search for: rounding_ep

...overflow. For example, i_small = dx < cell * (U * imax2(1,*ndiv) * DBL_EPSILON) *3; The part U = (1 + (h5 >= 1.5*h+.5)) ? 1/(1+h) : 1.5/(1+h5); is strange. Because (h5 >= 1.5*h+.5) is 1 or 0, (1 + (h5 >= 1.5*h+.5)) is never zero and 1/(1+h) will always be chosen. The comment for 'rounding_eps' says "1e-7 is consistent with seq.default()". Currently, seq.default() uses 1e-10 as fuzz.

...original code had been ? ?? ? ? > U = 1 + (h5 >= 1.5*h+.5) ? 1/(1+h) : 1.5/(1+h5); ? ? > and "of course" was intended to mean ? ? > U = 1 + ((h5 >= 1.5*h+.5) ? 1/(1+h) : 1.5/(1+h5)); ? ? > and this what I'll change it to, now. ? ? >> The comment for 'rounding_eps' says "1e-7 is consistent with seq.default()". Currently, seq.default() uses 1e-10 as fuzz. ? ? > Hmm, yes, thank you; this was correct when written, ? ? > but seq.default had been changed in the mean time, ? ? > namely in? svn r51095 | 2010-02-03 ? ? > Usually we are c...

...ut not what they intended!). The original code had been U = 1 + (h5 >= 1.5*h+.5) ? 1/(1+h) : 1.5/(1+h5); and "of course" was intended to mean U = 1 + ((h5 >= 1.5*h+.5) ? 1/(1+h) : 1.5/(1+h5)); and this what I'll change it to, now. > The comment for 'rounding_eps' says "1e-7 is consistent with seq.default()". Currently, seq.default() uses 1e-10 as fuzz. Hmm, yes, thank you; this was correct when written, but seq.default had been changed in the mean time, namely in svn r51095 | 2010-02-03 Usually we are cautious / reluctant to change such t...