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From: Jesse Perla on 27 May 2010 17:05
I have some functors that have more data that I might wish and
algorithms that execute them (e.g. some weird multidimensional
interpolators, etc.). I wrote some algorithms that accept these by
constant reference, with the assumption that the function supplies a
const operator(). i.e. stuff like:
template<typename F> double f(const F& f)
{
return f(.1);
};

I realize that the std algorithms always pass functors by value. What
is the reasoning for this with stateless algorithms like find
predicates, transform, etc. (I can see why std::generate might want a
copy by value)? Is it just to allow the functors to have a non-const
operator()? Are there any other reasons, such as lack of inlining
with modern compilers, that I can't pass const&?

-Jesse

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