Re: use of copy constructor with default arguments
* Andrey Vul:
On Oct 19, 6:51 pm, "Alf P. Steinbach" <al...@start.no> wrote:
* Andrey Vul:
Given the following:
class Bar {};
class Foo {
Bar *parent;
Foo(Bar *p, const Foo& f = 0) {
parent = p;
if (f != 0) {
/* do something with f */
}
}
};
Why does the compiler die while Foo& =0 is used, but works when Foo*
=0 is used?
What you wrote is equivalent to
Foo( Bar* p, Foo const& f = Foo( 0 ) )
which if it was valid would be equivalent to
Foo( Bar* p, Foo const& f = Foo( 0, Foo( 0 ) ) )
which if it was valid would be equivalent to
Foo( Bar* p, Foo const& f = Foo( 0, Foo( 0, Foo( 0 ) ) ) )
and so on.
So null [pointer] / pure virtual semantic defaults don't work with
references?
Formally there is no nullvalue for a reference.
That's much of the point of references.
Re "pure virtual semantics defaults" I think you're referring to the notation "=
0". There's no such thing for references. It can be meaningful and accepted, but
for a T& reference it then means to construct a T object with 0 as constructor
argument, T(0), since a reference must refer to an object.
Or do they only work when the first argument to the constructor is not
a pointer?
Or does the C++ standard specify nasal demons?
If you happen to dereference a nullpointer you get what in practice would be a
null-reference, except that you've then already crossed the border to Undefined
Behavior land (nasal demons), yes. :-)
Cheers & hth.,
- Alf
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