There has to be an easier way!

[SteveB]

Chris,
I think you are missing the point of exactly how PB and the PIC store any variable. When a variable is stored, it is stored as a binary number in a byte-wide file register.
Simple example:

B1  VAR  BYTE
B1 = 172

Now, this is oversimplified, but this is what is happening. The compiler :
1)Allocates a register in memory for B1
2)Generates Assemby code to Store "10101100" in that register
3)Remembers that any time you reference that variable again, it will actually be referencing the allocated register, and the value therein.


Now, lets add some more code:

R0  VAR  BIT
R1  VAR  BIT
'...
R7 VAR BIT

R0 = B1.0
R1 = B1.1
'...
R7 = B1.7

Now the compiler looks for a specific bit in the file register for B1, and assigns the value of that bit to the variable.

That Help?

Steve

[Christopher4187]

Actually, Steve (Mister e) understood what I was referring to. Here is the long story. I have a 16F870 monitoring 16 inputs and I am sending the status of all inputs wirelessly to a 18F4550. Then, the 18F4550 outputs the status through USB into VB6. Anyway, because of my limitation in PBP, I can't send word variables wirelessly so I am sending 4 bytes which covers the 16 inputs. On the transmitter side, my code looks like this...it's a portion of the code...and don't forget, it may not be efficient because I am not very proficient in PBP.

if PORTA.5=0 THEN
B1=0
ELSE
B1=1
ENDIF
if PORTA.2=0 THEN
B2=0
ELSE
B2=1
ENDIF
if PORTA.1=0 THEN
B3=0
ELSE
B3=1
ENDIF
if PORTA.0=0 THEN
B4=0
ELSE
B4=1
ENDIF
C1=(B1*1)+(B2*2)+(B3*4)+(B4*8)

This was the easy part. One the receiver side, it looks like this:

in1=c1.0
in2=c1.1
in3=c1.2
in4=c1.3

And then from here it sends the data to VB6. I wasn't aware that PBP stored the numbers as binary and that is where my confusion was. When sending the numbers over the data link, it is simply sent as (C1=9) or (C1=15), not (c1=%00001011). Furthermore, I never knew that you can use a c1.0 or c1.3 to take a portion of the number.

Previously, I was sending 4 bytes for each specific number so it was taking 96 bytes (when you include the start and end bit) to update all 16 inputs. Now, it only takes 6 bytes to send that data.

Thanks to all who helped.

Chris

[Acetronics2]

if PORTA.5=0 THEN
B1=0
ELSE
B1=1
ENDIF
if PORTA.2=0 THEN
B2=0
ELSE
B2=1
ENDIF
if PORTA.1=0 THEN
B3=0
ELSE
B3=1
ENDIF
if PORTA.0=0 THEN
B4=0
ELSE
B4=1
ENDIF
C1=(B1*1)+(B2*2)+(B3*4)+(B4*8)

Chris

> Hi, Christopher

> Let's do :

> C1 = PORTA & % 00100111
> C1.3 = C1.5 : C1.5 = 0
> C1 = C1 REV 4

That's all !!!

See manual $ 4.10 and 4.17.11 ...

Alain

[paul borgmeier]

Another Option - although more lines of code, this compiles to less code space -

temp var byte
C1 var byte

C1=0 ; set C1 to 0
temp=PORTA ; in case PORTA changes while “processing”
C1.0=temp.5
C1.1=temp.2
C1.2=temp.1
C1.3=temp.0

Paul Borgmeier
Salt Lake City, Utah
USA

[paul borgmeier]

or this

temp var byte
C1 var byte

C1=0 ; set C1 to 0
temp=PORTA ; in case PORTA changes while “processing”
if temp.5=1 then C1.0=1
if temp.2=1 then C1.1=1
if temp.1=1 then C1.2=1
if temp.0=1 then C1.3=1

Paul