I would like to know if there is a way to delete a pointer array without touching the pointed objects in memory.
I'm writing a restriction routine for a HashSet I implemented a couple of days ago, so when the hash table is full it gets replaced by another double sized table. I'm representing the hash table using an array of pointers to an object (User), and the array itself is declared dynamically in my HashSet class, so it can be deleted after copying all its content to the new table using a hash function.
So basically I need to:
Declare another table with a size that equals the double of the original array size.
Copy every pointer to User objects from my original array to the new one applying my hash function (it gets the User object from memory and it calculates the index using a string that represents the user's name).
After inserting all the pointers from the original array to the new one, I will have to free the allocated memory for the original array and replace the pointer in my HashSet class (member private userContainer) with the location of the new one (array).
The problem is that if I use delete[] userContainer to free the allocated memory for it, it will also delete every object in memory so the newly created replacement array will point to freed positions in memory!
What you describe does not sound right.
Let's say you have a class A and you create an array of As with:
A** array1 = new A*[32];
Then fill it:
for(int i = 0; i < 32; ++i)
array1[i] = new A();
Doing a delete[] array1 does not free the elements of array1.
So this is safe:
A** array1 = new A*[32];
for(int i = 0; i < 32; ++i)
array1[i] = new A();
A** arary2 = new A*[64];
for(i = 0; i < 32; ++i)
array2[i] = array1[i];
delete [] array1;
for(i = 0; i < 32; ++i)
// do something with array2[i]
In general, when you delete an array of pointers, whatever objects the pointers pointed to remain in existence. In fact, this is a potential source of large memory leaks.
But in some sort of reference-counted environment (eg, Objective-C or Qt), when you delete an array OBJECT (vs a simple [] array) then the reference counts are decremented and the objects will be deleted if the count goes to zero.
But if you're restructuring a hash table you'd better have somehow saved the pointer values before you delete the array, or else all the addressed objects will be lost. As you save them you can increment their reference counts (if you do it right).
(It would help to know what language you're dealing with, and what you mean by "array".)
I don't think your problem exists. Here's a baby example to show that there's nothing to worry about:
Foo * brr[10];
{
Foo * arr[10];
// This is not touching the objects!
for (Foo * it = arr; it != arr + 10; ++it) *it = new Foo;
std::copy(arr, arr + 10, brr);
} // no more arr
for (Foo * it = brr; it != brr + 10; ++it) delete *it; // fine
You can copy the pointers around freely as much as you like. Just remember to delete the object to which the pointers point when they're no longer needed.
A perhaps trivial reminder: Pointers don't have destructors; in particular, when a pointer goes out of scope, nothing happens.
Do you know the difference between malloc/free, new/delete and new[]/delete[]?
I figure that you might want to not use new[]/delete[] in your situation, as you don't want destructors to be called I guess?
Related
I am writing a program which needs to have an array of pointers to a struct SMesh, specifically
SMesh **mesh_arr;
When I work with an instance that has only one pointer in the array, everything is OK. But when I try to work with instance that has got two (or more) of them, Valgrind goes crazy with "Invalid write of size...".
I initialize it like this (with counter being the number of SMeshes that I need pointers to) :
SMesh **mesh_arr = new SMesh*;
for (int i = 0; i < counter; i++) mesh_arr[i] = new SMesh;
And delete it like this :
delete mesh_arr;
Do I delete it the wrong way or have I missed something?
You have to allocate enough elements to have the entire array, not only 1 element.
SMesh **mesh_arr = new SMesh*[counter];
for (int i = 0; i < counter; i++) mesh_arr[i] = new SMesh;
Using new[], you have to use delete[] to delete it.
delete[] mesh_arr;
Also don't forget to delete the individual SMeshs allocated.
You can avoid pointer trouble with a vector. No need to explicitly delete either.
#include <vector>
auto mesharr{std::vector<std::vector<SMesh>>(rows, std::vector<SMesh>(cols))};
You have only allocated one pointer to your structure in your new statement. Try
SMesh ** mesh_arr = new (SMesh*)[counter];
Let's say I have structure
struct Foo {
int whatever;
int whatever2;
...
};
And I have this type of array
Foo **array = new Foo*[128];
Then, to every pointer I assign new structure
for (int i = 0; i < 128; i++)
array[i] = new Foo;
Then, after a while I deallocate:
delete[] array;
How memory will be deallocated?
Only the space for pointers will be freed
Space for pointer + the objects pointed by those pointers themselves?
Your code leaks memory. You need to call delete once for every new call (assuming the new doesn't fail and either return NULL or throw bad_alloc). So, in this case, you'll need to have a similar loop to the one that calls array[i] = new Foo; that does delete array[i];
I think this is one of points in Scott Meyers book, Effective C++ have read long back. But basic idea is, to avoid memory leak when you use new. make sure to call delete as many times as new is called and map new with delete and new[] with delete[] accordingly. so in your case new is called 128 times in for loop and new[] is called once, so similarly delete should be called same way to array[i] 128 times and delete[] should be called once for array. Hope this helps.
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.
I have made for school purposes my own take on a dynamically allocated array using templates.
While what I'm about to ask works, I don't know how and why and I've reached the point where I need to know.
template <typename TElement>
DynamicArray<TElement>::ensureCapacity () {
if (capacity >= elemNumb) {
return; //we have space to store the values
}
//we need to allocate more space for the values
TElement *auxArray = myArray;
//create space to hold more numbers
capacity = capacity * 2;
myArray = new TElement[capacity];
//copy the values
for (int i = 0; i < size; i++) {
myArray[i] = auxArray[i];
}
//release the memory
delete[] auxArray;
}
I need to know: TElement *auxArray = myArray; How does this work ? is it using pointers, are elements copied one by one ? I need to understand how it works so that I can figure out the complexity of my algorithm. I don't mind if some one tells me the complexity but the real answer I'm looking for is how does that work ?
Also myArray = new TElement[capacity]; I do this before deleting the old myArray does this delete the old one ? or is it still floating somewhere in memory in one form or another ?
This
TElement *auxArray = myArray;
just means that auxArray points to whatever myArray is pointing to. There is no copying of anything else, it is just a pointer copy.
This
myArray = new TElement[capacity];
means that myArray now points to a new, dynamically allocated TElement array. The expression doesn't delete anything. But auxArray is pointing to what myArray was pointing before this assignment, so when you delete auxArray, you release the resources originally pointed to by myArray.
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.