1 LR Parsers  The most powerful shift-reduce parsing (yet efficient) is: LR(k) parsing. LR(k) parsing. left to right right-most k lookhead scanning derivation (k is omitted  it is 1)  LR parsing is attractive because:  LR parsing is most general non-backtracking shift-reduce parsing, yet it is still efficient.  The class of grammars that can be parsed using LR methods is a proper superset of the class of grammars that can be parsed with predictive parsers. LL(1)-Grammars  LR(1)-Grammars  An LR-parser can detect a syntactic error as soon as it is possible to do so a left-to-right scan of the input.

2 LR Parsers  LR-Parsers  covers wide range of grammars.  SLR – simple LR parser  LR – most general LR parser  LALR – intermediate LR parser (look-head LR parser)  SLR, LR and LALR work same (they use the same algorithm), only their parsing tables are different.

3 LR Parsing Algorithm SmSmSmSm XmXmXmXm S m- 1 X m- 1.. S1S1S1S1 X1X1X1X1 S0S0S0S0 a1a1a1a aiaiaiai anananan$ Action Table terminals and $ terminals and $s t four different a actions tes Goto Table non-terminal non-terminals t each item is a a state number tes LR Parsing Algorithm stack input output

4 A Configuration of LR Parsing Algorithm  A configuration of a LR parsing is: ( S o X 1 S 1... X m S m, a i a i+1... a n $ ) StackRest of Input  S m and a i decides the parser action by consulting the parsing action table. (Initial Stack contains just S o )  A configuration of a LR parsing represents the right sentential form: X 1... X m a i a i+1... a n $

5 Actions of A LR-Parser 1. shift s -- shifts the next input symbol and the state s onto the stack ( S o X 1 S 1... X m S m, a i a i+1... a n $ )  ( S o X 1 S 1... X m S m a i s, a i+1... a n $ ) 2. reduce A  (or reduce n, where n is a production number)  pop 2*|  | (=r) items from the stack;  then push A and s where s=goto[s m-r,A] ( S o X 1 S 1... X m S m, a i a i+1... a n $ )  ( S o X 1 S 1... X m-r S m-r A s, a i... a n $ )  Output is the reducing production reduce A  3. Accept – Parsing successfully completed 4. Error -- Parser detected an error (an empty entry in the action table)

6 Reduce Action  pop 2*|  | (=r) items from the stack; let us assume that  = Y 1 Y 2...Y r  then push A and s where s=goto[s m-r,A] ( S o X 1 S 1... X m-r S m-r Y 1 S m-r...Y r S m, a i a i+1... a n $ )  ( S o X 1 S 1... X m-r S m-r A s, a i... a n $ )  ( S o X 1 S 1... X m-r S m-r A s, a i... a n $ )  In fact, Y 1 Y 2...Y r is a handle. X 1... X m-r A a i... a n $  X 1... X m Y 1...Y r a i a i+1... a n $

7 (SLR) Parsing Tables for Expression Grammar stateid+*()$ETF 0s5s4123 1s6acc 2r2s7r2r2 3r4r4r4r4 4s5s4823 5r6r6r6r6 6s5s493 7s5s410 8s6s11 9r1s7r1r1 10r3r3r3r3 11r5r5r5r5 Action TableGoto Table 1) E  E+T 2) E  T 3) T  T*F 4) T  F 5) F  (E) 6) F  id

8 Actions of A (S)LR-Parser -- Example stackinputactionoutput 0id*id+id$shift 5 0id5*id+id$reduce by F  id F  id 0F3*id+id$reduce by T  F T  F 0T2*id+id$shift 7 0T2*7id+id$shift 5 0T2*7id5+id$reduce by F  id F  id 0T2*7F10+id$ reduce by T  T*FT  T*F 0T2+id$reduce by E  T E  T 0E1+id$shift 6 0E1+6id$shift 5 0E1+6id5$reduce by F  id F  id 0E1+6F3$reduce by T  F T  F 0E1+6T9$reduce by E  E+TE  E+T 0E1$accept

9 Constructing SLR Parsing Tables – LR(0) Item  An LR(0) item of a grammar G is a production of G a dot at the some position of the right side.  Ex:A  aBb Possible LR(0) Items:A . aBb (four different possibility) A  a. Bb (four different possibility) A  a. Bb A  aB. b A  aB. b A  aBb. A  aBb.  Sets of LR(0) items will be the states of action and goto table of the SLR parser.  A collection of sets of LR(0) items (the canonical LR(0) collection) is the basis for constructing SLR parsers.  Augmented Grammar: G’ is G with a new production rule S’  S where S’ is the new starting symbol.

10 The Closure Operation  If I is a set of LR(0) items for a grammar G, then closure(I) is the set of LR(0) items constructed from I by the two rules: 1. Initially, every LR(0) item in I is added to closure(I). 2. If A  . B  is in closure(I) and B  is a production rule of G; then B .  will be in the closure(I). We will apply this rule until no more new LR(0) items can be added to closure(I).

11 The Closure Operation -- Example E’  E closure({E’ . E}) = E  E+T { E’ . Ekernel items E  TE . E+T T  T*FE . T T  FT . T*F F  (E)T . F F  idF . (E) F . id }

12 GOTO function  Definition. Goto(I, X) = closure of the set of all items  Definition. Goto(I, X) = closure of the set of all items A  X.  where A .X  belongs to I   Intuitively: Goto(I, X) set of all items that “reachable” from the items of I once X has been “seen.”   E.g. consider I={E’  E., E  E.+T} and compute Goto(I, +) Goto(I, +) = { E  E+.T, T .T * F, T .F, F .( E ), F .id }

13 Construction of The Canonical LR(0) Collection  To create the SLR parsing tables for a grammar G, we will create the canonical LR(0) collection of the grammar G’.  Algorithm: C is { closure({S’ . S}) } repeat the followings until no more set of LR(0) items can be added to C. for each I in C and each grammar symbol X if goto(I,X) is not empty and not in C add goto(I,X) to C  goto function is a DFA on the sets in C.

14 The Canonical LR(0) Collection -- Example I 0 : E’ .EI 1 : E’  E.I 6 : E  E+.T I 9 : E  E+T. E .E+T E  E.+T T .T*F T  T.*F E .E+T E  E.+T T .T*F T  T.*F E .T T .F E .T T .F T .T*FI 2 : E  T. F .(E) I 10 : T  T*F. T .T*FI 2 : E  T. F .(E) I 10 : T  T*F. T .F T  T.*F F .id T .F T  T.*F F .id F .(E) F .(E) F .id I 3 : T  F.I 7 : T  T*.FI 11 : F  (E). F .id I 3 : T  F.I 7 : T  T*.FI 11 : F  (E). F .(E) F .(E) I 4 : F  (.E) F .id E .E+T E .E+T E .T I 8 : F  (E.) E .T I 8 : F  (E.) T .T*F E  E.+T T .T*F E  E.+T T .F T .F F .(E) F .(E) F .id F .id I 5 : F  id.

15 Transition Diagram (DFA) of Goto Function I0I0 I1I2I3I4I5I1I2I3I4I5 I 6 I 7 I 8 to I 2 to I 3 to I 4 I 9 to I 3 to I 4 to I 5 I 10 to I 4 to I 5 I 11 to I 6 to I 7 id ( F * E E + T T T ) F F F ( ( * (

16 Constructing SLR Parsing Table (of an augumented grammar G’) 1. Construct the canonical collection of sets of LR(0) items for G’. C  {I 0,...,I n } 2. Create the parsing action table as follows If a is a terminal, A .a  in I i and goto(I i,a)=I j then action[i,a] is shift j. If A . is in I i, then action[i,a] is reduce A  for all a in FOLLOW(A) where A  S’. If S’  S. is in I i, then action[i,$] is accept. If any conflicting actions generated by these rules, the grammar is not SLR(1). 3. Create the parsing goto table for all non-terminals A, if goto(I i,A)=I j then goto[i,A]=j 4. All entries not defined by (2) and (3) are errors. 5. Initial state of the parser contains S’ .S

17 Parsing Tables of Expression Grammar stateid+*()$ETF 0s5s4123 1s6acc 2r2s7r2r2 3r4r4r4r4 4s5s4823 5r6r6r6r6 6s5s493 7s5s410 8s6s11 9r1s7r1r1 10r3r3r3r3 11r5r5r5r5 Action TableGoto Table 1-2 E  E + T | T 3-4 T  T * F | F 5-6 T  ( E ) | id

18 SLR(1) Grammar  An LR parser using SLR(1) parsing tables for a grammar G is called as the SLR(1) parser for G.  If a grammar G has an SLR(1) parsing table, it is called SLR(1) grammar (or SLR grammar in short).  Every SLR grammar is unambiguous, but every unambiguous grammar is not a SLR grammar.

19 shift/reduce and reduce/reduce conflicts  If a state does not know whether it will make a shift operation or reduction for a terminal, we say that there is a shift/reduce conflict.  If a state does not know whether it will make a reduction operation using the production rule i or j for a terminal, we say that there is a reduce/reduce conflict.  If the SLR parsing table of a grammar G has a conflict, we say that that grammar is not SLR grammar.

20 Conflict Example S  L=R I 0 : S’ .S I 1 :S’  S. I 6 :S  L=.R S  R S .L=RR .L L  *R S .R I 2 :S  L.=RL .*R L  id L .*RR  L.L .id R  L L .id R .L I 3 :S  R. R .L I 3 :S  R. I 4 :L  *.R I 7 :L  *R. I 4 :L  *.R I 7 :L  *R. ProblemR .L ProblemR .L FOLLOW(R)={=,$}L .*R I 8 :R  L. = shift 6L .id reduce by R  L shift/reduce conflict I 5 :L  id. I 9 : S  L=R.

21 Conflict Example2 S  AaAb I 0 :S’ .S S  BbBaS .AaAb A   S .BbBa B   A . B . ProblemFOLLOW(A)={a,b}FOLLOW(B)={a,b} areduce by A   breduce by A   reduce by B   reduce by B   reduce/reduce conflictreduce/reduce conflict