WinCUPL question(s)

[Jokse]

I'm doing a FIELD in WinCUPL for a 22V10, but as soon as I want to use the .d flip flop part for synchronizing with a master clock on pin one, using something like out.d = input:[0..2FF] doesn't work... Do I really have to spell out the full logic equation or is there some way to get around that?

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Name Test;
PartNo 01;
Date 04-04-2018;
Revision 01;
Designer Flaf;
Company Flaf;
Assembly Flaf;
Location Flaf;
Device p22v10;

PIN 1 = clk;
PIN [2,3,4,5,6,7,8,9,10,11] = [in0..9];
PIN 22 = output;
FIELD input = [in9..0];
output.ar = 'b'0;
output.sp = 'b'0;
output.d = input:[0..2FF];

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  17:output.d = input:[0..2FF];
                           ^
[002ca] invalid number: 2FF
[002ca] invalid number: 2FF
[002ca] invalid number: 2FF
Total time: 0 secs 

[Martin A]

It would appear to work if you use 'h'2ff instead of just 2ff. Not sure why the explicit hex definition is needed though.

[Jokse]

Martin A wrote:

It would appear to work if you use 'h'2ff instead of just 2ff. Not sure why the explicit hex definition is needed though.

That certainly did the trick! Thanks!

[BigDumbDinosaur]

Martin A wrote:

It would appear to work if you use 'h'2ff instead of just 2ff. Not sure why the explicit hex definition is needed though.

The hex reference is required because decimal is the default in CUPL. Without the 'h' prefix, the compiler might try to evaulate the 2ff part as a variable.

[Martin A]

I always thought hex was the default, the following code from one of my projects compiles correctly without any explicit 'h' declarations.

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Name     AtomGAL ;
PartNo   00 ;
Date     09/02/2018 ;
Revision 01 ;
Designer Engineer ;
Company  None ;
Assembly None ;
Location  ;
Device   p22v10 ;

/* *************** INPUT PINS *********************/
PIN   1  = A12                      ; /*                                 */ 
PIN   2  = A13                      ; /*                                 */ 
PIN   3  = A14                      ; /*                                 */ 
PIN   4  = A15                      ; /*                                 */ 
PIN   5  = A11                      ; /*                                 */ 
PIN   6  = A10                      ; /*                                 */ 
PIN   7  = A9                       ; /*                                 */ 
PIN   8  = rw                       ; /*                                 */ 
PIN   9  = phi2                     ; /*                                 */ 



/* *************** OUTPUT PINS *********************/
PIN  14   = wr                      ; /*                                 */ 
PIN  15   = rd                      ; /*                                 */ 
PIN  16   = ma2                     ; /*                                 */ 
PIN  17   = vram                    ; /*                                 */ 
PIN  18   = io6                     ; /*                                 */ 
PIN  19   = io4                     ; /*                                 */ 
PIN  20   = io2                     ; /*                                 */ 
PIN  21   = io0                     ; /*                                 */ 
PIN  22   = ramcs                   ; /*                                 */ 
PIN  23   = romcs                   ; /*                                 */ 


Field address = [A15..0];

!ramcs = address:[0000..7FFF] ;

!io0 =address:[B000..B3FF] ;

!io2 =address:[B400..B7FF] ;

!io4 =address:[B800..BBFF] ;

!io6 =address:[BC00..BFFF] ;

!romcs = address:[C000..FFFF] ;

!vram = address:[8000..97FF] ;

!ma2 = address:[A000..AFFF] ;

!rd =  rw & phi2  ;

!wr = !rw & phi2  ;

[Tor]

Maybe WinCUPL will understand [02FF] as hex, but not [2FF] (without the 'h')?

[BigEd]

Maybe a 10 bit number, not being an even number of nibbles, throws it.

Is it worth padding the zero to make the numbers the same length?

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output.d = input:[000..2FF];

[cbscpe]

The default is hex. But it is overwritten when using a prefix, like in your case you say

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output.ar = ´b´0

after that the default is binary.