From: David Abrahams (dave_at_[hidden])
Date: 2004-07-28 10:23:21
Anthony Williams <anthony_w.geo_at_[hidden]> writes:
> The default implementation of swap is generally along the lines of
> template<typename T>
> void swap(T& lhs,T& rhs)
> T temp(lhs);
> The initialization of temp creates a genuine copy of the data.
How so? Didn't you say that *i is a reference to a tuple of
references? That makes T a tuple of references, and the
initialization of a tuple of references just copies a bunch of
> The two assignments then use the assignment operator of boost::tuple
> to assign through the references.
> Does that make it clearer?
-- Dave Abrahams Boost Consulting http://www.boost-consulting.com
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