Hi,
> On Jan 29, 2019, at 7:04 PM, Konrad Hinsen <konrad.hin...@fastmail.net> wrote:
>
> Tudor Girba <tu...@tudorgirba.com> writes:
>
>> But, does it solve your problem?
>
> Not yet. Not sure it will.
Would this not work:
p := (#digit asPParser separatedBy: ($+ asPParser / $- asPParser) trim) ==>
[:tokens |
| result |
result := tokens.
2 to: tokens size - 1 by: 2 do: [ :index |
(tokens at: index) = (tokens at: 2)
ifFalse: [result := PP2Failure message: 'expected ',
(tokens at: 2) asString, ' but got ', (tokens at: index) asString ]].
result ].
p end parse: '1+3+2 - 6' "a PP2Failure: 'expected + but got -‘"
p end parse: '1+3+2 + 6' "#($1 $+ $3 $+ $2 $+ $6)”
?
Please note that I used PetitParser2 for this one because it also addresses the
failure message issue you raised below. However, the logic would work in PP1 as
well.
Cheers,
Doru
>>> More generally, there seems to be a bug in how PetitParser handles
>>> PPFailure return values. The actually reported error is always different
>>> from what is passed to PPFailure. I love software ;-)
>>
>> What do you mean?
>
> If I return
>
> PPFailure message: 'whatever' at: 0
>
> from my parser, then it does fail but it reports a different error.
>
> After some more experiments, it looks like this is perhaps not a bug,
> but a feature that makes returning failures not so useful after all. I
> suspect that PetitParser simply backtracks after my failure and tries a
> different rule that then fails as well.
>
> A simple example:
>
> | p |
> p := #digit asParser plus flatten
> ==> [ :value | value asSet size = 1 ifTrue: [ value ] ifFalse:
> [ PPFailure message: 'not equal' at: 0 ]].
> p parse: '22233'
>
> The reported failure is "digit expected at 5".
>
> On the other hand, if there is no path for backtracking, the reported
> failure is the expected one:
>
> | p |
> p := (#digit asParser, #digit asParser) flatten
> ==> [ :value | value first = value second ifTrue: [ value ]
> ifFalse: [ PPFailure message: 'not equal' at: 0 ]].
> p parse: '23'
>
> Konrad.
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