5 Aug
2008
5 Aug
'08
1:35 a.m.
On Mon, Aug 4, 2008 at 6:08 PM, Steven Watanabe <watanabesj@gmail.com> wrote:
Emil Dotchevski wrote:
On Mon, Aug 4, 2008 at 4:35 PM, Steven Watanabe <watanabesj@gmail.com> 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