Re: [boost] variant invariant

From: Peter Dimov (pdimov_at_[hidden])
Date: 2008-08-04 19:11:18

Steven Watanabe:

> Emil Dotchevski 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?
>>
>
> You won't just end up with an int in a.
> You will also end up with an exception.
> Yes.  I think 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.

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.