[CLN-list] Operations with different precisions

[Bruno Haible]

Hi,

Sebastian Macke:
> following code snippet
> 
>  cln::cl_F x = cl_float(0.5, cln::float_format(50));
>  x = cos(x);
>  cout << "\n" << x << "\n";
>  x = x + cln::cl_F(1.);
>  cout << x << "\n";
> 
> gives the output
> 
>  0.8775825618903727161162815826038296519916451971097440529976L0
>  1.8775825618903728d0

This is normal, expected, and a feature.

> Is there a feature that the float with the highest precision
> is used as the result precision?

No, as this would be a misfeature.

Consider this:

  cln::cl_F x = cl_float(0.5, cln::float_format(50));
  x = cos(x);
  cout << "\n" << x << "\n";
  x = x + cln::cl_F(1./3.);
  cout << x << "\n";

Currently CLN gives the output

  0.8775825618903727161162815826038296519916451971097440529976L0
  1.210915895223706d0

With your suggestion, it would give the output

  0.8775825618903727161162815826038296519916451971097440529976L0
  1.2109158952237060309458978388512206449311173955960721779976L0  <===

but the correct result of cos(0.5) + 1/3 is

  1.210915895223706049449614915937162985324978530443077386331L0

So
  1) The result marked with "<===" would have way more wrong digits than
     correct ones, and you have no way to know it just by looking at the
     result!
  2) If you continue to do computations with that result, two third of the
     computation time will be wasted with treatment of garbage digits.

In other words, your proposed result would have an accuracy that is much lower
than the precision. However, the expectation for a final result is that the
the accuracy is equal or nearly equal to the precision.

Bruno