delete and delete[] are two different operators. delete is for delete an instance created by new and delete[] is for delete an array of elements created by new[].
In C++, arrays and vectors does not care about what they store. For both array and vector, you need to delete each item (if they were allocated using new) before deleting the array and the vector.
If you create the array like this
Sample* arr[100];
for (int i = 0; i < 100; ++i) {
arr[i] = new Sample();
}
you need to delete them like this
for (int i = 0; i < 100; ++i) {
delete arr[i];
}
If you create the vector like this
vector<Sample*> arr;
for (int i = 0; i < 100; ++i) {
arr.push_back(new Sample());
}
you need to delete like this
for (int i = 0; i < 100; ++i) {
delete arr[i];
}
Deleting an array of pointers delete[] array is suitable, but for an vector of pointers i would use a construct like this:
std::vector<Data*> m_data;
while(!m_data.empty()) delete m_data.back(), m_data.pop_back();
Calling delete[] on a std::vector is not correct C++. The vector is destroyed when it goes out of scope or when you delete a pointer to it if allocated on the heap with new. In both cases, however, destroying the array/vector does not call delete on the pointers it contains - you need to iterate through and do this yourself if you need to free that memory.
Related
I'm using a std::forward_list in C++ to store dynamically allocated elements. I'm wondering what the correct way to 'delete' the elements in list, or free the memory.
std::forward_list<int> integerList;
for (int i = 0; i < 5; i++) {
int* intPtr = new int(i);
integerList.push_front(*intPtr);
}
After this code executes, I think what should happen is the forward_list holds a dynamically allocated int. Now, how would I go about properly calling delete on it? I noticed the forward_list container has a clear() function, but does that call delete on all of the elements in each node?
I think what should happen is the forward_list holds a dynamically allocated int
No, it doesn't hold dynamically allocated int. The type of element is int, when integerList.push_front(*intPtr);, the value will be copied.
On the other hand, you get a memory leak in the for loop, you need to delete what you newed.
for (int i = 0; i < 5; i++) {
int* intPtr = new int(i);
integerList.push_front(*intPtr);
delete intPtr;
}
If you want to save dynamically allocated elements, you should declare the type of element as pointer, and delete the elements by yourself at last.
std::forward_list<int*> integerList;
for (int i = 0; i < 5; i++) {
int* intPtr = new int(i);
integerList.push_front(intPtr);
}
...
for (auto e : integerList) {
delete e;
}
And you could use smart pointers to avoid this kind of memory management, such as std::unique_ptr or std::shared_ptr.
Yes, you will have to iterate through the list and call delete each pointer element.
Say I want to declare an array of pointers to a specific object like
Object *objects = new Object[size];
and then add some new objects to my dynamically allocated array
for (int i = 0; i < size; i++){
Object* obj = new obj;
objects[i] = *obj;
}
is it enough to deallocate all memory by just calling delete[] on the array, or do I have to loop through the array first and call delete on each object? Or is this stupid to do in practice?
You always have to delete everything you new in some way. That means, you would need to delete obj in every iteration to avoid leaks. Note that you never really store the Object obj points to, but a copy of it.
The way you do it right know is quite unusual and not very handy anyway: The loop you showed does nothing useful since the new Objects[size] already default constructed your Objects. In particular, it does not add any elements, they are already there. You could just leave that out. If you want to change the content of your array, e.g. do more initialization, your loop would usually look more like this
for (int i = 0; i < size; i++){
objects[i] = newValue;
}
Remember, the Objects are already there after new Objects[size]!
In practice, you would be much better of with a
std::vector<Object> objects(size);
(or maybe, if the above does not fit your usecase, a
std::vector<std::unique_ptr<Object>> objects(size);
, but I feel like that is unlikely to be the case). For a good overview of what a std::vector can do and how to use it, read the documentation. Probably most importantly for you: You can index it just like an array.
First, you have to follow the deallocation is reverse order of allocation.
So number of times new that many times delete , similarly number of times new [] that many times delete []
// Assuming objects as array of pointers [ Object **objects ]
for (int i = 0; i < size; i++)
{
delete objects[i] ;
}
delete [] objects;
And your correct allocation should go as:
Object **objects= new Object*[size];
for (int i = 0; i < size; i++) {
objects[i] = new Object;
}
Also , you should use smart pointers std::unique_ptr for hassle free memory management
std::vector< std::unique_ptr<Object> > objects ;
You are storing an array of Objects and not pointers to Object. This means that they won't be deleted when you delete the array and the memory will leak.
If you want to store pointers to Objects in your array then you need to declare it as Object **objects = new Object *[size]; and then you will have to delete all the pointers in the array.
So the better code will look like:
//Allocation:
Object **objects = new Object *[size];
for (int i = 0; i < size; i++){
objects[i] = new Object;
}
//Deletion
for (int i = 0; i < size; i++){
delete objects[i];
}
delete [] objects;
An even better solution would be to use vectors and smart pointers and then you don't need the deletion code at all.
//Allocation:
vector<unique_ptr<Object>> objects;
for (int i = 0; i < size; i++){
objects.push_back(make_unique(new Object));
}
//Deletion
//Hey where did all the code go.
I know that if I use new[] I have to delete[], but this is my doubt:
Declaration:
char **data; // This will be a bi-dimensional array
Construct:
data = new char*[10];
for (int i=0; i<10; ++i)
{
data[i] = new char[128];
}
Destroy:
for (int i=0; i<10; ++i)
{
delete data[i]; // <--- delete or delete[] ???
}
delete[] data;
Another case where I have an array of objects:
Declaration:
myClass **object;
Construct:
object = new myClass*[10];
for (int i=0; i<10; ++i)
{
object[i] = new myClass();
}
Destroy:
for (int i=0; i<10; ++i)
{
delete object[i]; // <--- this object[] is not an array, right?
}
delete[] object; // <--- Is this correct?
You basically answered your own question. Since data[i] is an array, you need to delete[]. object[i] was allocated with new so you would delete it. Since both object and data are arrays, you would need to delete[] them.
The first dimension of your pointer data is an array. The second dimension is an array as well. Therefore you must use delete[] to properly deallocate memory reserved from your new[].
For your object, the first dimension was allocated using new. So you must use delete. Then use delete[] on object because it's an array.
The mnemonic that you mentioned (delete [] for new [] and delete for new) is a a good practice, and should be enforced as mandatory.
However, there is a huge difference between delete and delete[] when it comes to base types (int, char, void*,etc) and classes.
The difference between new and malloc is that one calls the constructor while the other doesn't. Same relation happens with delete and free when it comes to destructor.
In your example you gave two situations : one when an array of simple type is allocated and one with an array of objects allocated.
The behavior is different.
For the simple type, delete and delete[] will produce the same result .
Reason ? well, you basically allocated a block of 10 pointers of char* , so deleting it is no problem (the allocated memory paged will be freed).
In other words, if you have
char* a = new char[100];
then
delete a;
and
delete[] a;
will produce the same results (note that you used a new []);
This behavior is different when it comes to objects that have constructors and destructors.
Let's take the following class :
class SimpleClass
{
public:
SimpleClass(){printf("constructed");
~SimpleClass()(printf("destructed");
}
and the following code :
SimpleClass* arr1 = new SimpleClass[5];
SimpleClass* arr2 = new SimpleClass[5];
char* arr3 = new char[5];
char* arr4 = new char[5];
delete[] arr4;
delete arr3;
delete[] arr2;
delete arr1; //notice the surprise ?
Run the above code (if you can) in Visual Studio and open up a memory viewer. You will see that the memory allocated at ptrs arr4 and arr3 is correctly invalidated, same for arr2.
However when attempting to delete arr1 without the [] call you will get a SigSev error .
Why ?
Because in case of arr2 you have an array of allocated objects, and the delete[] statement means "pass through each destructor from the array"".
In case of arr1 it means : call the destructor of the object allocated at pointer arr1... a bit unfortunate if you take into account that there is more then 1 object allocated at that address, and you end up attempting to release a page which is lesser then the initial allocated size.
Bottom line :
delete[] for new[] and delete for new !
Works every time !
I have an array of pointers (that I created by calling new ptr*[size]). All of these pointers point to an object that was also put on the heap.
What is the proper way to delete the array and all new'd ptr's?
This is what I do now:
for (int i = 0; i < size; i++) delete array[i];
delete[] array; // Not sure since this double deletes array[0]
Does this do what I think it should?
Thanks
Every pointer allocated with new gets a corresponding delete. Every pointer allocated with new [] gets a corresponding delete []. That's really all you need to know. Of course, when you have a dynamically allocated array which contains dynamically allocated pointers the deallocation must occur in reverse order.
So it follows that the correct idiom would be...
int main()
{
int **container = new int*[n];
for(int i = 0; i < n; ++i)
container[i] = new int[size];
// ... and to deallocate...
for(int i = 0; i < n; ++i)
delete [] container[i];
delete [] container;
}
And then of course I say "stop doing that" and recommend you use a std::array or std::vector (and the template type would be unique_ptr<int>).
Yes, that does what you think it should. Since you did new for each element, you have to delete each element. And since you did new[] for the entire array, you need to delete[] the entire array.
As #djechlin rightly says in the comments, there's not really enough information to go on, but I'm presuming your prior code is something like this:
int** array = new int*[5];
for (int i = 0; i < 5; i++) {
array[i] = new int;
}
Note that array is not actually an array type. It is a "pointer to pointer to int" and the array of pointers it points to was allocated with new[]. That's why you need to delete[] it.
Yes. First you have to free the object each pointer in the array points to, then you have to free the array itself. In that order. If you reverse the order you'll have no reference to the objects and will leak a lot of memory.
Yes, first you delete each object to which elements of array point, and then you delete array of pointers itself. If you want to check your memory management, you can use tools like valgrind, they will be able to spot most errors.
In my class I have a dynamically allocated array of pointers. My declaration:
array = new Elem* [size];
for (int i = 0; i < size; i++) {
array[i] = NULL;
}
So there is an array of pointers, where each pointer points to a simple Elem struct.
The main question is, how should I properly deallocate the array. If I use only:
for (int i = 0; i < size; i++) {
delete array[i];
}
Valgrind reports 1 not-freed block, which is traced to the line where 'array = new Elem* [size];' states.
On the other hand if I add to the previous code:
delete array;
Which I thought is correct, valgrind reports 0 not-freed blocks, which is perfect, BUT it reports
Mismatched free() / delete / delete []
exactly on the line where 'delete array;' is. I tried 'delete []array' too, but that's just "1 not-freed blocks" too then! If somebody could explain me the proper way it would be much appreciated.
EDIT:
So using:
for (int i = 0; i < size; i++) {
delete array[i];
}
delete[] array;
is working probably fine. It is working in one of my classes (I have two similar) the other still reports some small leak. I would think it's just a minor bug somewhere, but valgrind still points to the line where
array = new Elem* [size];
stands.
EDIT2:
I solved this as well, thank you for your exhausting contribution!!
You need:
delete [] array;
Because it's an array.
I just noticed your note that you tried this too - it's the proper thing to do so I don't know why you'd still be getting an error.
Edit: This deserves a more thorough explanation.
When you create a pointer using new, the pointer may be to a single element or an array of elements depending on the syntax you use. But the pointer type is the same in both cases! The compiler relies on you to know what the pointer points to and treat it accordingly.
Elem ** single = new Elem*; // pointer to one pointer
single[0] = new Elem; // OK
single[1] = new Elem; // runtime error, but not compile time
Elem ** array = new Elem* [2]; // pointer to array of pointers
array[0] = new Elem; // OK
array[1] = new Elem; // OK
When you delete a pointer, the destructor is called for the object it points to or for each element of the array. But since the pointer type is the same in each case, the compiler relies on you to give it the proper syntax so it knows what to do.
delete single;
delete [] array;
In your case the elements of the array are pointers also, and pointers don't have destructors. That means those pointers won't be deleted and will become memory leaks if you don't delete them first. You were correct to have a loop to delete them individually before the final delete.
You should free everything in the array (if dynamically allocated) and then free the array itself.
for (int i = 0; i < size; i++) { // only free inside if dynamically allocated - not if just storing pointers
delete array[i];
}
delete[] array; // necesarry
The syntax for deleting an array is like this:
delete[] array;
Your for loop to delete the objects pointed to by the elements of the array is fine. The deletion of the array itself is the only problem. You need both the for loop and then the delete[] to dispose of the array itself.
for (int i = 0; i < size; i++) {
delete array[i];
}
delete[] array;
I suspect that you have tried using the for loop, or the delete[], but not both together. And if when you do that you still have leaks or errors, then you would need to show us the code that allocates the pointers that are elements of the array.
Using std::vector<> instead of an array would mean that you could stop worrying about these nitty gritty details and move to higher level of abstraction.
In this case, you need both.
for (int i = 0; i < size; i++) {
delete array[i];
}
delete[] array;
You call delete exactly once for each time you called new.
Note that although you need to call delete[] array here (because you allocated it with new[]), the delete[] operator does not call the destructors on the objects pointed to by elements of the array. This is because the delete[] operator calls destructors on objects in the array, and your array contains pointers but not objects. Pointers do not themselves have destructors.