CS606 Short Notes Lecture 16

Lecture 16

Here is the output trace for the expression : 2+4*6

>> Expr::isPresent() >> Term::isPresent() >> Factor::isPresent() token: 2 (257) << Factor::isPresent() return true >> Tprime::isPresent() token: + (267) << Tprime::isPresent() return false << Term::isPresent() return true >> Eprime::isPresent() token: + (267) >> Term::isPresent() >> Factor::isPresent() token: 4 (257) << Factor::isPresent() return true >> Tprime::isPresent() token: * (269) >> Factor::isPresent() token: 6 (257) << Factor::isPresent() return true >> Tprime::isPresent() token: (0) << Tprime::isPresent() return false << Tprime::isPresent() return true << Term::isPresent() return true >> Eprime::isPresent() token: (0) << Eprime::isPresent() return false << Eprime::isPresent() return true << Expr::isPresent() return true

** AST ** · (2+(4*6)) · Non-recursive Predictive Parsing

It is possible to build a non-recursive predictive parser. This is done by maintaining an explicit stack and using a table. Such a parser is called a table-driven parser. The nonrecursive LL(1) parser looks up the production to apply by looking up a parsing table. The LL(1) table has one dimension for current non-terminal to expand and another dimension for next token. Each table cell contains one production.

Consider the expression grammar

1 E ?

T E' 2 E' ? + T E'

ε

4 T ?

F T' 5 T' ? * F T'

ε

7 F ?

( E ) id

Using the table construction algorithm that will be discussed later, we have the predictive parsing table

id E E ? TE' E ? TE'

E'

E' ? +TE' E' ? ε E' ? ε

T T ? FT' T ? FT'

T'

T' ? ε T ? *FT' T' ? ε T' ? ε F F ? id F ? (E )

The rows are non-terminals and the columns are the terminals of the expression grammar. The predictive parser uses an explicit stack to keep track of pending non-terminals. It can thus be implemented without recursion. a b

Predictive parser stack

X Y Z

Parsing table M input

output

LL(1) Parsing Algorithm

The input buffer contains the string to be parsed; $ is the end-of-input marker. The stack contains a sequence of grammar symbols. Initially, the stack contains the start symbol of the grammar on the top of $. The parser is controlled by a program that behaves as follows:

The program considers X, the symbol on top of the stack, and a, the current input symbol. These two symbols, X and a determine the action of the parser. There are three possibilities.

1. X = a = $, the parser halts and annouces successful completion.
2. X = a ≠ $ the parser pops X off the stack and advances input pointer to next input symbol.

Q3. If X is a nonterminal, the program consults entry M[X,a] of parsing table M.

1. If the entry is a production M[X,a] = {X ? UVW }, the parser replaces X on top of the stack by WVU (with U on top). As output, the parser just prints the production used: X ? UVW. However, any other code could be executed here.
2. If M[X,a] =error, the parser calls an error recovery routine

Example: let’s parse the input string

id+id∗id

using the non-recursive LL(1) parser