Parser Hints

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Parser Hints
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The Stack

• To turn a Recognizer into a Parser, we need
  the use of a Stack
• All boolean Recognizer methods should continue to
  return boolean values, but in addition, if they
  succeed they should leave a Tree on the stack
• The stack should be an instance variable of the
  Parser
• Heres one easy way to do it
• Create a new Parser(String s) for each String you
  want to use, and call some parsing method (term,
  command, ...) for that particular String

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Recognizer helper methods

• private boolean number() return
  nextTokenMatches(Token.NUMBER)
• private boolean name() return
  nextTokenMatches(Token.NAME)
• private boolean keyword(String expectedKeyword)
  return nextTokenMatches(Token.KEYWORD,
  expectedKeyword)
• private boolean symbol(String expectedSymbol)
  return nextTokenMatches(Token.SYMBOL,
  expectedSymbol)

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nextTokenMatches

• private boolean nextTokenMatches(int type)
  Token t tokenizer.next() if (t.getType()
  type) return true else
  tokenizer.pushBack(t) return false
• private boolean nextTokenMatches(int type, String
  value) Token t tokenizer.next() if
  (type t.getType() value.equals(t.getValue())
  ) return true else
  tokenizer.pushBack(t) return false

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Revised nextTokenMatches

• private boolean nextTokenMatches(int type)
  Token t tokenizer.next() if (t.getType()
  type) stack.push(new Tree(t))
  return true else tokenizer.pushBack(t
  ) return false
• private boolean nextTokenMatches(int type, String
  value) Token t tokenizer.next() if
  (type t.getType() value.equals(t.getValue())
  ) stack.push(new Tree(t))
  return true else tokenizer.pushBack(t)
  return false

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comparator

• ltcomparatorgt "lt" "" "gt
• public boolean comparator() if
  (symbol("lt")) return true if (symbol(""))
  return true if (symbol("gt")) return true
  return false
• Whatever a comparator is found, one Tree node
  (with the Token representing that comparator as
  its value) is left on the stack
• Since we recognized one thing, and we leave one
  thing on the stack, this method does not need to
  be changed in any way

7
while command

• public boolean whileCommand() if
  (keyword("while")) if (condition())
  if (block())
  return true
  error("Error in \"while\" statement")
  return false

makeTree(3, 2, 1)
The makeTree method will take three things off
the stack and replace them with one thing
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Counting

• I like to count this way 1 2
  3 while condition block
• This way is easier to implement 2
  1 0 while condition block
• I prefer the first way because, although it means
  trickier arithmetic in makeTree, it simplifies
  counting everywhere else

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Accessing the Stack

• A stack, as a stack, has no methods for accessing
  the nth thing from the top, but...
• class Stack extends Vector
• ...and Vector has an elementAt(int index) operator

• This represents a stack with five things in it
• The e is at the top of the stack
• This means that, with some annoying arithmetic,
  you can access any element of the stack you like,
  counting elementAt(vector.size() - 1) as the top

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Recursion in BNF

• lttermgt ltfactorgt lttermgt ltfactorgtis bad,
  because of the left recursion
• Implemented in the obvious way, it would cause
  our program to go into an infinite recursion
• lttermgt ltfactorgt ltfactorgt lttermgthas
  another problem
• Implemented in the obvious way, it will build a
  tree of the wrong shape
• lttermgt ltfactorgt ltfactorgt works well
• Implemented in the obvious wayiteratively, not
  recursivelyit builds a tree of the correct shape

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term()

• public boolean term() if (!factor()) return
  false while (multiplyOperator())
  if (!factor()) error("No term after '' or
  '/'") makeTree(2, 3, 1)
  return true

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The End