4 Aug
2008
4 Aug
'08
11:35 p.m.
AMDG Peter Dimov wrote:
Look, if I have:
variant<foo,int> a=foo(....); variant<foo,int> b=foo(....);
It is my understanding that if I now do:
a=b;
I may end up with an int in a. Do you think that this behavior is reasonable?
My naive expectation would be for the above code to call foo::operator=. This would of course imply that variant::operator= would require the types to be assignable.
It does call operator= when both operands have the same type. Thanks.
That aside, I think that it would be reasonable to expect the type to remain foo at least when foo has a nothrow default constructor.
Yes. It shouldn't add runtime overhead to try to fall back to the original type, if possible. In Christ, Steven Watanabe