czardas Posted August 17, 2014 Share Posted August 17, 2014 In an imaginary language. I declare a variable called CONFLICT. I later use a function which expects the string parameter CONFLICT. There is no way to differentiate between the variable and the string parameter. What should the interpreter do? 1) Prevent the user from declaring the variable CONFLICT 2) Allow the declaration and only throw an error if something actually breaks. 3) Something else - please say what. operator64 ArrayWorkshop Link to comment Share on other sites More sharing options...
orbs Posted August 17, 2014 Share Posted August 17, 2014 There is no way to differentiate between the variable and the string parameter. i read "there is no way to differentiate between variable name and variable value." do i understand correctly? if so, the imaginary language may have some very real problems. i think the imaginary developer should push his/her imagination a bit further and have this resolved, for this is the cause of the issue, and your suggested options of resolution are handling one possible symptom only. Signature - my forum contributions: Spoiler UDF: LFN - support for long file names (over 260 characters) InputImpose - impose valid characters in an input control TimeConvert - convert UTC to/from local time and/or reformat the string representation AMF - accept multiple files from Windows Explorer context menu DateDuration - literal description of the difference between given dates Apps: Touch - set the "modified" timestamp of a file to current time Show For Files - tray menu to show/hide files extensions, hidden & system files, and selection checkboxes SPDiff - Single-Pane Text Diff Link to comment Share on other sites More sharing options...
Mat Posted August 17, 2014 Share Posted August 17, 2014 With naming conflicts, variables in the innermost scope take preference. So in this case using CONFLICT inside the function refers to the argument. czardas 1 AutoIt Project Listing Link to comment Share on other sites More sharing options...
trancexx Posted August 17, 2014 Share Posted August 17, 2014 The way I read czardas, the problem here is conflict of literal value of the argument and its identifier.PRINT(CONFLICT) // prints CONFLICTCONFLICT = ABCD PRINT(CONFLICT) // prints ABCD ♡♡♡ . eMyvnE Link to comment Share on other sites More sharing options...
JohnOne Posted August 17, 2014 Share Posted August 17, 2014 Would the language not expect "CONFLICT" as a parameter and CONFLICT as a variable. AutoIt Absolute Beginners Require a serial Pause Script Video Tutorials by Morthawt ipify Monkey's are, like, natures humans. Link to comment Share on other sites More sharing options...
czardas Posted August 17, 2014 Author Share Posted August 17, 2014 i read "there is no way to differentiate between variable name and variable value." do i understand correctly? Assume this to be true. With naming conflicts, variables in the innermost scope take preference. So in this case using CONFLICT inside the function refers to the argument. Unlike the following scenario. The way I read czardas, the problem here is conflict of literal value of the argument and its identifier. PRINT(CONFLICT) // prints CONFLICT CONFLICT = ABCD PRINT(CONFLICT) // prints ABCD This scenario!. Would the language not expect "CONFLICT" as a parameter and CONFLICT as a variable. That would be cheating. operator64 ArrayWorkshop Link to comment Share on other sites More sharing options...
czardas Posted August 17, 2014 Author Share Posted August 17, 2014 (edited) Here's an idea! Preventing the variable CONFLICT form being passed to the function PRINT, but allowing it in other situations, is possibly a solution. ; CONFLICT = ABCD ABCD = CONFLICT PRINT(CONFLICT) // prints CONFLICT (always) PRINT(ABCD) // prints ABCD (because this value was assigned) : It seems to be more or less what Mat was saying. It's still a bit tricky. Actually! ; Local $sCode = "distance (Floor( au + distance(distance, au, angst )) , au , angst ) + angst" ; Imaginary syntax $sCode = StringRegExpReplace($sCode, "(\bdistance)(\s*\()", "_$1(") Local $sDistParam = '(au|angst)' ; Potential conflicts Local $sRegExpDistParam = '(_distance\()(.+)(\(.+\))*(\s*,\s*)' & $sDistParam & '(\s*,\s*)' & $sDistParam & '\s*\)' Local $iWorking = -1 While $iWorking $sCode = StringRegExpReplace($sCode, $sRegExpDistParam, '$1$2$3$4"$5","$7")') $iWorking = @extended WEnd MsgBox(0, "Interpreter", $sCode) Piece of cake! I wish. I thought I had it. I know I can fix this. Edited August 17, 2014 by czardas operator64 ArrayWorkshop Link to comment Share on other sites More sharing options...
Mat Posted August 17, 2014 Share Posted August 17, 2014 Solutions are: start variables with a letter, use quotes for strings, or start string literals with a character. Anything else would just be confusing. AutoIt Project Listing Link to comment Share on other sites More sharing options...
czardas Posted August 17, 2014 Author Share Posted August 17, 2014 (edited) Sorry, strings are to be reserved for undefined numeric base variants. 'ACE'. It's a kind of numbers only language with minimal syntax. So 100 is not necessarily the same as '100' and string variables may only contain the letters A-F. Edited August 17, 2014 by czardas operator64 ArrayWorkshop Link to comment Share on other sites More sharing options...
czardas Posted August 17, 2014 Author Share Posted August 17, 2014 (edited) I couldn't find a working regexp and gave up. If someone can figure it out, I'd love to see a more elegant solution. I eventually resorted to brute force. ; expandcollapse popupLocal $sCode = "Distance (Floor( au + Random(Distance(Distance, au, angst ), au, angst )) , au , angst ) + angst + Distance(1, au , angst )" ; Imaginary syntax $sCode = StringRegExpReplace($sCode, "(\s*,\s*)", ",") $sCode = StringRegExpReplace($sCode, "(Random|Distance|Floor)(\s*\()", "_$1(") Local $aFunctions = StringRegExp($sCode, "(_\w+[^_]+)",3) Local $iCharCount, $sParentheses, $iCurrentFunc, _ $sSearch = "_Distance(", $sDistParam = '(au|angst)' ; Potential conflicts For $i = 0 To UBound($aFunctions) -1 If Not StringInStr($aFunctions[$i], $sSearch) Then ContinueLoop $iCharCount = 1 $sParentheses = 1 $iCurrentFunc = $i For $j = $iCurrentFunc To UBound($aFunctions) -1 For $k = 1 To StringLen($aFunctions[$j]) If $iCharCount > Stringlen($sSearch) Then If StringMid($aFunctions[$j], $k, 1) = ")" Then $sParentheses -= 1 ElseIf StringMid($aFunctions[$j], $k, 1) = "(" Then $sParentheses += 1 EndIf EndIf If $sParentheses = 0 Then $aFunctions[$j] = StringRegExpReplace(StringLeft($aFunctions[$j], $k), "(.*)" & $sDistParam & '(,)' & $sDistParam & "(\s*\)\z)",'$1"$2"$3"$4"$5') _ & StringTrimLeft($aFunctions[$j], $k) ExitLoop 2 EndIf $iCharCount += 1 Next Next Next $sCode = "" For $i = 0 To UBound($aFunctions) -1 $sCode &= $aFunctions[$i] Next MsgBox(0, "", $sCode) ; It appears to be parsing correctly now, even though the unfinished code looks somewhat unsophisticated - it's just a proof of concept for the time being. Identical variable names can be used, but will never be recognised by any function in which they appear as parameters. The same goes for function names, which will never be mistaken for variables. Disambiguation ; Before Distance (Floor( au + Random(Distance(Distance, au, angst ), au, angst )) , au , angst ) + angst + Distance(1, au , angst ) ; After _Distance(_Floor( au + _Random(_Distance(Distance,"au","angst" ),au,angst )),"au","angst" ) + angst + _Distance(1,"au","angst" ) ; I don't actually expect anyone to write such a messy expression as the one above. Nor do I expect anyone to convert a random number of astronomical units to angstroms. Edited August 17, 2014 by czardas operator64 ArrayWorkshop Link to comment Share on other sites More sharing options...
TypeIt Posted August 18, 2014 Share Posted August 18, 2014 I couldn't find a working regexp and gave up. If someone can figure it out, I'd love to see a more elegant solution.It's impossible to accept every context-free language with a finite-state automaton. The language you're describing is a context-free language, but isn't a regular language. That means that there is no regular expression that can do that for you.A simplified grammar of your language could be:expression -> expression + term | term term -> term * factor | factor factor -> ( expression ) | string | number | identifierYou can write a simple recursive descent parser which will accept your language. You simply write these rules in AutoIt:expandcollapse popup#include <Array.au3> #Region Parser Func ParseExpression (Const ByRef $charArray, ByRef $index, ByRef $errors) Local $expression = ParseTerm ($charArray, $index, $errors) While $charArray [$index] == "+" $index += 1 Local $newExpression = ["+", $expression, ParseTerm ($charArray, $index, $errors)] $expression = $newExpression WEnd Return $expression EndFunc Func ParseTerm (Const ByRef $charArray, ByRef $index, ByRef $errors) Local $term = ParseFactor ($charArray, $index, $errors) While $charArray [$index] == "*" $index += 1 Local $newTerm = ["*", $term, ParseFactor ($charArray, $index, $errors)] $term = $newTerm WEnd Return $term EndFunc Func ParseFactor (Const ByRef $charArray, ByRef $index, ByRef $errors) If $charArray [$index] == "(" Then $index += 1 Local $factor = ["()", ParseExpression ($charArray, $index, $errors)] If $charArray [$index] == ")" Then $index += 1 Else _ArrayAdd ($errors, "syntax error: missing ')' at " & $index) EndIf Return $factor EndIf If $charArray [$index] == '"' Then Return ParseString ($charArray, $index, $errors) If NextIsNumber ($charArray, $index) Then Return ParseNumber ($charArray, $index, $errors) Return ParseIdentifier ($charArray, $index, $errors) EndFunc Func NextIsNumber (Const ByRef $charArray, ByRef $index) Local $charCode = Asc ($charArray [$index]) Return $charCode >= Asc ("0") And $charCode <= Asc ("9") EndFunc Func ParseNumber (Const ByRef $charArray, ByRef $index, ByRef $errors) Local $number = 0 Do $number *= 10 $number += Asc ($charArray [$index]) - Asc ("0") $index += 1 Until Not NextIsNumber ($charArray, $index) Local $result = ["number", $number] Return $result EndFunc Func ParseString (Const ByRef $charArray, ByRef $index, ByRef $errors) Local $string = "" $index += 1 ; Keep it simple and do not allow escape sequences. While $charArray [$index] <> '"' And $index < UBound ($charArray) - 1 $string &= $charArray [$index] $index += 1 WEnd If $charArray [$index] == '"' Then $index += 1 Else _ArrayAdd ($errors, "syntax error: missing '""' at " & $index) EndIf Local $result = ["string", $string] Return $result EndFunc Func NextIsIdentifierStart (Const ByRef $charArray, ByRef $index) Return $charArray [$index] == "$" Or $charArray [$index] == "@" Or $charArray [$index] == "_" Or StringRegExp ($charArray [$index], "\p{L}") EndFunc Func NextIsIdentifierPart (Const ByRef $charArray, ByRef $index) Return StringRegExp ($charArray [$index], "\w") EndFunc Func ParseIdentifier (Const ByRef $charArray, ByRef $index, ByRef $errors) Local $identifier = $charArray [$index] $index += 1 While NextIsIdentifierPart ($charArray [$index]) $identifier &= $charArray [$index] WEnd Local $result = ["identifier", $identifier] Return $result EndFunc Func Parse (Const ByRef $expression, ByRef $errors) Local $charArray = StringSplit (StringRegExpReplace ($expression & Chr (0), "\s", ""), "") Local $index = 1 ; StringRegExpReplace doesn't return a zero-based array. Local $result = ParseExpression ($charArray, $index, $errors) If $index < UBound ($charArray) - 1 Then _ArrayAdd ($errors, "syntax error: unexpected character at " & $index) If UBound ($errors) > 0 Then SetError (1, UBound ($errors)) Return $result EndFunc #EndRegion #Region Visitor Func VisitExpression (ByRef $expression) Switch $expression [0] Case "+" Return VisitAddition ($expression) Case "*" Return VisitMultiplication ($expression) Case "()" Return VisitParentheses ($expression) Case "number" Return VisitNumber ($expression) Case "string" Return VisitString ($expression) Case "identifier" Return VisitIdentifier ($expression) EndSwitch EndFunc Func VisitAddition (ByRef $expression) Return VisitExpression ($expression [1]) & " + " & VisitExpression ($expression [2]) EndFunc Func VisitMultiplication (ByRef $expression) Return VisitExpression ($expression [1]) & " * " & VisitExpression ($expression [2]) EndFunc Func VisitParentheses (ByRef $expression) Return "(" & VisitExpression ($expression [1]) & ")" EndFunc Func VisitNumber (ByRef $expression) Return $expression [1] EndFunc Func VisitString (ByRef $expression) Return '"' & $expression [1] & '"' EndFunc Func VisitIdentifier (ByRef $expression) Return $expression [1] EndFunc #EndRegion Local $defaultExpression = '0*1 + 2*(34 + "5") + 6' Local $expression = InputBox ("", "Please enter an expression.", $defaultExpression) If @error Then Exit Local $output = "Before: " & $expression & @CRLF & @CRLF Local $errors [0] Local $abstractSyntaxTree = Parse ($expression, $errors) If @error Then $output &= UBound ($errors) & " error(s):" & @CRLF For $i = 1 To UBound ($errors) $output &= $i & ". " & $errors [$i - 1] & @CRLF Next $output &= @CRLF EndIf Local $transformedExpression = VisitExpression ($abstractSyntaxTree) $output &= "After: " & $transformedExpression & @CRLF & @CRLF Func _Sin ($argument) Return Sin ($argument) EndFunc Local $result = Execute ($transformedExpression) $output &= "Result: " & $result MsgBox (0, "", $output)A general visitor (VisitExpression($visitAddition,$visitMultiplication,...,ByRef $expression) with VisitAddition(ByRef $expression,$visitExpression)) isn't possible (without first-class functions). czardas, jvanegmond and Mobius 3 A link to a random TVTropes page to make this the last post you'll read. Link to comment Share on other sites More sharing options...
czardas Posted August 18, 2014 Author Share Posted August 18, 2014 (edited) Whoa - TypeIt, that's brilliant! I very much appreciate you spending the time to write such a detailed script for me. I can see it will be very helpful - not only for me. I will study it and try to make the best use of it. Thank you. Edited August 18, 2014 by czardas operator64 ArrayWorkshop Link to comment Share on other sites More sharing options...
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