float point

[enrico]

I use the PIC16F877.
I must do the following expression:

long i,ppm;

ppm=(i-205)*0.244200244*10;

but this operation take up so much FLASH program.
How can I do this operation with float point and take up less FLASH program?

thank you
___________________________
This message was ported from CCS's old forum
Original Post ID: 7730

[Sherpa Doug]

:=I use the PIC16F877.
:=I must do the following expression:
:=
:=long i,ppm;
:=
:=ppm=(i-205)*0.244200244*10;
:=
:=but this operation take up so much FLASH program.
:=How can I do this operation with float point and take up less FLASH program?
:=
:=thank you

How about
ppm = ((int32)(i-205)*2442)/1000

Can you product really tell .244200244 from .2442 ?

___________________________
This message was ported from CCS's old forum
Original Post ID: 7734

[Mark]

How accurate does ppm need to be?


How much of an error can you live with?

with a 0\% error, then
ppm=(i-205)*0.244200244*10;

ROM used: 263 (3\%)
Largest free fragment is 2048
RAM used: 15 (9\%) at main() level
28 (16\%) worst case

with a 2.3\% error, then
ppm=(i-205)*10/4;
ROM used: 64 (1\%)
Largest free fragment is 2048
RAM used: 12 (7\%) at main() level
16 (9\%) worst case

with a 0.08\% error, then
ppm=(i-205)*61/25;
ROM used: 108 (1\%)
Largest free fragment is 2048
RAM used: 12 (7\%) at main() level
17 (10\%) worst case

with a 1.7\% error, then
ppm=(i-205)*12/5;
ROM used: 108 (1\%)
Largest free fragment is 2048
RAM used: 12 (7\%) at main() level
17 (10\%) worst case



You will have to make sure that (i-205) does not overflow if you use one of the examples. Multiplying by 61 or 12 could cause you to overflow. Also note that your final result is an integer so the error might be less than you expect.

Regards,
Mark

:=I use the PIC16F877.
:=I must do the following expression:
:=
:=long i,ppm;
:=
:=ppm=(i-205)*0.244200244*10;
:=
:=but this operation take up so much FLASH program.
:=How can I do this operation with float point and take up less FLASH program?
:=
:=thank you
___________________________
This message was ported from CCS's old forum
Original Post ID: 7757

[TSchultz]

:=How accurate does ppm need to be?
:=
:=
:=How much of an error can you live with?
:=
:=with a 0\% error, then
:= ppm=(i-205)*0.244200244*10;
:=
:= ROM used: 263 (3\%)
:= Largest free fragment is 2048
:= RAM used: 15 (9\%) at main() level
:= 28 (16\%) worst case
:=
:=with a 2.3\% error, then
:= ppm=(i-205)*10/4;
:= ROM used: 64 (1\%)
:= Largest free fragment is 2048
:= RAM used: 12 (7\%) at main() level
:= 16 (9\%) worst case
:=
:=with a 0.08\% error, then
:= ppm=(i-205)*61/25;
:= ROM used: 108 (1\%)
:= Largest free fragment is 2048
:= RAM used: 12 (7\%) at main() level
:= 17 (10\%) worst case
:=
:=with a 1.7\% error, then
:= ppm=(i-205)*12/5;
:= ROM used: 108 (1\%)
:= Largest free fragment is 2048
:= RAM used: 12 (7\%) at main() level
:= 17 (10\%) worst case
:=
:=
:=
:=You will have to make sure that (i-205) does not overflow if you use one of the examples. Multiplying by 61 or 12 could cause you to overflow. Also note that your final result is an integer so the error might be less than you expect.
:=
:=Regards,
:=Mark
:=
:=:=I use the PIC16F877.
:=:=I must do the following expression:
:=:=
:=:=long i,ppm;
:=:=
:=:=ppm=(i-205)*0.244200244*10;
:=:=
:=:=but this operation take up so much FLASH program.
:=:=How can I do this operation with float point and take up less FLASH program?
:=:=
:=:=thank you

You could also get a 0.18\% error and drop the division if you tried;

ppm=(i-205)*39/16

The divide by 16 could be done using a shift saving precious code space and CPU time.

Like before though this does assume that the value of i is alsways small enough to be multiplied by 39 without causing an overflow.

- Troy
___________________________
This message was ported from CCS's old forum
Original Post ID: 7836