What is wrong with using delete
instead of delete[]
?
Is there something special happening under the covers for allocating and freeing arrays?
Why would it be different from malloc
and free?
What is wrong with using delete
instead of delete[]
?
Is there something special happening under the covers for allocating and freeing arrays?
Why would it be different from malloc
and free?
Objects created with new[]
must use delete[]
. Using delete
is undefined on arrays.
With malloc and free you have a more simple situation. There is only 1 function that frees the data you allocate, there is no concept of a destructor being called either. The confusion just comes in because delete[]
and delete look similar. Actually they are 2 completely different functions.
Using delete won't call the correct function to delete the memory. It should call delete[](void*)
but instead it calls delete(void*)
. For this reason you can't rely on using delete
for memory allocated with new[]
[16.13] Can I drop the
[]
when deleteing array of some built-in type (char, int, etc)?No!
Sometimes programmers think that the
[]
in thedelete[] p
only exists so the compiler will call the appropriate destructors for all elements in the array. Because of this reasoning, they assume that an array of some built-in type such aschar
orint
can bedelete
d without the[]
. E.g., they assume the following is valid code:void userCode(int n) { char* p = new char[n]; ... delete p; // ← ERROR! Should be delete[] p ! }
But the above code is wrong, and it can cause a disaster at runtime. In particular, the code that's called for
delete p
isoperator delete(void*)
, but the code that's called fordelete[] p
isoperator delete[](void*)
. The default behavior for the latter is to call the former, but users are allowed to replace the latter with a different behavior (in which case they would normally also replace the corresponding new code in operatornew[](size_t)
). If they replaced thedelete[]
code so it wasn't compatible with the delete code, and you called the wrong one (i.e., if you saiddelete p
rather thandelete[] p
), you could end up with a disaster at runtime.
Why does delete[]
exist in the first place?
Whether you do x or y:
char * x = new char[100];
char * y = new char;
Both are stored in char *
typed variables.
I think the reason for the decision of delete
, and delete[]
goes along with a long list of decisions that are in favor of efficiency in C++. It is so that there is no enforced price to do a lookup of how much needs to be deleted for a normal delete operation.
Having 2 new
and new[]
seems only logical to have delete
and delete[]
anyway for symmetry.