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From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2008-08-04 21:08:56


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

This behavior is stated here:

The discussion here
distinguishes between same-type and different-type assignments, too.

In Christ,
Steven Watanabe

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