OPERATOR PRECEDENCE PARSING

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Presentation transcript:

OPERATOR PRECEDENCE PARSING Shubham Chauhan (11CS10043) Date: 3rd Sep

OPERATOR GRAMMAR No Ɛ-transition. No two adjacent non-terminals. Eg. E  E op E | id op  + | * The above grammar is not an operator grammar but: E  E + E | E* E | id

OPERATOR PRECEDENCE If a has higher precedence over b; a .> b If a has lower precedence over b; a <. b If a and b have equal precedence; a =. b Note: id has higher precedence than any other symbol $ has lowest precedence. if two operators have equal precedence, then we check the Associativity of that particular operator.

PRECEDENCE TABLE Example: w= $id + id * id$ .> <. Example: w= $id + id * id$ $<.id.>+<.id.>*<.id.>$

BASIC PRINCIPLE Scan input string left to right, try to detect .> and put a pointer on its location. Now scan backwards till reaching <. String between <. And .> is our handle. Replace handle by the head of the respective production. REPEAT until reaching start symbol.

ALGORITHM w  input a  input symbol b  stack top Repeat { if(a is $ and b is $) return if(a .> b) push a into stack move input pointer else if(a <. b) c  pop stack until(c .> b) else error() }

EXAMPLE STACK INPUT ACTION/REMARK $ id + id * id$ $ <. Id $ id $ <. + $ + id * id$ + <. Id $ + id * id$ id .> * + <. * $ + * id$ * <. Id $ + * id id .> $ * .> $ + .> $ accept

Algorithm for Constructing Precedence Functions Operator precedence parsers use precedence functions that map terminal symbols to integers. Algorithm for Constructing Precedence Functions Create functions fa for each grammar terminal a and for the end of string symbol. Partition the symbols in groups so that fa and gb are in the same group if a =· b (there can be symbols in the same group even if they are not connected by this relation). Create a directed graph whose nodes are in the groups, next for each symbols a and b do: place an edge from the group of gb to the group of fa if a <· b, otherwise if a ·> b place an edge from the group of fa to that of gb. If the constructed graph has a cycle then no precedence functions exist. When there are no cycles collect the length of the longest paths from the groups of fa and gb respectively.

Consider the following table: Resulting graph: gid fid f* g* g+ f+ f$ g$

From the previous graph we extract the following precedence functions: id + * $ f 4 2 5 1 3