16C internal floating point format?

[ThomasF]

(04-03-2021 08:48 PM)billf Wrote:  So, if someone has access to a real calculator, could you do me a favor and let me know what you get in (56 bit) hex when you store the following numbers into registers? 0.0 42.0 -42.0 1e42 1e-42 0.0001e-99 9.999999999e99, and finally, the results of (99 <ent> 9e99 *), (-99 <ent> 9e99 *) and (0 1/x). I think that covers all the bases.

Thanks!

Hi Bill,

This is what I get:

0.0 --> 0h
42.0 --> 04200000000001h
-42.0 --> 94200000000001h
1e42 --> 1000000000042h
1e-42 --> 1000000000958h
0.0001e-99 --> not possible to enter
9.999999999e99 --> 09999999999099h
(99 <ent> 9e99 *) --> 09999999999099h
(-99 <ent> 9e99 *) --> 99999999999099h
(0 1/x) --> Just gives "Error 0"

Note that 99 <ent> 9e99 gives overflow - i.e. it becomes 9.999999999e99
I also used 64 bit wordsize when in hex-mode (FLOAT 8 - <enter number> STO - hex 0 WSIZE RCL), since I expect that registers are in 64 bits length (to be able to store a 64 bit integer), correct me if I am wrong.

Edit: was partly wrong. Used the I-register which is always full size!

Hope this helps you!

Cheers,
Thomas