Re: [CLN-list] Operations with different precisions

12 Oct 2009

      Hi,

Sebastian Macke wrote:
...
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
Is there a feature that the float with the highest precision
is used as the result precision?
I suppose a typo: Of course, in operators taking two floats, the 
precision of the float with the *least* precision is used as the result 
precision.

Regards
    -richy.
-- 
Richard B. Kreckel
<http://www.ginac.de/~kreckel/>