Re: [boost] variant invariant

From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2008-08-04 19:35:30

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