Michael Gong schrieb:
> Hi, all,
>
> Thanks for the explanation. After google it, it looks like the explicit
> type conversion also works on user-defined type.
Yes, it is basically creating a new instance of some type by
constructing it. No more, no less.
Unless a constructor has been declared with the "explicit" keyword C++
applies this type conversion even implicitly if required:
struct Bar{};
struct Baz{};
struct Foo{
Foo(const Bar&);
explicit Foo(const Baz&);
};
void f( Foo );
int main() {
Bar bar;
Baz baz;
f( bar ); // works, as compiler implicitly generates f(Foo(bar))
f( baz ); // does not work, compiler is forbidden to use constructor
// for implicit type conversions
f( Foo( baz ) ); // works, we invoke constructor explicitly
}
> Meanwhile, to be more correct, I think it is not a "copy-constructor" ,
> but just a constructor call.
In the int('A') example it is the copy-constructor. The compiler
provides automatically generated default- and copy-constructors for any
type that has no user-defined constructor.
If some constructor is defined explicitly, copy- and default-constructor
are no longer generated by the compiler. Your Point class therefore has
*no* default-constructor.
> For example:
>
> class Point {
> private :
> int x, y;
> public:
> Point(int x_, int y_) {
> cout << "default constructor" << "\n";
> }
>
> Point(const Point& p) {
> cout << "copy constructor" << "\n";
> }
> };
>
> int main() {
> Point(3, 4); /* it will call the default constructor */
> }
>
Daniel
