8 January 2004 Department of Software & Media Technology 1 Top Down Parsing Recursive Descent Parsing Top-down parsing: –Build tree from root symbol –Each production corresponds to one recursive procedure –Each procedure recognizes an instance of a non-terminal, returns tree fragment for the non-terminal

8 January 2004 Department of Software & Media Technology 2 General model Each right-hand side of a production provides body for a function Each non-terminal on the right hand side is translated into a call to the function that recognizes that non-terminal Each terminal in the right hand side is translated into a call to the lexical scanner. If the resulting token is not the expected terminal error occurs. Each recognizing function returns a tree fragment.

8 January 2004 Department of Software & Media Technology 3 Example: parsing a declaration FULL_TYPE_DECLARATION ::= type DEFINING_IDENTIFIER is TYPE_DEFINITION; Translates into: –get token type –Find a defining_identifier -- function call –get token is –Recognize a type_definition -- function call –get token semicolon In practice, we already know that the first token is type, that’s why this routine was called in the first place! Predictive parsing is guided by the next token

8 January 2004 Department of Software & Media Technology 4 Example: parsing a loop FOR_STATEMENT ::= ITERATION_SCHEME loop STATEMENTS end loop; Node1 := find_iteration_scheme; -- call function get token loop List1 := Sequence of statements -- call function get token end get token loop get token semicolon; Result := build loop_node with Node1 and List1 return Result

8 January 2004 Department of Software & Media Technology 5 Problem: If there are multiple productions for a non-terminal, mechanism is required to determine which production to use: IF_STAT ::= if COND then Stats end if; IF_STAT ::= if COND then Stats ELSIF_PART end if; When next token is if, so which production to use ?

8 January 2004 Department of Software & Media Technology 6 One Solution: factorize grammar If several productions have the same prefix, rewrite as single production: IF_STAT ::= if COND then STATS [ELSIF_PART] end if; –Problem now reduces to recognizing whether an optional –Component (ELSIF_PART) is present

8 January 2004 Department of Software & Media Technology 7 Second Problem of Recursion Grammar should not be left-recursive: E ::= E + T | T Problem: to find an E, start by finding an E… –Original scheme leads to infinite loop –Grammar is inappropriate for recursive-descent

8 January 2004 Department of Software & Media Technology 8 Solution to left-recursion E ::= E + T | T means that eventually E expands into T + T + T …. Rewrite as: –E ::= TE’ –E’ ::= + TE’ | epsilon Informally: E’ is a possibly empty sequence of terms separated by an operator

8 January 2004 Department of Software & Media Technology 9 Recursion can involve multiple productions A ::= B C | D B ::= A E | F –Can be rewritten as: A ::= A E C | F C | D –Now apply previous method –General algorithm to detect and remove left-recursion

8 January 2004 Department of Software & Media Technology 10 Further Problem Transformation does not preserve associativity : – E ::= E + T | T – Parses a + b + c as (a + b) + c – E ::= TE’, E’ ::= + TE’ | epsilon – Parses a + b +c as a + (b + c) –Incorrect for a - b – c : must rewrite tree

8 January 2004 Department of Software & Media Technology 11 In practice: use loop to find sequence of terms Node1 := P_Term; -- call function that recognizes a term loop exit when Token not in Token_Class_Binary_Addop; Node2 := New_Node (P_Binary_Adding_Operator); Scan; -- past operator Set_Left_Opnd (Node2, Node1); Set_Right_Opnd (Node2, P_Term); -- find next term Set_Op_Name (Node2); Node1 := Node2; -- operand for next operation end loop;

8 January 2004 Department of Software & Media Technology 12 LL (1) Parsing LL (1) grammars If table construction is successful, grammar is LL (1): left-to right, leftmost derivation with one-token lookahead. If construction fails, can conceive of LL (2), etc. Ambiguous grammars are never LL (k) If a terminal is in First for two different productions of A, the grammar cannot be LL (1). Grammars with left-recursion are never LL (k) Some useful constructs are not LL (k)

8 January 2004 Department of Software & Media Technology 13 Building LL (1) parse tables Table indexed by non-terminal and token. Table entry is a production: for each production P: A  loop for each terminal a in First (  ) loop T (A, a) := P; end loop; if  in First (  ), then for each terminal b in Follow (  ) loop T (A, b) := P; end loop; end if; end loop; All other entries are errors. If two assignments conflict, parse table cannot be built.

8 January 2004 Department of Software & Media Technology 14 Left Recursion Removal & Left Factoring Left Recursion Removal: Left Factoring:

8 January 2004 Department of Software & Media Technology 15 Synatx Tree Construction in LL(1) First and Follow Sets LL(k) Parsers (Extending the Lookahead Error Recovery in Top Down Parsers Error Recovery in LL(1) Parsers