Re: [boost] [review] Multiprecision review scheduled for June 8th - 17th, 2012

Subject: Re: [boost] [review] Multiprecision review scheduled for June 8th - 17th, 2012
From: Marc Glisse (marc.glisse_at_[hidden])
Date: 2012-06-02 12:24:02

On Fri, 1 Jun 2012, John Maddock wrote:

>>> * the example about using 0 vs 11 temporaries doesn't mention what happens 
>>> to the boost package if we disable expression template. Does it use a 
>>> single temporary (using rvalue-references) or plenty?
>> 
>> Plenty, haven't measured how many, but I'd expect it to be the same as for 
>> mpclass.  I guess I should cover that.
>> 
>> I think you really want to investigate using rvalue references, in that 
>> case. Which reminds me: there is some overlap with boost/operators.hpp (I 
>> heard someone was rewriting it?), so the rvalue reference techniques could 
>> be the same, and the documentation could say something about the relation 
>> between the 2.
>
> rvalue references and move semantics are already supported.

You have a move constructor, but I haven't seen anything like:

number operator+(number&& a, number const&b){return move(a+=b);}

which would bring the number of allocations down to 1 for types with a 
good move constructor (I may have missed it, I can't easily test anything 
these days).

(returning a number&& is tempting, especially for types with a bad move 
constructor, but is not very safe)

> Nod.  Support for fused multiply-add is on my TODO list.

;-) My point was actually that FMA shouldn't have anything special. What 
makes it interesting in GMP is that malloc is replaced with alloca for the 
temporary. Otherwise, it is a plain mul and add. The same optimization 
could just as well have been done for addadd (a+b+c) instead. With 
expression templates, there should be enough information to do that for 
any combination, not just a+b*c. It may get quite complicated though, and 
not so useful for a backend like cpp_int.

What happens when you try to mix numbers, say z+q where z is an integer 
and q a rational? I imagine that z is converted to a rational and the user 
can't provide an optimized version of the operation (maybe for a future 
version?).

-- 
Marc Glisse