Exercise for the students - get it to work with UDT ;-) Darwin long double is really a doubledouble - I think, so really a UDT, for which getting the righ type is critical. NTL quad_float is the same - I think. First try gave 0 for epsilon, after changing to use ALL quad_float I've got as far as: quad_float epsilon_quad_float() { using NTL::to_quad_float; quad_float one; one = to_quad_float("1.0"); quad_float two; one = to_quad_float("2.0"); quad_float eps = one; quad_float onepluseps; do { eps /= two; onepluseps = one + eps; } while (onepluseps != one); return eps + eps; } // quad_float epsilon_quad_float() but it hangs! Will look at this further.... Paul | -----Original Message----- | From: boost-bounces@lists.boost.org | [mailto:boost-bounces@lists.boost.org] On Behalf Of Matthias Troyer | Sent: 04 December 2005 19:45 | To: boost@lists.boost.org | Subject: Re: [boost] [math] floating point classification - | testinghelpwanted | | | On Dec 3, 2005, at 8:09 PM, John Maddock wrote: | | >> ld = hi + low as above is not among the bit patterns he lists as | >> valid. If you apply the definition of epsilon to the "normal" long | >> doubles only, then you get a value like the one Paul | Bristow computed | >> for NTL's quad_float. | > | > I agree: that explanation is quite explicit when it says it | follows | > Kahan's | > "double double" and the IEEE spec. So the low part must be | > normalised so | > that it's bit's "follow on" from those in the high part. As it | > stands, | > 1+numeric_limits<>::epsilon() should evaluate to 1, but we really | > need to | > check this out. Does anyone of a numerical inclination, want to | > run some | > tests on Darwin? | | Since i asked the students in my class to write a program | calculating | the machine epsilon as a homework, I just ran our solution | (posted at | http://www.itp.phys.ethz.ch/lectures/PT/Exercises/Uebung1/Solution/ | eps.cpp). Please overlook that the code is ugly, and does not | use any | templates - they just had one week of C++ training. Anyway, this code: | | | long double epsilon_long_double() { | long double eps = 1.; | long double onepluseps; | | do { | eps /= 2; | onepluseps = 1. + eps; | } while ( onepluseps != 1.); | | return 2*eps; | | } | | | returns 0, hence there is really the problem that | numeric_limits<long | double>::epsilon() == numeric_limits<long double>::min() | | Matthias | | _______________________________________________ | Unsubscribe & other changes: | http://lists.boost.org/mailman/listinfo.cgi/boost |