CODEWITHSUNDEEP
c++language-designdynamic-memory-allocationdelete-operator
Michael0x2a
Michael0x2a

Reputation: 64328

Why does the compiler require `delete [] p` versus `delete p[]`?

In C++, if you want to dynamically allocate an array, you can do something like this:

int *p;
p = new int[i];  // i is some number

However, to delete the array, you do...

delete[] p;

Why isn't it delete p[]? Wouldn't that be more symmetrical with how it was originally created? What is the reason (if there is any) why the language was designed this way?

Upvotes: 16

Views: 1004

Answers (2)

user396672
user396672

Reputation: 3166

Because array decays to pointer when passing as parameter to function (or operator). So delete p[] would be simply equivalent to delete p.

[edit] We may think of delete as of special template operator. The operator should be able to distingish between p and p[] to pick the right "specialization" (non-array or array deletion). However, template argument deduction rules make this choice impossible (due to array decaying we can't distingish between p[] and p when deducing the argument).

So we can't use operator with name delete for both casees and need to introduce another operator delete[] with diffrent name (suffix [] can be treated as part of operator's name) for array case.

[edit 2] Note. delete p[] is not valid sintax at all according to the current standard. The reasoning above only shows problems that could araise if we would try to interpret delete p[] using existing c++ concepts.

Upvotes: 2

Michael Anderson
Michael Anderson

Reputation: 73580

One reason could be to make these cases more distinct.

int ** p;
delete[] p
delete p[1];

If it were delete p[] then a one character error would have pretty nasty consquences.

Upvotes: 29

Related Questions