Top-Down Parsign

• Construct a parse tree for the input string.
• OR finding the leftmost derivation for an input string.
• Consider id +id*id with the following grammar

• E	→	TE'
  E'	→	+TE' | ε'
  T	→	FT'
  T'	→	*FT' | ε'
  F	→	(E) | id

  
  E =>lm TE'                id[+id*id] (E → TE')
    =>lm FT'E'              id[+id*id] (T → FT')
    =>lm idT'E'             id[+id*id] (F → id) 
    =>lm idE'               id+[id*id] (T' → ε)  (Note we could have made a poor choice here)
    =>lm id+TE'             id+[id*id] (E' → +TE') (Note, we could have made a poor choice here)
    =>lm id+FT'E'           id+id[*id] (T → FT')
    =>lm id+idT'E'          id+id[*id] (F → id)
    =>lm id+id*FT'E'        id+id*[id] (T' → *FT') (Note, we could have made a poor choice here)
    =>lm id+id*idT'E'       id+id*id[] (F → id)
    =>lm id+id*idE'         id+id*id[] (T' → ε) (Note, we could have made a poor choice here)
    =>lm id+id*id           id+id*id[] (E' → ε) (Note, we could have made a poor choice here)
      

• At each step, determine the production to be applied.
• Then process the rest of the input string.
• We may have to backtrack in this process
• In the above grammar, any E', T' or F production may lead to backtracking
• But that was too much of a pain to type up, so none of that required.
• LL(1) is a parse from left to right, using a leftmost derivation with 1 lookhead to tokens

Recursive Descent Parsing

• Each production needs a routine similar to

• Assume A → A1|A2| ... |An 
  and  Ai =  Xi1Xi2 ... Xik
  bool A(symbol) 
  
   save ← symbol
   for j ← 1  to  n do
     for i ← 1 to k do
          if Xji is a nonterminal
               Call procedure Xji()
          else if Xji = symbol
               symbol ← Get_Next_Token()
          else
               RESET(symbol, save) //backtrack, this may be complex
               break
    return sumbol == save
  
  
      

• Using the above, how could parsing cad be problematic with

  S → c A d 
  A → ab | a