'New' operator

[Bit byte]

I have written a trivial class that overloads the new operator:

class A {
public:

A();
void* operator new(size_t size);
void operator delete(void* p);

int getX(void);
void setX(int i);
private:
int x ;
};
I set a breakpoint in my implementation for the new operator. Some where
in my code, I have a line like this:

A *a_obj = new A ;

The size argument passed to my 'new' function was 4 (presumably bytes?)
I am guessing that this is the sizeof(A)?

Is this the general case?. I mean what if I had a much more complecated
class (say with nested pointers and structures etc?).

Is it really sizeof(A) that is being passed to my 'new' function?

[Michiel.Salters]

Bit byte wrote:

I have written a trivial class that overloads the new operator:

class A {
public:

A();
void* operator new(size_t size);
void operator delete(void* p);

int getX(void);
void setX(int i);
private:
int x ;
};
I set a breakpoint in my implementation for the new operator. Some where
in my code, I have a line like this:

A *a_obj = new A ;

The size argument passed to my 'new' function was 4 (presumably bytes?)
Yes. In C++, sizeof(char) is 1 byte, by definition, and every size is a
multiple
of that.
I am guessing that this is the sizeof(A)?
Yes, since that's what you're constructing.
Is this the general case?. I mean what if I had a much more complecated
class (say with nested pointers and structures etc?).

Is it really sizeof(A) that is being passed to my 'new' function?

If A was more complicated, it would still hold. However, there is an
important
exception. If someone derives class B from A, and adds members, you
might see sizeof(B). This happens if class B doesn't override
A::operator new.
In that case A::operator new is inherited like any other function.

I haven't ever seen an exception to sizeofB) >= sizeof(A), but there
might one.

HTH,
Michiel Salters