Re: [boost] variant invariant

From: Emil Dotchevski (emil_at_[hidden])
Date: 2008-08-04 20:43:17

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? If that's the case, then maybe
I missed something in the documentation.

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