Question about code snippet for standard hex

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1

Hello,

I’m curious why there is a need to set two constants to define one pin. Could you not just assign a servo to a particular pin such as FRHH con “8”, instead of FRHH con “0” then FRHH con “8”. The code example is below for the hexapod:

[code]RRHH con “0” ;Rear Right leg : Hip Horizontal : pin 0
RRHH2 con “0”
RRHV con “0” ;Rear Right leg : Hip Vertical : pin 1
RRHV2 con “1”
RRK con “0” ;Rear Right leg : Knee : pin 2
RRK2 con “2”

MRHH con “0” ;Middle Right leg : Hip Horizontal : pin 4
MRHH2 con “4”
MRHV con “0” ;Middle Right leg : Hip Vertical : pin 5
MRHV2 con “5”
MRK con “0” ;Middle Right leg : Knee : pin 6
MRK2 con “6”

FRHH con “0” ;Front Right leg : Hip Horizontal : pin 8
FRHH2 con “8”
FRHV con “0” ;Front Right leg : Hip Vertical : pin 9
FRHV2 con “9”
FRK con “1” ;Front Right leg : Knee : pin 10
FRK2 con “0”

RLHH con “1” ;Rear Left leg : Hip Horizontal : pin 16
RLHH2 con “6”
RLHV con “1” ;Rear Left leg : Hip Vertical : pin 17
RLHV2 con “7”
RLK con “1” ;Rear Left leg : Knee : pin 18
RLK2 con “8”

MLHH con “2” ;Middle Left leg : Hip Horizontal : pin 20
MLHH2 con “0”
MLHV con “2” ;Middle Left leg : Hip Vertical : pin 21
MLHV2 con “1”
MLK con “2” ;Middle Left leg : Knee : pin 22
MLK2 con “2”

FLHH con “2” ;Front Left leg : Hip Horizontal : pin 24
FLHH2 con “4”
FLHV con “2” ;Front Left leg : Hip Vertical : pin 25
FLHV2 con “5”
FLK con “2” ;Front Left leg : Knee : pin 26
FLK2 con "6[/code]

2

supposedly it saved space.

Alan KM6VV

3

You obviously could do this several different ways. Obviously you could simply do it with one if your pin number is less than 10. Personally I doubt it saves space, but it might be a hair faster than doing the conversion.
To verify the space, I had a simple program:

bar var long
dpin con 10
DP1	con	"1"
DP2 con "2"
serout s_out, i9600, [DP1, DP2, dec dpin]

If yo ulook at the generated assembly language you will find:

... .stabs "0 7 0",32,0,0,_DEBUGLABEL0 .word _CONPCMD .long 0x21 ;S_OUT .word _CLRWPCMD .word _CONPCMD .long 0x40d0 ;I9600 .word _CLRWP3CMD .word _CONSTROUTCMD .word _SEROUTFUNCCMD .long 0x2 ;COUNT .long 0x31 ;DP1 .long 0x32 ;DP2 .word _CONCMD .long 0xa ;DPIN .word _DECOUTCMD .long 0xa ;FLAG .word _SEROUTFUNCCMD ...
So you will see that for each character you output, it is actually passed as a LONG (4 bytes). As for doing a decimal it looks like it passed a WORD, a Long and a word, so again 8 bytes. But: the code did not have to take the time to do the Decimal to Ascii conversion…

Kurt

4

Anytime I ever find something that looks more complicated than it needs to be, there is always an underlying reason. :wink: