Delete element of array type pointer cause error?

Question

I have an example:

int* arr = new int[1];
//First Time
arr[0] = 5;
cout << &arr[0] << " " << arr[0] << endl;
delete &arr[0];
cout << &arr[0] << " " << arr[0] << endl;

//Second time
arr[0] = 5;
cout << &arr[0] << " " << arr[0] << endl;
delete &arr[0]; //Error from here
cout << &arr[0] << " " << arr[0] << endl;

The result from the code above:

00529EF0 5
00529EF0 -572662307
00529EF0 5
//Where the code stop

I don't know why it causes me the error? It doesn't let me delete arr[0] the second time, even though the first time successfully deleted and I assign a new value for it.

Just because it looks like the first delete worked, doesn't mean it did. What are you trying to accomplish? — Stephen Newell
– Stephen Newell, Commented Mar 24, 2020 at 23:55
You delete something and then proceed as if that something is still valid. Basic recipe for things blowing up. I think it's on you to explain why you think this should work. (Forget the second deletion; why should something of the form delete &X; followed by X = 5; ever work?) — JaMiT
– JaMiT, Commented Mar 25, 2020 at 0:35

user9254539 · Accepted Answer · 2020-03-24 23:58:24Z

4

If you dynamically allocate an array with new[], you have to delete it with delete[], which will free the entire array. You can't free individual elements. Your code has undefined behavior because you used delete instead of delete[], so anything can happen. Just because it appeared to work doesn't mean it's correct.

answered Mar 24, 2020 at 23:58

user9254539

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1 Comment

user4581301 Over a year ago

Undefined Behaviour. There is likely nothing in programming that is more fearsome.

Remy Lebeau · Accepted Answer · 2020-03-25 00:34:48Z

Undefined Behavior.

&arr[0] is the same memory address that arr decays to. So you are really calling delete arr;. This is undefined behavior because you are using delete when you need to use delete[] instead. Memory allocated with new MUST be freed with delete. Memory allocated with new[] MUST be freed with delete[].

But, lets assume your compiler's runtime has enough info in the allocated array's metadata to free the array correctly. delete'ing an array free the entire array. You can't free individual elements of the array. And accessing any element of of array after freeing the entire array is undefined behavior. You are accessing arr[0] after calling delete arr; to free the array. You are not allocating a new array for arr to point at. You are lucky your code did not just crash outright.

That is the definition of undefined behavior - ANYTHING could happen. You might read the old memory before it is overwritten. You could read from random memory. You could corrupt memory. You could crash.

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