Re: [boost] variant invariant

AMDG 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. In Christ, Steven Watanabe