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ML-YACC

ML-YACC. David Walker COS 320. Outline. Last Week Introduction to Lexing, CFGs, and Parsing Today: More parsing: automatic parser generation via ML-Yacc Reading: Chapter 3 of Appel. Parser Implementation. Implementation Options: Write a Parser from scratch

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ML-YACC

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  1. ML-YACC David Walker COS 320

  2. Outline • Last Week • Introduction to Lexing, CFGs, and Parsing • Today: • More parsing: • automatic parser generation via ML-Yacc • Reading: Chapter 3 of Appel

  3. Parser Implementation • Implementation Options: • Write a Parser from scratch • not as boring as writing a lexer, but not exactly a weekend in the Bahamas • Use a Parser Generator • Very general & robust. sometimes not quite as efficient as hand-written parsers. Nevertheless, good for lazy compiler writers. Parser Specification

  4. Parser Implementation • Implementation Options: • Write a Parser from scratch • not as boring as writing a lexer, but not exactly a weekend in the Bahamas • Use a Parser Generator • Very general & robust. sometimes not quite as efficient as hand-written parsers. Nevertheless, good for lazy compiler writers. Parser Specification Parser parser generator

  5. Parser Implementation • Implementation Options: • Write a Parser from scratch • not as boring as writing a lexer, but not exactly a weekend in the Bahamas • Use a Parser Generator • Very general & robust. sometimes not quite as efficient as hand-written parsers. Nevertheless, good for lazy compiler writers. stream of tokens Parser Specification Parser parser generator abstract syntax

  6. ML-Yacc specification • three parts: User Declarations: declare values available in the rule actions %% ML-Yacc Definitions: declare terminals and non-terminals; special declarations to resolve conflicts %% Rules: parser specified by CFG rules and associated semantic action that generate abstract syntax

  7. ML-Yacc declarations (preliminaries) • specify type of positions %pos int * int • specify terminal and nonterminal symbols %term IF | THEN | ELSE | PLUS | MINUS ... %nonterm prog | exp | op • specify end-of-parse token %eop EOF • specify start symbol (by default, non terminal in LHS of first rule) %start prog

  8. Simple ML-Yacc Example grammar symbols %% %term NUM | PLUS | MUL | LPAR | RPAR %nonterm exp | fact | base %pos int %start exp %eop EOF %% exp : fact () | fact PLUS exp () fact : base () | base MUL factor () base : NUM () | LPAR exp RPAR () semantic actions (currently do nothing) grammar rules

  9. attribute-grammars • ML-Yacc uses an attribute-grammar scheme • each nonterminal may have a semantic value associated with it • when the parser reduces with (X ::= s) • a semantic action will be executed • uses semantic values from symbols in s • when parsing is completed successfully • parser returns semantic value associated with the start symbol • usually a parse tree

  10. attribute-grammars • semantic actions typically build the abstract syntax for the internal language • to use semantic values during parsing, we must declare symbol types: • %terminal NUM of int | PLUS | MUL | ... • %nonterminal exp of int | fact of int | base of int • type of semantic action must match type declared for LHS nonterminal in rule

  11. ML-Yacc with Semantic Actions grammar symbols with type declarations %% %term NUM of int | PLUS | MUL | LPAR | RPAR %nonterm exp of int | fact of int | base of int %pos int %start exp %eop EOF %% exp : fact (fact) | fact PLUS exp (fact + exp) fact : base (base) | base MUL base (base1 * base2) base : NUM (NUM) | LPAR exp RPAR (exp) computing integer result via semantic actions grammar rules with semantic actions

  12. ML-Yacc with Semantic Actions datatype exp = Int of int | Add of exp * exp | Mul of exp * exp %% ... %% exp : fact (fact) | fact PLUS exp (Add (fact, exp)) fact : base (base) | base MUL exp (Mul (base, exp)) base : NUM (Int NUM) | LPAR exp RPAR (exp) computing abstract syntax via semantic actions

  13. A simpler grammar datatype exp = Int of int | Add of exp * exp | Mul of exp * exp %% ... %% exp : NUM (Int NUM) | exp PLUS exp (Add (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | LPAR exp RPAR (exp) why don’t we just use this simpler grammar?

  14. A simpler grammar datatype exp = Int of int | Add of exp * exp | Mul of exp * exp %% ... %% exp : NUM (Int NUM) | exp PLUS exp (Add (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | LPAR exp RPAR (exp) this grammar is ambiguous! E E * E E E E + NUM E E E E * + NUM NUM + NUM * NUM NUM NUM NUM NUM

  15. a simpler grammar datatype exp = Int of int | Add of exp * exp | Mul of exp * exp %% ... %% exp : NUM (Int NUM) | exp PLUS exp (Add (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | LPAR exp RPAR (exp) But it is so clean that it would be nice to use. Moreover, we know which parse tree we want. We just need a mechanism to specify it! E E * E E E E + NUM E E E E * + NUM NUM + NUM * NUM NUM NUM NUM NUM

  16. Recall how LR parsing works: desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E + E elements of desired parse parsed so far We have a shift-reduce conflict. What should we do to get the right parse?

  17. Recall how LR parsing works: desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E + E * elements of desired parse parsed so far We have a shift-reduce conflict. What should we do to get the right parse? SHIFT

  18. Recall how LR parsing works: desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E + E * NUM elements of desired parse parsed so far SHIFT SHIFT

  19. Recall how LR parsing works: desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E + E * E elements of desired parse parsed so far REDUCE

  20. Recall how LR parsing works: desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E + E elements of desired parse parsed so far REDUCE

  21. Recall how LR parsing works: desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E elements of desired parse parsed so far REDUCE

  22. The alternative parse exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E + NUM NUM yet to read Input from lexer: NUM + NUM * NUM elements parsed so far State of parse so far: E + E We have a shift-reduce conflict. Suppose we REDUCE next

  23. The alternative parse exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + NUM NUM yet to read Input from lexer: NUM + NUM * NUM elements parsed so far State of parse so far: E REDUCE

  24. The alternative parse exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR * E E NUM E E + NUM NUM yet to read Input from lexer: NUM + NUM * NUM elements parsed so far State of parse so far: E * E Now: SHIFT SHIFT REDUCE

  25. The alternative parse E exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR * E E NUM E E + NUM NUM yet to read Input from lexer: NUM + NUM * NUM elements parsed so far State of parse so far: E REDUCE

  26. Summary desired parse tree: exp ::= NUM | exp PLUS exp | exp MUL exp | LPAR exp RPAR E E E + E E * NUM yet to read NUM NUM Input from lexer: NUM + NUM * NUM State of parse so far: E + E elements of desired parse parsed so far We have a shift-reduce conflict. We have E + E on stack, we see *. We want to shift. We ALWAYS want to shift since * has higher precedence than + ==> symbols to the right on the stack get processed first

  27. Example 2 exp ::= NUM | exp PLUS exp | exp MUL exp | exp MINUS exp | LPAR exp RPAR E E - NUM NUM yet to read Input from lexer: NUM - NUM - NUM elements parsed so far State of parse so far: E - E We have a shift-reduce conflict. We have E - E on stack, we see -. We want “-” to be a left-associative operator. ie: NUM – NUM – NUM == ((NUM – NUM) – NUM) What do we do?

  28. Example 2 exp ::= NUM | exp PLUS exp | exp MUL exp | exp MINUS exp | LPAR exp RPAR E E E - NUM NUM yet to read Input from lexer: NUM - NUM - NUM elements parsed so far State of parse so far: E We have a shift-reduce conflict. We have E - E on stack, we see -. What do we do? REDUCE

  29. Example 2 exp ::= NUM | exp PLUS exp | exp MUL exp | exp MINUS exp | LPAR exp RPAR - E E NUM E E - NUM NUM yet to read Input from lexer: NUM - NUM - NUM elements parsed so far State of parse so far: E - E SHIFT SHIFT REDUCE

  30. Example 2 E exp ::= NUM | exp PLUS exp | exp MUL exp | exp MINUS exp | LPAR exp RPAR - E E NUM E E - NUM NUM yet to read Input from lexer: NUM - NUM - NUM elements parsed so far State of parse so far: E REDUCE

  31. Example 2: Summary E exp ::= NUM | exp PLUS exp | exp MUL exp | exp MINUS exp | LPAR exp RPAR - E E NUM E E - NUM NUM yet to read Input from lexer: NUM - NUM - NUM elements parsed so far State of parse so far: E We have a shift-reduce conflict. We have E - E on stack, we see -. What do we do? REDUCE. We ALWAYS want to reduce since – is left-associative.

  32. precedence and associativity • three solutions to dealing with operator precedence and associativity: 1) let Yacc complain. • its default choice is to shift when it encounters a shift-reduce error • BAD: programmer intentions unclear; harder to debug other parts of your grammar; generally inelegant 2) rewrite the grammar to eliminate ambiguity • can be complicated and less clear 3) use Yacc precedence directives • %left, %right %nonassoc

  33. precedence and associativity • given directives, ML-Yacc assigns precedence to each terminal and rule • precedence of terminal based on order in which associativity is specified • precedence of rule is the precedence of the right-most terminal • eg: precedence of (E ::= E + E) == prec(+) • a shift-reduce conflict is resolved as follows • prec(terminal) > prec(rule) ==> shift • prec(terminal) < prec(rule) ==> reduce • prec(terminal) = prec(rule) ==> • assoc(terminal) = left ==> reduce • assoc(terminal) = right ==> shift • assoc(terminal) = nonassoc ==> report as error yet to read ....................T E input: terminal T next: RHS of rule on stack: ........E % E

  34. precedence and associativity datatype exp = Int of int | Add of exp * exp | Sub of exp * exp | Mul of exp * exp | Div of exp *exp %% %left PLUS MINUS %left MUL DIV %% exp : NUM (Int NUM) | exp PLUS exp (Add (exp1, exp2)) | exp MINUS exp (Sub (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | exp DIV exp (Div (exp1, exp2)) | LPAR exp RPAR (exp)

  35. precedence and associativity precedence directives: %left PLUS MINUS %left MUL DIV yet to read prec(MUL) > prec(PLUS) ....................MUL E input: terminal T next: RHS of rule on stack: ...E PLUS E

  36. precedence and associativity precedence directives: %left PLUS MINUS %left MUL DIV yet to read prec(MUL) > prec(PLUS) ....................MUL E input: terminal T next: RHS of rule on stack: ... E PLUS E SHIFT

  37. precedence and associativity precedence directives: %left PLUS MINUS %left MUL DIV yet to read prec(PLUS) = prec(SUB) ....................SUB E input: terminal T next: RHS of rule on stack: ...E PLUS E

  38. precedence and associativity precedence directives: %left PLUS MINUS %left MUL DIV yet to read prec(PLUS) = prec(SUB) ....................SUB E input: terminal T next: RHS of rule on stack: ...E PLUS E REDUCE

  39. one more example datatype exp = Int of int | Add of exp * exp | Sub of exp * exp | Mul of exp * exp | Div of exp *exp | Uminus of exp %% %left PLUS MINUS %left MUL DIV %% exp : NUM (Int NUM) | MINUS exp (Uminus exp) | exp PLUS exp (Add (exp1, exp2)) | exp MINUS exp (Sub (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | exp DIV exp (Div (exp1, exp2)) | LPAR exp RPAR (exp) yet to read ....................MUL E ...MINUS E what happens?

  40. one more example datatype exp = Int of int | Add of exp * exp | Sub of exp * exp | Mul of exp * exp | Div of exp *exp | Uminus of exp %% %left PLUS MINUS %left MUL DIV %% exp : NUM (Int NUM) | MINUS exp (Uminus exp) | exp PLUS exp (Add (exp1, exp2)) | exp MINUS exp (Sub (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | exp DIV exp (Div (exp1, exp2)) | LPAR exp RPAR (exp) yet to read ....................MUL E ...MINUS E what happens? prec(*) > prec(-) ==> we SHIFT

  41. the fix datatype exp = Int of int | Add of exp * exp | Sub of exp * exp | Mul of exp * exp | Div of exp *exp | Uminus of exp %% %left PLUS MINUS %left MUL DIV %left UMINUS %% exp : NUM (Int NUM) | MINUS exp %prec UMINUS (Uminus exp) | exp PLUS exp (Add (exp1, exp2)) | exp MINUS exp (Sub (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | exp DIV exp (Div (exp1, exp2)) | LPAR exp RPAR (exp) yet to read ....................MUL E ...MINUS E

  42. the fix datatype exp = Int of int | Add of exp * exp | Sub of exp * exp | Mul of exp * exp | Div of exp *exp | Uminus of exp %% %left PLUS MINUS %left MUL DIV %left UMINUS %% exp : NUM (Int NUM) | MINUS exp %prec UMINUS (Uminus exp) | exp PLUS exp (Add (exp1, exp2)) | exp MINUS exp (Sub (exp1, exp2)) | exp MUL exp (Mul (exp1, exp2)) | exp DIV exp (Div (exp1, exp2)) | LPAR exp RPAR (exp) yet to read ....................MUL E ...E MINUS E changing precedence of rule alters decision: prec(UMINUS) > prec(MUL) ==> we REDUCE

  43. the dangling else problem • Grammar: S ::= if E then S else S | if E then S | ... • Consider: if a then if b then S else S • parse 1: if a then (if b then S else S) • parse 2: if a then (if b then S) else S • Parser reports shift-reduce error • in default behavior: shift (what we want)

  44. the dangling else problem • Grammar: S ::= if E then S else S | if E then S | ... • Alternative solution is to rewrite grammar: S ::= M | U M ::= if E then M else M | ... U ::= if E then S | if E then M else U

  45. default behavior of ML-Yacc • Shift-Reduce error • shift • Reduce-Reduce error • reduce by first rule • generally considered unacceptable • for assignment 3, your job is to write a grammar for Fun such that there are no conflicts • you may use precedence directives tastefully

  46. Note: To enter ML-Yacc hell, use a parser to catch type errors • when doing assignment 3, your job is to catch parse errors • there are lots of programming errors that will slip by the parser: • eg: 3 + true • catching these sorts of errors is the job of the type checker • just as catching program structure errors was the job of the parser, not the lexer • attempting to do type checking in the parser is impossible (in general) • why? Hint: what does “context-free grammar” imply?

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