Re: [boost] variant invariant

From: Emil Dotchevski (emil_at_[hidden])
Date: 2008-08-04 21:35:26

On Mon, Aug 4, 2008 at 6:08 PM, Steven Watanabe <watanabesj_at_[hidden]> wrote:
> Emil Dotchevski wrote:
>>
>> On Mon, Aug 4, 2008 at 4:35 PM, Steven Watanabe <watanabesj_at_[hidden]>
>> wrote:
>>
>>>
>>> It does call operator= when both operands have the same type.  Thanks.
>>>
>>
>> Is it then true that even foo::operator= throws, in the code below:
>>
>> variant<foo,int> a=foo(....);
>> variant<foo,int> b=foo(....);
>> a=b;
>>
>> we will never end up with an int in a?
>
> correct.

Cool, thanks!

Emil Dotchevski
Reverge Studios, Inc.
http://www.revergestudios.com/reblog/index.php?n=ReCode