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From: Joel de Guzman (joel_at_[hidden])
Date: 2006-05-03 18:25:22

John Christopher wrote:
> Joel,
> I was wondering if "at_c" is only a convenience version of "at" or there is
> a deeper difference?
> In other words, is
> std::cout << at_c<0>(v) << std::endl;
> fully equivalent to
> std::cout << at<boost::mpl::int_<0> >(v) << std::endl;

Yes, they are fully equivalent. Yes, at_c is only a convenience
version of at. at is more suitable for MPL style metaprogramming
when the computation of the index is involved.

> Thanks for this very nice library

Most welcome! :)


Joel de Guzman

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