Re: [Boost-users] converting to float128 from double

Subject: Re: [Boost-users] converting to float128 from double
From: John Maddock (jz.maddock_at_[hidden])
Date: 2016-05-04 07:29:31

On 03/05/2016 20:58, Cooper, Bridgette R D wrote:
> Hi John,
>
> No, my code is several hundreds of lines long. I extracted what I thought would be quite self contained, but yeah you are right about the return type mis-match.
>
> I am not sure that this helps. What I really want is to do a lot of computation in float128 precision, and for some reason when I cast up a double vector to a float128 vector  all the operations on the float128 vector are happening a double precision, and the cast up only happens at the final stage.
>
> I've changed the initiation of epoint and gpoint vectors to the same as your code snippet and still the same problem.
>
> So for example, if I accumulate the original double vector without casting the values to float128 type and accumulate into a float128 variable, the cast up only happens to the result. This is the same number that I see as the result of accumulating the float128vector values.
>
> Hope this clarifies things a bit

Not really :(

I think you're going to have to try to produce a self contained test 
case - my guess is you'll spot the error in the process of doing that?

Best, John.

> ________________________________________
> From: Boost-users <boost-users-bounces_at_[hidden]> on behalf of John Maddock <jz.maddock_at_[hidden]>
> Sent: Tuesday, May 3, 2016 6:27:39 PM
> To: boost-users_at_[hidden]
> Subject: Re: [Boost-users] converting to float128 from double
>
> On 03/05/2016 17:35, Cooper, Bridgette R D wrote:
>> Hi John,
>>
>> Thanks for the response.
>>
>> Here are some more snippets from my code.
> I don't see any obvious errors, but in any case you must have extracted
> that from something else but it doesn't compile!
>
> Not least you can't implicitly return from a float128 to a double in
>
>       return(bcoef.at(0))
>
> So if that's compiling for you, then I suspect your declaration of bcoef.
>
> If I re-write and simplify in a slightly more C++ idiom how does it
> compare to your fortran code now:
>
> double stieltjes_gamma(const std::vector<double> & e8, const
> std::vector<double> & g8, int printlevel)
> {
>      assert(e8.size() == g8.size());  // precondition?
>      std::vector<float128> epoint(e8.begin(), e8.end()),
> gpoint(g8.begin(), g8.end()), bcoef(e8.size());
>
>      bcoef[0] = 0.0Q;
>      for(int i = 0; i < e8.size(); i++)
>      {
>         bcoef[0] += gpoint[i];
>      }
>      std::cout <<
> std::setprecision(std::numeric_limits<float128>::max_digits10) <<
> bcoef.at(0) << std::endl;
>      return static_cast<double>(bcoef[0]);
> }
>
> And finally... note that summing at higher precision can not actually
> protect you from cancellation error (in any language) since the error is
> inherent in the input values.
>
> HTH, John.
>
>
>> #include <boost/multiprecision/float128.hpp>
>> #include <boost/math/cstdfloat/cstdfloat_types.hpp>
>> #include <boost/math/cstdfloat/cstdfloat_limits.hpp>
>> #include <boost/math/cstdfloat/cstdfloat_cmath.hpp>
>> #include <boost/math/cstdfloat/cstdfloat_iostream.hpp>
>> #include <boost/multiprecision/detail/float_string_cvt.hpp>
>> #include <boost/multiprecision/detail/generic_interconvert.hpp>
>> extern "C"{
>> #include <quadmath.h>
>> }
>> #include <iostream>
>> #include <sstream>
>> #include <vector>
>> #include <iterator>
>>
>>           using namespace boost::multi precision;
>>
>>
>>    double stieltjes_gamma(int num, const std::vector<double> & e8, const std::vector<double> & g8, int printlevel )
>> {
>>
>>          std::vector<float128> epoint(num), gpoint(num), bcoef(num);
>>
>>
>>
>>          for (int i=0; i < num; i++)
>>                   {
>>                   epoint.at(i)=(float128(e8.at(i)));
>>                   gpoint.at(i)=(float128(g8.at(i)));
>>                   }
>>
>>    bcoef.at(0)=0.0Q;
>>           for (int i = 0; i < num; i++)
>>                   {
>>                   bcoef.at(0)+=gpoint.at(i)
>>                   }
>>          std::cout << std::setprecision(std::numeric_limits<float128>::max_digits10)  << bcoef.at(0)<< std::endl;
>>       return(bcoef.at(0))
>> }
>>
>> When I compare this to fortran code that does the same I don't see the same value for bcoef.at(0).
>>
>> 0.0345217724606016853085032291117978629  <- from above c++ code
>> 3.452175962187407162907246287483402E-0002   <- Fortran correctly utilising quad precision on same input vector
>>
>> The input vectors in this case are 7976 so errors in precision are really quickly apparrant. If I print  the gpoint vector, it gets filled with the same garbage for additional precision from double  as is for the fortran.
>>
>> Thanks,
>> ________________________________________
>> From: Boost-users <boost-users-bounces_at_[hidden]> on behalf of John Maddock <jz.maddock_at_[hidden]>
>> Sent: Tuesday, May 3, 2016 5:06:47 PM
>> To: boost-users_at_[hidden]
>> Subject: Re: [Boost-users] converting to float128 from double
>>
>> On 03/05/2016 13:44, Cooper, Bridgette R D wrote:
>>> Hi,
>>>
>>>
>>> I have a function that takes as an argument a vector of doubles and
>>> tries to convert it to a vector of float128.
>>>
>>>
>>> When I accumulate the values in the float128 vector it looks like it
>>> does double additions and only casts up at the last step.
>>>
>>>
>>> The initial vector is defined as conststd::vector<double>
>>>
>>> The float128 vector is defined as std::vector<float128> epoint(num)
>>>
>>>
>>> And I try to do the casting (inside a loop) via:
>>> epoint.at(i)=(float128(e8.at(i)));
>>>
>>>
>>> If I do the accumulation on the original vector with no casting into a
>>> float128, I get the same as accumulating the vector that should be of
>>> float128 type.
>>>
>>>
>>> How should I be doing the casting?
>>>
>> Sorry but you'll have to post way more code than that to see where the
>> mistake is.
>>
>> John.
>>
>>> Thanks,
>>>
>>>
>>>
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