explicit copy constructor

[=?gb2312?B?wfXquw==?=]

Hi, folks,

I am trying to make my copy constructor explicit, but when the
scenario below comes into being, weird things happen.
Here is the code snippet:
#include <iostream>
using namespace std;

class Base {
public:
Base() {
}
Base(const Base& other) {
}
};

template<typename T>
class Derived: public Base {
public:

Derived() {
}

template<typename U>
Derived(const Derived<U>& other): Base(*(Base*)&other) {
cout << "Cast copy constructor\n";
}

//here is the trouble maker
explicit Derived(const Derived& other): Base(*(Base*)&other) {
cout << "Normal copy constructor\n";
}
};

int main() {

Derived<intd1;

Derived<intd2 = d1;

return 0;
}

Intuitively, I want "Normal copy constructor" printed, but what I saw
is "Cast copy constructor", while I remove "explicit" keyword,
everything is OK, the output is "Normal copy constructor", any
informative tips will receive my big thanks.

On 8 10 , 11 53 , "Alf P. Steinbach" <[e-mail]:

An explicit constructor must be called explictly.
You don't call it explicitly, so how can you expect it to be called?

So, you mean if I define a explicit copy constructor, there can not be
any temporary object in my code, right?

Apart from that, first, you really don't want to use reinterpret_cast as
you do above, and secondly, the templated constructor isn't a copy
constructor -- a "copy constructor" is by definition not templated.

The templated constructor is used for type casting from Derived<Uto
Derived<T>.

Thanks for your informative tips.
Big thanks goes to you.