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How does delete[] know it's an array?

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How does delete [] know how many elements to delete?

new[] also stores the number of elements it created in the memory block (independently of malloc ), so that later delete[] can retrieve and use that number to call the proper number of destructors.

What does delete [] array do in C++?

Delete is an operator that is used to destroy array and non-array(pointer) objects which are created by new expression. Delete can be used by either using Delete operator or Delete [ ] operator. New operator is used for dynamic memory allocation which puts variables on heap memory.

How delete [] is different from delete?

delete is used for one single pointer and delete[] is used for deleting an array through a pointer.

How does the compiler know that when we call the delete operator how many bytes of memory to delete?

The compiler doesn't allocate the memory nor does the compiler read it.


One question that the answers given so far don't seem to address: if the runtime libraries (not the OS, really) can keep track of the number of things in the array, then why do we need the delete[] syntax at all? Why can't a single delete form be used to handle all deletes?

The answer to this goes back to C++'s roots as a C-compatible language (which it no longer really strives to be.) Stroustrup's philosophy was that the programmer should not have to pay for any features that they aren't using. If they're not using arrays, then they should not have to carry the cost of object arrays for every allocated chunk of memory.

That is, if your code simply does

Foo* foo = new Foo;

then the memory space that's allocated for foo shouldn't include any extra overhead that would be needed to support arrays of Foo.

Since only array allocations are set up to carry the extra array size information, you then need to tell the runtime libraries to look for that information when you delete the objects. That's why we need to use

delete[] bar;

instead of just

delete bar;

if bar is a pointer to an array.

For most of us (myself included), that fussiness about a few extra bytes of memory seems quaint these days. But there are still some situations where saving a few bytes (from what could be a very high number of memory blocks) can be important.


The compiler doesn't know it's an array, it's trusting the programmer. Deleting a pointer to a single int with delete [] would result in undefined behavior. Your second main() example is unsafe, even if it doesn't immediately crash.

The compiler does have to keep track of how many objects need to be deleted somehow. It may do this by over-allocating enough to store the array size. For more details, see the C++ Super FAQ.


Yes, the OS keeps some things in the 'background.' For example, if you run

int* num = new int[5];

the OS can allocate 4 extra bytes, store the size of the allocation in the first 4 bytes of the allocated memory and return an offset pointer (ie, it allocates memory spaces 1000 to 1024 but the pointer returned points to 1004, with locations 1000-1003 storing the size of the allocation). Then, when delete is called, it can look at 4 bytes before the pointer passed to it to find the size of the allocation.

I am sure that there are other ways of tracking the size of an allocation, but that's one option.


This is very similar to this question and it has many of the details your are looking for.

But suffice to say, it is not the job of the OS to track any of this. It's actually the runtime libraries or the underlying memory manager that will track the size of the array. This is usually done by allocating extra memory up front and storing the size of the array in that location (most use a head node).

This is viewable on some implementations by executing the following code

int* pArray = new int[5];
int size = *(pArray-1);

delete or delete[] would probably both free the memory allocated (memory pointed), but the big difference is that delete on an array won't call the destructor of each element of the array.

Anyway, mixing new/new[] and delete/delete[] is probably UB.


It doesn't know it's an array, that's why you have to supply delete[] instead of regular old delete.


I had a similar question to this. In C, you allocate memory with malloc() (or another similar function), and delete it with free(). There is only one malloc(), which simply allocates a certain number of bytes. There is only one free(), which simply takes a pointer as it's parameter.

So why is it that in C you can just hand over the pointer to free, but in C++ you must tell it whether it's an array or a single variable?

The answer, I've learned, has to do with class destructors.

If you allocate an instance of a class MyClass...

classes = new MyClass[3];

And delete it with delete, you may only get the destructor for the first instance of MyClass called. If you use delete[], you can be assured that the destructor will be called for all instances in the array.

THIS is the important difference. If you're simply working with standard types (e.g. int) you won't really see this issue. Plus, you should remember that behavior for using delete on new[] and delete[] on new is undefined--it may not work the same way on every compiler/system.