Topic of the day: Ch. 4: Top-down parsing

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1 INF5110, Tuesday, February 10, 2015 Topic of the day: Ch. 4: Top-down parsing Slides also included here: The first 4 slides for Ch. 4 that also appeared at the end of the slides used at February 3 (about First, Follow etc.). 1

2 Ch. 4: Top-down parsing (also called LL(1) parsing) LL(1)-parsing: About the textbook and our curriculum: This slide should be read once more after we have been through Ch. 4: What in our textbook is called LL(1)-parsering (4.2.1, and 4.2.4) is a formal method for top down parsing that uses an explicit stack. This method is not much used, and is not included in our curriculum. From Ch. 4 we shall concentrate on the parsing method recursive descent. This is an intuitive method that many of you probably have already been in touch with, e.g. in INF2100. The recursive descent method is used with many types of descriptions of the grammar: Syntax diagrams, EBNF, or pure BNF. One usually work on an intuitive basis, and do not prove that things will work out correctly. We will discuss the technical requirements for rec. desc. to work correctly Our definition of an LL(1)-grammar: A pure BNF grammar that fullfils the requirements discussed in the previous point. 2

3 Example: «Recursive descent» from pure BNF Main idea: Write a method/procedure/function for each non-terminal Start each method by finding the actual alternative used Next token Global variable if a + b * ( c + d ) <= Then let the method do the parsing according to that «Typical» rec. dec. method for the last production in the above grammar. This is a very simple case. Checks that the expected terminal symbol really corresponds to the current token, and moves to next token. Is often used simply to move to the next input token (check is known to be OK) 3

4 Typical situation during recursive descent parsing Each triangle represents a method call S C B LL(1) philosophy: Each method can, by looking at only the current token, decide which alternative should be used for the current non-terminal (here A). A A α β γ... Input Which alternative for A? Must be decidable from the token Analysed and found OK token Should match the rest of the chosen alternative Resides in the variable «token» 4

5 5 Often rec. desc. with the given BNF won t work Performing left factoring sometimes helps Original version When A = if-stmt and token = «if» there are two alternatives that can be chosen: This is not an LL(1) grammar! The textbook proposes: Rewrite to: Rec. desc. method: BUT: We could also have used left factoring. Then we would get a separate procedure for the elsepart, corresponding to these productions: ifstmt if (exp) stmt elsepart elsepart ε else stmt

6 More problems with top-down parsing from pure BNF: Left recursion won t work! Here First(exp) will contain all terminals in First(X), where X is any of the the (other) alternatives. Thus both alternatives may start with «(» or «number», and we therefore don t know what alternative to use if one of therse is next token. In general, and by the same argument, recursive descent will never work with left recursion: If there are more than one alternative, there will be at leat two alternatives that have overlapping Firstsets. If there is only one alternative, we will end up in an infinite recursion. One possibility: We can remove left recursion from pure BNF, as we have learned earlier. We then get: 6

7 Problems after removing left recursion Example: «3 4 5» From previous slide: After removal of left recursion. The concrete syntax tree will be: abstract syntax tree that we really wanted to build: Rec.desc. methods The abstract syntax tree that we really wanted to build: 7

8 Rec.desc. after traditional removal of left recursion: The parse tree will now be right associative instead of instead of left recursive This must be corrected for, but this is not in the curriculum (page )! Expression: The abstract syntax tree that we really wanted to build: 8

9 Not curriculum: How to obtain a left recursive abstract syntax tree, even if the parse tree is right recursive Produce AST for (or compute the value of): The abstract syntax tree we want to build: 5-6 In the textbook: The parameter valuesofar to the procedure exp is the value of the expression to the left of current position. For tree-building this would be: rootoftreesofar 9

10 Another way to handler top-down parsing for pure BNF: When problems, rewrite to EBNF or syntax diagrams. exp exp addop term term Rewrite to EBNF: term term multop factor factor Rewrite to EBNF: We then get the program: We then get the program: exp Would have got the same program with this syntax diagram: term + - Would have got the same program with this syntax diagram: term factor *

11 How to produce something durig recursive-decent parsing? We want to «produce» an abstract syntax tree But the program below (from the textbook) instead produces the value of the expression: For it should return 4 Method calls! Can easily be extended to a full calculator

12 Building a syntax tree NB: Method call In an OO language: newtemp.leftchild NB: Method calls Important for AST: If there is only one «term», then no new node is produced! We deliver the one we got! 12

13 Method for abstract tree building: Production: factor (exp) number function factor: syntaxtree; var fact: syntaxtree; begin case token of ( : number : else error(...) ; end case; return fact; end factor;

14 Method for abstract tree building: Production: factor (exp) number function factor: syntaxtree; var fact: syntaxtree; begin case token of ( : number : match ( ( ) ; fact = exp ; match ( ) ) ; else error(...) ; end case; return fact; end factor; Gives dummy-test

15 Method for abstract tree building: Production: factor (exp) number function factor: syntaxtree; var fact: syntaxtree; begin case token of ( : match ( ( ) ; fact = exp ; match ( ) ) ; number : fact = makenumbernode(number) ; match (number ) ; else error(...) ; end case; return fact; end factor; Gives dummy-test

16 Parsing of an if-statement and building an AST if-stmt -> if (exp) stmt [else stmt] StmntNode (if) testchild thenchild elsechild := nil a b c + b 1 16

17 Important: What should occur in the variable «token» when entering and leaving the recurive methods? As before: The triangles are method calls. LL(1)-requirement: Each method (written for a specific nonterminal) can, by only looking at the current token, decide which of its alternatives is the correct one Input S C B A The rules saying what should be in «current» when entering or leaving a syntax method (say for A) When the method for A is called, the first term. symbol of the syntactic unit derived from A should reside in «token». When we leave the method for A, the first term. symbol after the unit derived from A should reside in «token» Det valgte alternativet for A Dette skal ligge i «token» når metoden for A returnerer Analysed, and found OK This should reside in «token when the method for A is called 17

18 LL(1) grammar 1. That is, if a First(α ) The LL(1) requirement for a pure BNF-grammar: What is required for a recursive descent parser should work directly from the grammar, without any rewriting To decide if a grammar is LL(1) : Set up a table M[N,T] and fill it in according to the following rules: then include A -> α in M[A, a] (Our) definition of LL(1): A pure BNF grammar is LL(1) if M[A,a] has only one production (or «error» ) in each entry. 2. That is, if: α is nullable, and a Follow(A ) Then add A -> α to M[A, a] 18

19 Filling in M[N, T] to check if G is LL(1) First Follow - Left factoring is done - The grammar is not left recursive statement other, if $, else if-stmt if $, else else-part else, ε $, else exp 0, 1 ) Note: Even if we: - Remove left rec. - Do left factoring. this is still not enough to guarantee that the grammar is LL(1)! Here: The table has two productions in M[else-part, else] 19

20 Earlier slide: The situation during rec. desc. parsing The triangles are method calls. S C Note: A grammar that is ambiguous can not be LL(1) B Therefore: An LL(1) grammar is unambiguous! A Which alternative for A? Input Analysed and found OK token Should be matched againt the chosen alternative for A 20

21 LL(1) table for an expression grammar We have earlier found First and Follow as an exercise: Original expr. grammar. Not LL(1)! With left rec. removed. Is it LL(1) now? We get the following First and Follow sets: 21

22 The M[N,T] table: Har fjernet venstre-rekursjon: Result: Only one choice in each entry! This expression grammar is LL(1). 22

23 When the parser finds an error The parser should all the way test that the program is OK, and if an error is found: At least do: Output an understandable error message, and may be stop the parsing. For «free-standing» compilers with error recovery : Give an understandable error message, as above. Read input symbols until it finds a point where it can resume the parsing to possible find more errors. Usually, if an error has been found, no machine code will be produced Observation: It is usually difficult to resume parsing after a syntax error. Resuming compilation after a semantic error is much easier. 23

24 Error messages in general: Important: Try to avoid error messages that only occur because of an already reported error! Report an error as early as possible, if possible at the first point where the program cannot be extended to a correct program. One has to make sure that, after an error, one don t end up in a infinite loop without reading any input symbols. What is a good error message? Assume that the method «factor» chooses the alternative «( exp )», but that it, when control returns from method «exp», do not find a «)». One could report: «left paranthesis is missing» But this can often be confusing, e.g. if what the program text is: «( a + b c )». Here the «exp» method will terminate after «( a + b» (as «c» cannot extend the expression). You should terefore rather give the message «error in expression or left paranthesis is missing» 24

25 Handling of sytax errors using recursive decent. Not included in curriculum! Method: «Panic mode» with use of «Synchronizing set» Synch-set (stack or parameter): <program> $ begin <deklsekv>; <setningssekv> end end <stmnt>; <stmnt>; <stmnt>;... if <expr> then <stmnt> [ else <stmnt> ] ; First(stmt) then First(stmt) else name if while for <term> ± <term> ± <term>... <factor> * <factor> *<factor>... (<expr>) <term> ± <term> ± <term> First(term) ( integer name * First(factor) ) + - ( tall navn 25

26 Syntax errors when using recursive descent, and a «synch stack» Not included in curriculum! From the sketch at the previous page we can easily find: - Which call should continue the execution? - What input symbol should this method search for before resuming? - - We assume that $ is added to the synch. stack only by the outermost method (for the start symbol) - The union of everything on the stack is called the synch. set, SS The algorithm for this goes is as follows: For each coming input symbol, test if it is a member of SS If so: Look through the SS stack from newest to oldest, and find the newest method that are willing to resume at one of these symbol This method will itself know how to resume after the actual input symbol What is not easy is to program this without destroing the nich program structure occuring from pure recursive descent. 26

27 Not included in the curriculum: Procedures for expressions with error recovery. The philosophy here is a little different than above, but not fully clear. Also { +, - }?? Main philosophy The method checkinput is called twice: First to check that the construction starts correctly, and secondly to check that the symbol after the construction is legal. Uses parameters, not a stack The procedures must themselves resume execution at the right place inside themselves when they get the control back, or it must terminate immediately if it cannot resume execution on the current symbol. if token in {(,number} then Why not the full synchset? 27

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