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Strange parse "bug"/behavior?

 [1/5] from: petr:krenzelok:trz:cz at: 17-Mar-2007 12:23


Hi, in my recent tries with REBOL parser, I mistakenly thought, that if I want to skip one char, I need to do "skip 1", instead of "1 skip" or skip . But follow following code: 1) rule: [any [copy char skip 1 (prin char)]] parse/all "this is a test string" rule result: this is a test string== true 2) rule: [any [copy char skip 1]] parse/all "this is a test string" rule result: ** Script Error: Invalid argument: ** Near: parse/all "this is a test string" rule And my question is - what is going on here? Should not case 1) fail too? How is that string is properly parsed, just because there is "(prin char)" added, if integer 1 contained in rule is not correct? Thanks, -pekr-

 [2/5] from: rebol-list2::seznam::cz at: 17-Mar-2007 15:26


rule: [any [copy char skip 2 (prin char)]] parse/all "this is a test string" rule ;== tthhiiss iiss aa tteesstt ssttrriinngg== true so it looks that the number means to repeat the action 2 times it's feature, not a bug d. PK> Hi, PK> in my recent tries with REBOL parser, I mistakenly thought, that if I PK> want to skip one char, I need to do "skip 1", instead of "1 skip" or PK> "skip". But follow following code: PK> 1) PK> rule: [any [copy char skip 1 (prin char)]] PK> parse/all "this is a test string" rule PK> result: PK> this is a test string== true PK> 2) PK> rule: [any [copy char skip 1]] PK> parse/all "this is a test string" rule PK> result: PK> ** Script Error: Invalid argument: PK> ** Near: parse/all "this is a test string" rule PK> And my question is - what is going on here? Should not case 1) fail too? PK> How is that string is properly parsed, just because there is "(prin PK> char)" added, if integer 1 contained in rule is not correct? PK> Thanks, PK> -pekr-

 [3/5] from: petr::krenzelok::trz::cz at: 17-Mar-2007 15:40


rebOldes wrote:
> rule: [any [copy char skip 2 (prin char)]] > parse/all "this is a test string" rule > ;> tthhiiss iiss aa tteesstt ssttrriinngg== true > > so it looks that the number means to repeat the action 2 times > it's feature, not a bug > > d. >
ahh, now I see - I wondered WHAT action should it repeat - now I see - the paren block action, because:
>> rule: [any [copy char skip 2]]
== [any [copy char skip 2]]
>> parse/all "this is a test string" rule
** Script Error: Invalid argument: ** Near: parse/all "this is a test string" rule -pekr-

 [4/5] from: pwawood:g:mail at: 17-Mar-2007 22:55


Hi Petr I'm no expert but perhaps the following examples will help a little
> 1) > rule: [any [copy char skip 1 (prin char)]] > parse/all "this is a test string" rule > > result: > this is a test string== true >> alphaorspace: charset [#"a" - #"z" #"A" - #"Z" #" "]
== make bitset! #{ 0000000001000000FEFFFF07FEFFFF0700000000000000000000000000000000 }
>> rule: [any [copy char alphaorspace (prin char) skip]]
== [any [copy char alphaorspace (prin char) skip]]
>> parse/all "this is a test string" rule
ti sats tig== false My understanding of this rule is: 1. any tells parse to look for zero or more occurrences of the pattern in the following block 2. the pattern in the block is a single alphaorspace followed by a skip. 3. if the pattern is found, parse will copy the character that matches alphaorspace into the "variable" char and execute the rebol code enclosed in parentheses. 4. It will then look for the next pattern to match that in the block until it doesn't find one of it reaches the end of the string. If we change the rule slightly to put the code after the skip, we get :
>> rule: [any [copy char alphaorspace skip (prin char)]]
== [any [copy char alphaorspace skip (prin char)]]
>> parse/all "this is a test string" rule
ti sats ti== false My understanding of the subtle change, the "g" of string is not printed is that with this rule parse reached the end of the string after the g so the skip didn't "match" so the code in parentheses was not executed. Putting skip before the copy char alphaorspace gives:
>> rule: [any [skip copy char alphaorspace (prin char)]]
== [any [skip copy char alphaorspace (prin char)]]
>> parse/all "this is a test string" rule
hsi etsrn== false Probably what we expected. If we only look for alpha characters instead of alpha or spaces, we get the following result:
>> alpha: charset [#"a" - #"z" #"A" - #"Z"]
== make bitset! #{ 0000000000000000FEFFFF07FEFFFF0700000000000000000000000000000000 }
>> rule: [any [copy char alpha (prin char) skip]]
== [any [copy char alpha (prin char) skip]]
>> parse/all "this is a test string" rule
ti== false As I understand: 1. parse matches "th" and prints "t" 2. parse matches "is" and prints "i" 3. parse cannot match the space so stops.
> And my question is - what is going on here? Should not case 1) fail > too? > How is that string is properly parsed, just because there is "(prin > char)" added, if integer 1 contained in rule is not correct?
I'm afraid I couldn't work out why case 1 worked either. Regards Peter

 [5/5] from: petr:krenzelok:trz:cz at: 17-Mar-2007 16:44


Peter, thanks for the input, but I already found out, why or why my case did not work: rule: [any [copy char skip 1 (prin char)]] ... simply means, that 0 or more times 1 char is copied into 'char word. What does 1 mean here is - perform 1-time what follows = (print char). Let's verify it: rule: [any [copy char skip 5 (prin char)]] parse/all "this is a test string" rule results in: ttttthhhhhiiiiisssss iiiiisssss aaaaa ttttteeeeesssssttttt ssssstttttrrrrriiiiinnnnnggggg== true Simply the rule will perform paren (rebol code) given number of times, which is 5 times here. I did not expect it being possible, as I did not consider paren being a typical part of parse dialect .... But thanks for the input. Petr