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April 2010

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[boost::numeric::functional::plus_base] Why the arguments are not const reference
by Kan Li 01 Apr '10

01 Apr '10

It seems like the operators define in boost::numeric::functional takes reference as argument instead of const reference. Which makes it compile error if I use boost::numeric::functional::plus<int, int> op; int n = op(1,1); The way to work around is boost::numeric::functional::plus<const int, const int> op; int n = op(1,1); This is very ugly and not makes much sense, compared to std::plus. So is it a design flaw or is there some other reason for designing it in this way?

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