Reading boost::variant's documentation: "If any bounded type is nothrow default-constructible (as indicated by boost::has_nothrow_constructor), the library guarantees variant will use only single storage and in-place construction for every bounded type in the variant. Note, however, that in the event of assignment failure, an unspecified nothrow default-constructible bounded type will be default-constructed in the left-hand side operand so as to preserve the never-empty guarantee." Do I read correctly that in an attempt to assign one variant object to another, the left-hand object _might_ silently change its type? This makes it very difficult for the user to enforce some useful invariants for types that have boost::variant members, which in my mind is very important. Wouldn't it be better to guarantee that operator= either succeeds or doesn't change the observable state of the left-hand side operand, even at the cost of memory allocation? In that case the other behavior may still be provided through a separate function. Emil Dotchevski Reverge Studios, Inc. http://www.revergestudios.com/reblog/index.php?n=ReCode