Hi!
_error is meant to produce an error whenever it is compiled. You can only use
it with constants:
if target_clock<12000000 then _error "Need at least 12MHz for this program"end
if
If there's a variable in the if statement, the content will have to be
compiled, and so the error is raised.
If you want to see characters depending on a value of a variable, you can only
check at runtime. So, connect some kind of a display and use print_string()
Greets,Kiste
Am Samstag, 13. November 2021, 09:56:29 MEZ hat Rob CJ <rob...@hotmail.com>
Folgendes geschrieben:
Hi Vasile,
Thanks for the update. I found something strange. Given the code below.
include 18f4550
var byte i =0
if (i == 100)then _error "Is 100"end if
When I compile this I get:
jal jalv25r6 (compiled Oct 29 2021)generating p-code16858 tokens, 151522 chars;
3221 lines; 3 filesgenerating PIC code pass 1generating PIC code pass 20 data
accesses checked, 0 errors1 skips checked, 0 errors writing
resultd:\PIC\Compiler\Test\2021\ADC_in_array\Test.jal:6: "Is 100"1 errors, 0
warningsThe terminal process "c:\jallib\compiler\jalv2_64.exe
'd:\PIC\Compiler\Test\2021\ADC_in_array\Test.jal', '-s', 'c:\jallib\lib'"
terminated with exit code: 1.
Which is not correct. Same happens when you replace it with _warn but then it
is generated as warning.
But when I have this program:
include 18f4550
var byte i =0
if !defined(i)then _error "i is not defined"end if
I get:
jal jalv25r6 (compiled Oct 29 2021)generating p-code16858 tokens, 151533 chars;
3221 lines; 3 filesgenerating PIC code pass 1generating PIC code pass 20 data
accesses checked, 0 errors0 skips checked, 0 errorswriting resultCode area: 6
of 32768 used (bytes)Data area: 1 of 2048 usedSoftware stack available: 2046
bytesHardware stack depth 0 of 310 errors, 0 warnings
Which is correct because when I change i to j the output is:
jal jalv25r6 (compiled Oct 29 2021)generating p-code16858 tokens, 151533 chars;
3221 lines; 3 filesgenerating PIC code pass 1generating PIC code pass 20 data
accesses checked, 0 errors0 skips checked, 0 errorswriting
resultd:\PIC\Compiler\Test\2021\ADC_in_array\Test.jal:6: "i is not defined"1
errors, 0 warningsThe terminal process "c:\jallib\compiler\jalv2_64.exe
'd:\PIC\Compiler\Test\2021\ADC_in_array\Test.jal', '-s', 'c:\jallib\lib'"
terminated with exit code: 1.
So not sure why this happens but is seems to be not OK.
Kind regards,
Rob
Van: jallib@googlegroups.com <jallib@googlegroups.com> namens vsurducan
<vsurdu...@gmail.com>
Verzonden: zaterdag 13 november 2021 08:08
Aan: jallib@googlegroups.com <jallib@googlegroups.com>
Onderwerp: Re: [jallib] adc average Hi Rob,I think you will not find any issue,
please don't bother. The code works now (without interrupts), I had a
multiplication error which is not part of the code.
My original test used tmr0 interrupt and USB_serial, the average measurements
should happen in 400us intervals and it seems something went wrong because the
values are not those expected.
About the test procedure, I've loaded the ADRESH and ADRESL with constant
values, without using any ADC_read procedure. Then tried to test the average
procedure using _ERROR. When the compiler finds the line containing _ERROR it
reports an error no matter if the condition is true or false. Did I use it in
the wrong way?A very old version of the JAL compiler used a pragma_test (or
something like that) previously to start a testing procedure. The result was
delivered at compile time.thank you,
On Fri, Nov 12, 2021 at 11:37 PM Rob CJ <rob...@hotmail.com> wrote:
Hi Vasile,
If I want to test this I need to create a program that does not use your
one_ch_ad_read routine. So if I use the standard adc library the program looks
like this. Is this program giving the same problem for you (I do not know which
PIC you are using but it seem to be an 18F)?
About the error and warning. They are only created at compile time and I am not
sure what voltage is at the time you compile the the program. I am also not
sure if this works for variables.
include 18f4550--pragma target clock 48_000_000-- magical statements: using a
20MHz Xtal, you can run @48MHz !pragma target PLLDIV P5pragma target CPUDIV
P1pragma target USBDIV P2pragma target OSC HS_PLL
pragma target WDT control -- watchdogpragma target LVP enabled
-- allow Low Voltage Programmingpragma target MCLR external -- reset
externally
WDTCON_SWDTEN = off -- no watchdog
-- We'll start to set all pins as digital-- then, using ADC lib, we'll
configure needed-- ones as analog.enable_digital_io()
const sample_nr = 8 ; buffer size dimensionvar word voltage1_a[sample_nr];
sample arrayvar word one_ch_ad_read, ch0_adc_valuevar word voltage1 =0 ; the
average value we needconst word c256 =256
-- Configure ADC-- Step 1: ADC input pin setup pin_AN0_direction = input
-- Set A0 to analog input ADCON1_PCFG = 0b000-- Step 2: Set VDD
and VSS as VrefADCON1_VCFG0 = FALSE -- Vref+ is
VDDADCON1_VCFG1 = FALSE -- Vref- is VSS-- Step 3:
Use Fosc/64 as ADC clock when using a 48 MHz target clock, see
datasheetADCON2_ADCS = 0b110 -- Now we can include the libraryinclude adc-- And
initialize the whole with our parametersadc_init()
var byte i for 8 using iloop ch0_adc_value = adc_read_high_res(0); compute
result voltage1_a [i] = ch0_adc_value ; store sample voltage1 = voltage1 +
voltage1_a[i] ; do a sum of all samples each i step end loop
Van:jallib@googlegroups.com <jallib@googlegroups.com> namens vsurducan
<vsurdu...@gmail.com>
Verzonden: vrijdag 12 november 2021 07:38
Aan: jallib@googlegroups.com <jallib@googlegroups.com>
Onderwerp: Re: [jallib] adc average Hi Rob,If I replace the ADRESH and ADRESL
content with fixed values, it looks like computation is ok, however this is not
the real situation when reading ADC.
How can a computing error be reported after compilation time?It seems _ERROR or
_WARN does not work... Like this:
if (voltage1 <= 490) | (voltage1 >= 510) then _ERROR "average computing error"
; this error to be reported after compilation, but only if it's true.
end if
On the other hand I'm using interrupts in usb_serial for debugging and STATUS
and FSR0L are also used...
; 243 voltage1 = voltage1 + voltage1_a[i]
bcf v__status, v__c,v__access
rlcf v___i_2,w,v__banked
movwf v____temp_66,v__banked
lfsr 0,v_voltage1_a
movf v____temp_66,w,v__banked
addwf v__fsr0l,f,v__access
movf v__ind,w,v__access
movwf v__pic_temp,v__access
incf v__fsr0l,f,v__access
movf v__ind,w,v__access
movwf v__pic_temp+1,v__access
movf v__pic_temp,w,v__access
addwf v_voltage1,f,v__banked
movf v__pic_temp+1,w,v__access
addwfc v_voltage1+1,f,v__banked
On Thu, Nov 11, 2021 at 8:07 PM Rob CJ <rob...@hotmail.com> wrote:
Hi Vasile,
Strange that this does not work.
Did you try to isolate the problem to a minimum program without using special -
other - procedures so that it can easily be reproduced by anybody that does not
have the library you are using?
That would make it easier to find the root cause.
Thanks.
Kind regards,
Rob
Van:jallib@googlegroups.com <jallib@googlegroups.com> namens vsurducan
<vsurdu...@gmail.com>
Verzonden: donderdag 11 november 2021 13:47
Aan: jallib@googlegroups.com <jallib@googlegroups.com>
Onderwerp: [jallib] adc average Hi all ( aka Kiste and Rob&Rob ?) :)
One possible solution to take 8 adc average values is below (and does not work):
const sample_nr = 8 ; buffer size dimension
var word voltage1_a[sample_nr] ; sample array
var word one_ch_ad_read
var word voltage1 = 0 ; the average value we need
const word c256 = 256
var byte i
for 8 using i loop
one_ch_ad_read (0, 0b10, 0b00, 1) ; read adc procedure
ch0_adc_value = c256*ADRESH + ADRESL ; compute result
voltage1_a [i] = ch0_adc_value ; store sample
voltage1 = voltage1 + voltage1_a[i] ; do a sum of all samples each i step
end loop
voltage1 = voltage1 >> 3 ; divide by 8 and get the adc average
The issue is at the line:
voltage1 = voltage1 + voltage1_a[i] ;
initially voltage1 = 0
for i=0 voltage1 = 0 + voltage1_a[0]for i=1 voltage1 = voltage1 (i=0) +
voltage1_a[1] ...etc
this is not working as presumed...
If I remove this line from for...loop...end loop and write as:voltage1 =
voltage1_a [0] + voltage1_a [1] + voltage1_a [2] + voltage1_a [3]
+ voltage1_a [4] + voltage1_a [5] + voltage1_a [6] + voltage1_a [7]
then everything is ok.
The question is: why it does not work?
thx
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