Re: diff between user defined delete and delete[]

From:

Sachin <sachinc.biradar@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 17 Nov 2008 02:51:55 -0800 (PST)

Message-ID:

<d05e32b2-d32b-444d-9ba9-888ccd9d613f@i20g2000prf.googlegroups.com>

On Nov 17, 2:44 pm, Salt_Peter <pj_h...@yahoo.com> wrote:

On Nov 17, 4:13 am, mail....@gmail.com wrote:

Suppose in a class we overload four operators:
operator new
operator delete
operator new[]
operator delete[]

class Test{
public:
        void * operator new (size_t t){
                cout<<"\nCalling... new";
                return malloc(t);
        }
        void operator delete (void *p){
                cout<<"\nCalling ... delete";
                free(p);
        }
        void * operator new [] (size_t t){
                cout<<"\nCalling ... new[]";
                return malloc(t);
        }

        void operator delete [] (void *p){
                cout<<"\nCalling ... delete[]";
                free(p);
        }

};

If we perform
Test *p=0; delete p;
It calls operator delete.

But if we perform
Test *p=0; delete []p;
It doesn't call operator delete[] until and unless we don't call
operator new[]. Means If we do like this;
Test *p=0; p=new Test[10]; delete[] p; It calls operator de=

lete[].

Why???

The FAQ covers the subject. Read the 2 common techniques used with
primitive arrays

[16.14] After p = new Fred[n], how does the compiler know there are n
objects to be destructed during delete[] p?http://www.parashift.com/c++-f=

aq-lite/freestore-mgmt.html#faq-16.14

Although in modern code unless there is a really good reason:
a) don't allocate on the heap manually, use smart pointers or RAII
b) prefer dynamic containers over primitive, fixed arrays
(std::vector, std::deque, etc)- Hide quoted text -

- Show quoted text -

VC++ Compiler does call delete[], even if you dont call new[]
explicitly.

Thx,
Sachin