Martin Kittel wrote:
// works fine
A a;
m_a = a;
// needs access to copy-constructor, why?
m_a = A();
This is about reference binding of parameters. 8.5.3 says how references are
initialized. in particular, 'a' is a non-temporary lvalue so the reference can
be bound directly to it -- no copy ctor required.
'A()' is a temporary rvalue, so we can either
a) bind the reference directly to the rvalue, or
b) create a temporary (with the copy ctor) and then bind the reference to that.
(footnote 93 points out that option b will recurse so eventually we have to
choose option a.)
Regardless of whether we immediately choose option a or not, the constructor we
would have called had we chosen option b must be callable. This rule means your
code will still compile on a lamer compiler that did actually copy the object,
even if you developed it on a better compiler that immediatly bound the reference.
This final check is the 'surprise', and g++ is correct in performing it.
nathan
--
Nathan Sidwell :: http://www.codesourcery.com :: CodeSourcery LLC
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