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


Question

Alright, I think we all agree that what happens with the following code is undefined, depending on what is passed,

void deleteForMe(int* pointer)
{
     delete[] pointer;
}

The pointer could be all sorts of different things, and so performing an unconditional delete[] on it is undefined. However, let's assume that we are indeed passing an array pointer,

int main()
{
     int* arr = new int[5];
     deleteForMe(arr);
     return 0;
}

My question is, in this case where the pointer is an array, who is it that knows this? I mean, from the language/compiler's point of view, it has no idea whether or not arr is an array pointer versus a pointer to a single int. Heck, it doesn't even know whether arr was dynamically created. Yet, if I do the following instead,

int main()
{
     int* num = new int(1);
     deleteForMe(num);
     return 0;
}

The OS is smart enough to only delete one int and not go on some type of 'killing spree' by deleting the rest of the memory beyond that point (contrast that with strlen and a non-\0-terminated string -- it will keep going until it hits 0).

So whose job is it to remember these things? Does the OS keep some type of record in the background? (I mean, I realise that I started this post by saying that what happens is undefined, but the fact is, the 'killing spree' scenario doesn't happen, so therefore in the practical world someone is remembering.)

1
133
10/2/2016 11:57:20 AM

Accepted Answer

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.

98
9/10/2015 5:19:13 PM

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.


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