There has to be an easier way!

[Christopher4187]

I re-read what I wrote and it may be a bit confusing. Here is the short story:

b=7 now I need to extrapolate to binary


r1=1
r2=1
r3=1
r4=0


If it's still not clear, let me know.

[mister_e]

maybe i don't understand what you need but lets assume
B1= 11110000 bin => F0 hex => 240 decimal
B2=00001111 bin => 0F hex => 15 decimal

even if you define
B1=$F0
OR
B1=%11110000
OR
B1=240

It do the same thing.

IF B1=$F0 then....
IF b1=%11110000 then ....
IF B1=240 then .....

Maybe i need to sleep, or it's really what you need

[mister_e]

HONK! i think i got it.

Let's say B1=%11110000

R0 should be Bit 0 of B1, R7 should be Bit7 of B1... i'm i right?
If so...
R0=B1.0
R1=B1.1
R2=B1.2
R3=B1.3
..
...
R7=B1.7

[Christopher4187]

Steve,

You solved another one of my problems! Yes, to answer your question, I want to have a decimal number and convert that to binary. In that binary number, take each bit and it represents another number. So, within one number I extract 8 different boolean numbers. I'll try your suggestion but I do think it will work.

Thanks,

Chris

[Christopher4187]

I see one small issue.


For example, I know that b1=9 which is 00001011.

If b1=9 (the decimal number), how can I let PBP know that it is actually 00001011 (the binary number).

Or, can I simply go

R0=B1.0
R1=B1.1
R2=B1.2
R3=B1.3

[BobK]

Hi Christopher,

Decimal 9 in binary is %00001001 not 00001011.

Binary representation is from right to left meaning the digit on the far right is the first is 1 then 2 then 4 then 8 then 16 then 32 then 64 then 128.

I have been reading your posts and just don't quite get what it is you are trying to do.

"I don't have much experience with the lookup command and I don't understand what the manual states. I will post the code here and someone let me know if I can clean it up please.

B1=8
B2=6
B3=11
B4=15

B1 in binary is %00001000
B2 in binary is %00000110
B3 in binary is %00001011
B4 in binary is %00001111

Hope this helps!

BobK

[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