# LL(k) Parsing. Predictive Parsers. LL(k) Parser Structure. Sample Parse Table. LL(1) Parsing Algorithm. Push RHS in Reverse Order 10/17/2012

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1 Predictive Parsers LL(k) Parsing Can we avoid backtracking? es, if for a given input symbol and given nonterminal, we can choose the alternative appropriately. his is possible if the first terminal of every alternative in a production is unique: A a B D b B B B c b c e D d parsing an input abced has no backtracking. LL(k) L left to right scan L leftmost derivation k k symbols of lookahead in practice, k = 1 It is table-driven and efficient. Left factoring to enable predication: A change to A A A For predicative parsers, must eliminate left recursion LL(k) Parser Structure Sample Parse able int * + ( ) Syntax Stack Parser Driver Output Syntax stack hold right hand side (RHS) of grammar rules M[A,b] an entry containing rule A or error Parser driver next action based on (current token, stack top) Implementation with 2-D parse table: A row for each non-terminal A column for all possible terminals and (the end of input marker) Every table entry contains at most one production Required for a grammar to be LL(1) No backtracking Fixed action for each (non-terminal, input symbol) combination Algorithm symbol at the top of the syntax stack a current input symbol Push RHS in Reverse Order symbol at the top of the syntax stack a current input symbol Parsing based on (, a): If = a =, then parser halts with success If = a, then pop from stack and advance input head If a, then Case (a): if, then parser halts with failed, input rejected Case (b): if N, M[,a] = RHS if M[,a] = B c D : B c D pop and push RHS to stack in reverse order 1

2 LL(1) Grammars Remove left recursive and perform left factoring Given the grammar: E + E int * int ( E ) int * int he grammar has no left recursion but requires left factoring. E After rewriting grammar, we have: E + E int ( E ) int * + ( ) int * int int * int E int * + ( ) int * + ( ) int * int int * int int int * + ( ) int * + ( ) 2

3 int * int int * int int * + ( ) int * + ( ) int * int int * int * int * + ( ) int * + ( ) int * int int * int int int * + ( ) int * + ( ) 3

4 int * int int * int int * + ( ) int * + ( ) int * int int * int Accept! int * + ( ) int * + ( ) Action List Stack Input Action E int * int E int * int int int int * int terminal * int * * int terminal int int int int terminal Halt and accept Constructing the Parse able We need to know what non-terminals to place our productions in the table? We know that we have restricted our grammars so that left recursion is eliminated and they have been left factored. hat means that each production is uniquely recognizable by the first terminal that production would derive. hus, we can construct our table from 2 sets: For each symbol A, the set of terminals that can begin a string derived from A. his set is called the FIRS set of A For each non-terminal A, the set of terminals that can appear after a string derived from A is called the FOLLOW set of A 4

5 First( ) First( ) = set of terminals that start string of terminals derived from. Apply following rules until no terminal or can be added 1. If t, then First( t ) = { t }. For example First( + ) = { + }. 2. If N and exists (nullable), then add to First( ). For example, First( ) = { *, }. 3. If N and m, where 1, 2, 3,... m are nonterminals, then: for each i from 1 to m if 1 i-1 are all nullable (or if i = 1) First( ) = First( ) First( i ) Follow( ) Follow( ) = { t S * t } Intuition: if A B, then First( B ) Follow( A ) However, B may be i.e., Apply following rules until no terminal or can be added 1. Follow( S ), where S is the start symbol. e.g., Follow( E ) = {... }. 2. Look at the occurrence of a non-terminal on the right hand side of a production which is followed by something If A B, then First( ) - { } Follow( B ) 3. Look at N on the RHS that is not followed by anything, if (A B) or (A B and First( )), then Follow( A ) Follow( B ) Algorithm to Compute FIRS, FOLLOW, and nullable Initialize FIRS and FOLLOW to all empty sets, and nullable to all false. foreach terminal symbol Z FIRS[Z] {Z} do foreach production 1 2 k if 1 k are all nullable (or if k = 0) then nullable[] true foreach i from 1 to k, each j from i + 1 to k if 1 i 1 are all nullable (or if i = 1) then FIRS[] FIRS[] FIRS[ i ] if i+1 k are all nullable (or if i = k) then FOLLOW[ i ] FOLLOW[ i ] FOLLOW[] if i+1 j 1 are all nullable (or if i + 1 = j ) then FOLLOW[ i ] FOLLOW[ i ] FIRS[ j ] until FIRS, FOLLOW, and nullable did not change in this iteration. Example E + E int ( E ) First Set: E + E int ( E ) Follow Set: E + E int ( E ) int Int *,, ), + +,, ) (, int, ), + E (, int, ) Constructing LL(1) Parse able Constructing LL(1) Parse able o construct the parse table, we check each A For each terminal a First( ), add A to M[A, ]. If First( ), then for each terminal b Follow(A), add A to M[A, ]. If First( ) and Follow(A), then add A to M[A, ]. For each terminal a First( ), add A to M[A, ]. E + E int ( E ) int int *,, ), + +,, ) (, int, ), + E (, int, ) int * + ( ) E E E + E 5

6 Constructing LL(1) Parse able Constructing LL(1) Parse able If First( ), then for each terminal b Follow(A), add A to M[A, ]. If First( ) and Follow(A), then add A to M[A, ]. E + E int ( E ) int int *,, ), + +,, ) (, int, ), + E (, int, ) E + E int ( E ) int int *,, ), + +,, ) (, int, ), + E (, int, ) int * + ( ) E E E + E int * + ( ) E E E + E Is a Grammar LL(1)? Observation If a grammar is LL(1), then each of its LL(1) table entries contain at most one rule. Otherwise, it is not LL(1) wo methods to determine if a grammar is LL(1) or not: 1. Construct LL(1) table, and check if there is a multi-rule entry or 2. Check each rule as if the table were being constructed: G is LL1(1) iff for a rule A a) First( ) First( ) = b) at most one of and can derive c) If derives, then First( ) Follow( ) = Ambiguous Grammars Some grammars may need more than one token of lookahead (k). However, some grammars are not LL regardless of how the grammar is changed. S if C then S if C then S else S C b change to S if C then S else S C b problem sentence: if b then if b then a else a else First() First()- Follow(S) else... else Follow() Removing Ambiguity o remove ambiguity, it is possible to rewrite the grammar. For the if-then-else example, how to rewrite? May not even need to rewrite in this case, we can just use the else S production over the LL(1) Summary LL(1) parsers operate in linear time and at most linear space relative to the length of input because: ime each input symbol is processed constant number of times Space stack is smaller than the input (in case we remove ) However, by changing the grammar, It might make the other phases of the compiler more difficult It becomes harder to determine semantics and generate code It is less appealing to programmers 6

7 Summary First and Follow sets are used to construct predictive parsing tables Intuitively, First and Follow sets guide the choice of rules: For non-terminal A and input t, use a production rule A where t First( ) For non-terminal A and input t, if A and t Follow(A), use the production A where First( ) What is LL(0)? Why are LL(2)... LL(k) are not widely used? 7

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