Re: Question on auto_ptr, Which function will call first?

From:
=?iso-8859-2?B?Smn47SBQYWxl6GVr?= <jpalecek@web.de>
Newsgroups:
comp.lang.c++
Date:
Fri, 03 Oct 2008 20:21:58 +0200
Message-ID:
<op.uignywzcu2flwt@localhost>
On Fri, 03 Oct 2008 18:00:28 +0200, asm23 <asmwarrior@gmail.com> wrote:

Hi, everyone, I'm studying the <<Thinking in C++>> volume Two. In
Chapter One, the example code : Auto_ptr.cpp
//-------------------------------------------------------
#include <memory>
#include <iostream>
#include <cstddef>
using namespace std;

class TraceHeap {
    int i;
public:
    static void* operator new(size_t siz) { //*****NOTE A
        void* p = ::operator new(siz);
        cout << "Allocating TraceHeap object on the heap "
            << "at address " << p << endl;
        return p;
    }
    static void operator delete(void* p) {
        cout << "Deleting TraceHeap object at address "
            << p << endl;
        ::operator delete(p);
    }
    TraceHeap(int i) //*******NOTE B
        : i(i)
    {
        ;
    }
    int getVal() const { return i; }
};

int main() {
    auto_ptr<TraceHeap> pMyObject(new TraceHeap(5));
    cout << pMyObject->getVal() << endl; // Prints 5
}
//------------------------------------------------------------------

My question is :

In My code: which code will be called first? The *NOTE A* or *NOTE B* ?
And Why?

I'm debugging through this code and found that *NOTE A* will called
first. Can someone explained it?


I haven't looked up the standard, but common sense tells the constructor
cannot be called before memory allocation, as it needs to construct an
object in the memory allocated by the new operator.

When I trace into the *new* function, the *siz* value is 4. I don't know
why it will be 4?


That's sizeof(TraceHeap)

Regards
     Jiri Palecek

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