LR parsing techniques

LR parsing techniques • SLR (not in the book) • Simple LR parsing • Easy to implement, not strong enough • Uses LR(0) items • Canonical LR • Larger parser but powerful • Uses LR(1) items • LALR (not in the book) • Condensed version of canonical LR • May introduce conflicts • Uses LR(1) items

Finding handles • As a shift/reduce parser processes the input, it must keep track of all potential handles. • For example, consider the usual expression grammar and the input string x+y. • Suppose the parser has processed x and reduced it to E. Then, the current state can be represented by E • +E where • means • that an E has already been parsed and • that +E is a potential suffix, which, if found, will result in a successful parse. • Our goal is to eventually reach state E+E•, which represents an actual handle and should result in the reduction EE+E

LR parsing • Typically, LR parsing works by building an automaton where each state represents what has been parsed so far and what we hope to parse in the future. • In other words, states contain productions with dots, as described earlier. • Such productions are called items • States containing handles (meaning the dot is all the way to the right end of the production) lead to actual reductions depending on the lookahead.

SLR parsing • SLR parsers build automata where states contain items (a.k.a. LR(0) items) and reductions are decided based on FOLLOW set information. • We will build an SLR table for the augmented grammar S'S S  L=R S  R L  *R L  id R  L

SLR parsing • When parsing begins, we have not parsed any input at all and we hope to parse an S. This is represented by S'S. • Note that in order to parse that S, we must either parse an L=R or an R. This is represented by SL=R and SR • closure of a state: • if AaBb represents the current state and B is a production, then add B   to the state. • Justification: aBb means that we hope to see a B next. But parsing a B is equivalent to parsing a , so we can say that we hope to see a  next

SLR parsing • Use the closure operation to define states containing LR(0) items. The first state will be: • From this state, if we parse, say, an id, then we go to state • If, after some steps we parse input that reduces to an L, then we go to state S' S S   L=R S   R L   *R L   id R   L L  id  S  L =R R  L 

SLR parsing • Continuing the same way, we define all LR(0) item states: I1 S R I6 S  L=  R R   L L   *R L   id S' S S   L=R S   R L   *R L   id R   L S' S  S  L=R  I0 I9 id L I3 S  L =R R  L  = I2 L * * L  * R R   L L   id L   * R id R I5 L R  L  I7 I3 L  id  R id L  *R  I8 * I4 S  R 

SLR parsing • The automaton and the FOLLOW sets tell us how to build the parsing table: • Shift actions • If from state i, you can go to state j when parsing a token t, then slot [i,t] of the table should contain action "shift and go to state j", written sj • Reduce actions • If a state i contains a handle A, then slot [i, t] of the table should contain action "reduce using A", for all tokens t that are in FOLLOW (A). This is written r(A) • The reasoning is that if the lookahead is a symbol that may follow A, then a reduction A should lead closer to a successful parse. • continued on next slide

SLR parsing • The automaton and the FOLLOW sets tell us how to build the parsing table: • Reduce actions, continued • Transitions on non-terminals represent several steps together that have resulted in a reduction. • For example, if we are in state 0 and parse a bit of input that ends up being reduced to an L, then we should go to state 2. • Such actions are recorded in a separate part of the parsing table, called the GOTO part.

SLR parsing • Before we can build the parsing table, we need to compute the FOLLOW sets: S' S S  L=R S  R L  *R L  id R  L FOLLOW(S') = {$} FOLLOW(S) = {$} FOLLOW(L) = {$, =} FOLLOW(R) = {$, =}

SLR parsing state action goto id = * $ S L R 0 s3 s5 1 2 4 1 accept 2 s6/r(RL) 3 r(Lid) r(Lid) 4 r(SR) 5 s3 s5 7 8 6 s3 s5 7 9 7 r(RL) r(RL) 8 r(L*R) r(L*R) 9 r(SL=R) Note the shift/reduce conflict on state 2 when the lookahead is an =

Conflicts in LR parsing • There are two types of conflicts in LR parsing: • shift/reduce • On some particular lookahead it is possible to shift or reduce • The if/else ambiguity would give rise to a shift/reduce conflict • reduce/reduce • This occurs when a state contains more than one handle that may be reduced on the same lookahead.

Conflicts in SLR parsing • The parser we built has a shift/reduce conflict. • Does that mean that the original grammar was ambiguous? • Not necessarily. Let's examine the conflict: • it seems to occur when we have parsed an L and are seeing an =. A reduce at that point would turn the L into an R. However, note that a reduction at that point would never actually lead to a successful parse. In practice, L should only be reduced to an R when the lookahead is EOF ($). • An easy way to understand this is by considering that L represents l-values while R represents r-values.

Conflicts in SLR parsing • The conflict occurred because we made a decision about when to reduce based on what token may follow a non-terminal at any time. • However, the fact that a token t may follow a non-terminal N in some derivation does not necessarily imply that t will follow N in some other derivation. • SLR parsing does not make a distinction.

Conflicts in SLR parsing • SLR parsing is weak. • Solution : instead of using general FOLLOW information, try to keep track of exactly what tokens many follow a non-terminal in each possible derivation and perform reductions based on that knowledge. • Save this information in the states. • This gives rise to LR(1) items: • items where we also save the possible lookaheads.

Canonical LR(1) parsing • In the beginning, all we know is that we have not read any input (S'S), we hope to parse an S and after that we should expect to see a $ as lookahead. We write this as: S'S, $ • Now, consider a general item A, x. It means that we have parsed an , we hope to parse  and after those we should expect an x. Recall that if there is a production , we should add  to the state. What kind of lookahead should we expect to see after we have parsed ? • We should expect to see whatever starts a . If  is empty or can vanish, then we should expect to see an x after we have parsed  (and reduced it to B)

Canonical LR(1) parsing • The closure function for LR(1) items is then defined as follows:For each item A, x in state I, each production  in the grammar,and each terminal b in FIRST(x),add , b to IIf a state contains core item  with multiple possible lookaheads b1, b2,..., we write , b1/b2 as shorthand for , b1and , b2

Canonical LR(1) parsing I1 I9 S R I6 S L=  R, $ R   L, $ L   *R, $ L   id, $ S' S, $ S   L=R, $ S   R, $ L   *R, =/$ L   id, =/$ R   L, $ S' S , $ SL=R, $ I0 id L Lid, $ I3' S  L =R, $ R  L , $ = I2 * L R L, $ I7' * L *R, $ R  L, $ L  id, $ L  *R, $ L *R, =/$ R  L, =/$ L  id, =/$ L  *R, =/$ L id R I5 I5' L *R , $ I3 L  id , =/$ R id I8' * L R * I4 S  R, =/$ L *R , =/$ I8 R L, =/$ I7

Canonical LR(1) parsing • The table is created in the same way as SLR, except we now use the possible lookahead tokens saved in each state, instead of the FOLLOW sets. • Note that the conflict that had appeared in the SLR parser is now gone. • However, the LR(1) parser has many more states. This is not very practical.

LALR(1) parsing • This is the result of an effort to reduce the number of states in an LR(1) parser. • We notice that some states in our LR(1) automaton have the same core items and differ only in the possible lookahead information. Furthermore, their transitions are similar. • States I3 and I3', I5 and I5', I7 and I7', I8 and I8' • We shrink our parser by merging such states. • SLR : 10 states, LR(1): 14 states, LALR(1) : 10 states

Canonical LR(1) parsing I1 I9 S R I6 S L=  R, $ R   L, $ L   *R, $ L   id, $ S' S, $ S   L=R, $ S   R, $ L   *R, =/$ L   id, =/$ R   L, $ S' S , $ SL=R, $ I0 id L I3 S  L =R, $ R  L , $ = I2 * L * L *R, =/$ R  L, =/$ L  id, =/$ L  *R, =/$ id R I5 I3 L  id , =/$ R L, =/$ I7 id L R * I4 S  R, =/$ L *R , =/$ I8

Conflicts in LALR(1) parsing • Note that the conflict that had vanished when we created the LR(1) parser has not reappeared. • Can LALR(1) parsers introduce conflicts that did not exist in the LR(1) parser? • Unfortunately YES. • BUT, only reduce/reduce conflicts.

Conflicts in LALR(1) parsing • LALR(1) parsers cannot introduce shift/reduce conflicts. • Such conflicts are caused when a lookahead is the same as a token on which we can shift. They depend on the core of the item. But we only merge states that had the same core to begin with. The only way for an LALR(1) parser to have a shift/reduce conflict is if one existed already in the LR(1) parser. • LALR(1) parsers can introduce reduce/reduce conflicts. • Here's a situation when this might happen: A  B , x A  C , y A  B  , y A  C , x A  B  , x/y A  C , x/y merge with to get:

Error recovery in LR parsing • Errors are discovered when a slot in the action table is blank. • Phase-level recovery • associate error routines with the empty table slots. Figure out what situation may have cause the error and make an appropriate recovery. • Panic-mode recovery • discard symbols from the stack until a non-terminal is found. Discard input symbols until a possible lookahead for that non-terminal is found. Try to continue parsing.

Error recovery in LR parsing • Phase-level recovery • Consider the table for grammar EE+E | id + id $ E 0 e1 s2 e1 1 1 s3 e2 accept 2 e3e3 r(Eid) 3 e1 s2 e1 4 4 s3 e2 r(EE+E) Error e1: "missing operand inserted". Recover by inserting an imaginary identifier in the stack and shifting to state 2. Error e2: "missing operator inserted". Recover by inserting an imaginary operator in the stack and shifting to state 3 Error e3: "extra characters removed". Recover by removing input symbols until $ is found.

LR(1) grammars • Does right-recursion cause a problem in bottom-up parsing? • No, because a bottom-up parser defers reductions until it has read the whole handle. • Are these grammars LR(1)? How about LL(1)? SAa | Bb Ac Bc SAca | Bcb Ac Bc SAa | Bb AcA | a BcB | b LR(1): YES LL(1): NO LL(2): YES LR(1): NO LL(1): NO LL(2): NO LR(2): YES LR(1) : YES LL(k) : NO

LR parsing techniques