29. Can we define new operation?

Is this example more or less complete?

#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;
class MyClass
{
public:
        MyClass(){ cout << "MyClass::MyClass() is called" << endl;}
        void doSmth() { cout << "MyClass::doSmth() is called" << endl;}
        void* operator new(size_t);
        void operator delete(void*);
};

void* MyClass::operator new(size_t size)
{
    void *storage = malloc(size);


    cout << "Placement new operator is called" << endl;

    if(NULL == storage) {
            throw "allocation fail : no free memory";
    }
    return storage;
}

void MyClass::operator delete(void *obj)
{
    cout << "Delete operator is called" << endl;
    free(obj);

    obj = NULL;
}

int main()
{

    MyClass *mc = new MyClass;

    mc->doSmth();

    delete mc;

}

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>>> There is also a so-called 'placement new' that allows you to create an object on a certain memory area that you can choose.

How we can create an object on a certain memory area?

thank you, will  review that a bit later

Yes, I have checked obj=NULL  does not affect the original pointer.
When I tried to change to void** , it started to blame:

new_del.cpp:13:30: error: ‘operator delete’ takes type ‘void*’ as first parameter
   void operator delete(void**);

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Yes, we may overload the new and delete, but are those operator's prototypes are "carved in stone"?

int buffer[256]; // providing 'sizeof(MyClass)' only would work with gcc/g++

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I don't get this comment. PLease, explain.