Recursive descent parsing Abstract Syntax Trees ASTs n

Recursive descent parsing

Abstract Syntax Trees (ASTs) n An AST is a way of representing a computer program n n It is abstract because it throws away unnecessary information (comments, whitespace, punctuation used only for disambiguation, etc. ) It represents syntax, not semantics n However, well-written syntax can sometimes help with the semantics n For example, <expr> : : = <integer> | <expr> + <expr> | <expr> * <expr> is technically correct, but gives us no help in recognizing that multiplication has precedence over addition n The program is represented as a tree 2

The Stack n One easy way to do recursive descent parsing is to have each parse method take the tokens it needs, build a parse tree, and put the parse tree on a global stack n Write a parse method for each nonterminal in the grammar n n n Each parse method should get the tokens it needs, and only those tokens n Those tokens (usually) go on the stack Each parse method may call other parse methods, and expect those methods to leave their results on the stack Each (successful) parse method should leave one result on the stack

From Recognizer to Parser n First, make a Recognizer n n Create a Stack<Tree<Token>> as a globally available instance variable n n You will use this to hold the trees as you build them The methods from the Recognizer all return a boolean; do not change this n n The Recognizer code will form the “skeleton” of your Parser Your new results will go onto the Stack Each time you recognize something, also build a Tree to represent it, and put this Tree onto the Stack n Most of the time, you will assemble the new Tree from the two Trees you most recently put onto the Stack

Example: Unary minus n n A simple example of an <expression> is -5 As the program steps through the code for <expression>, it puts – (minus) on the stack, then it puts 5 on the stack n n n Each of these is in the form of a Tree consisting of a single node with a Token as its value Note that the most recently parsed item, 5, is on the top of the stack These can be combined into a new Tree, with the minus as the root and the 5 as its child

make. Tree n /** * Removes two Trees from the stack, makes a new Tree and * puts it on the stack. The element on the top of the stack * (the most recently seen element) is made the child of the * element beneath it (which was seen earlier). */ private void make. Tree() { Tree<Token> child = stack. pop(); Tree<Token> parent = stack. pop(); parent. add. Child(child); stack. push(parent); }

Example: while statement n n <while statement> : : = “while” <condition> <block> The parse method for a <while statement> does this: n n n Calls the Tokenizer, which returns a “while” token Makes the “while” into a Tree, which it puts on the stack Calls the parser for <condition>, which parses a condition and puts a Tree representation of that condition on the stack n n Calls the parser for <block>, which parses a block and puts a Tree representation of that block on the stack n n Stack now contains: “while” <condition> (stack “top” is on the right), where <condition> stands for some created Tree Stack now contains: “while” <condition> <block> Pops the top three things from the stack, assembles them into a Tree representing a while statement, and pushes this Tree onto the stack

Sample Java code n public boolean while. Command() { if (keyword("while")) { if (condition()) { if (block()) { make. Tree(3, 2, 1); return true; } } error("Error in "while" statement"); } return false; } while condition block <block> <condition> while <while stmt>

make. Tree n private void make. Tree(int root. Index, int. . . child. Indices) { // Get root from stack Tree<Token> root = get. Stack. Item(root. Index); // Get other trees from stack and add them as children of root for (int i = 0; i < child. Indices. length; i++) { root. add. Child(get. Stack. Item(child. Indices[i])); } // Pop root and all children from stack for (int i = 0; i <= child. Indices. length; i++) { stack. pop(); } // Put the root back on the stack. push(root); } private Tree<Token> get. Stack. Item(int n) { return stack. get(stack. size() - n); }

Fancier error messages n public boolean while. Command() { if (keyword("while")) { if (condition()) { if (block()) { make. Tree(3, 2, 1); // or some such return true; } error("Error in "while" block"); } error("Error in "while" condition"); } return false; }

Alternative code n n public boolean while. Command() { if (keyword("while") && condition() && block()) { make. Tree(3, 2, 1); // or some such return true; } return false; } No room for an error condition in this code

Alternative code with one message n public boolean while. Command() { if (keyword("while")) { if (condition()) && (block()) { make. Tree(3, 2, 1); // or some such return true; } error("Error in "while" statement"); } return false; }

Tricky problem: Defining <term> n <term> : : = <factor> “*” <term> | <factor> n n n <term> : : = <term> “*” <factor> | <factor> n n n (For simplicity, I’m ignoring the “/” and “%” operators) This is logically correct, but it defines the wrong tree for expressions such as x * y * z -- treats it as x * (y * z) This is equally correct, and correctly defines x * y * z as meaning (x * y) * z However, the left recursion can’t be programmed: “To recognize a term, first recognize a term” Solution: <term> : : = <factor> { “*” <factor> } n n This turns the recursion into an iteration The result is easy to program correctly

Code for term() n n public boolean term() { if (!factor()) return false; while (multiply. Operator()) { if (!factor()) { error("No term after '*' or '/'"); } make. Tree(2, 3, 1); // *, first factor, second factor } return true; } Here’s a snippet of code from my JUnit test methods: n n use("x * y * z"); assert. True(parser. term()); assert. Tree("*(*(x, y), z)"); use(String) and assert. Tree(String) are helper methods that I’ve written; you should be able to figure out what they do

An is. Command() method n public boolean command() { if (action()) return true; if (thought()) return true; }

My helper methods n n I wrote a number of helper methods for the Parser and for the Parser. Test classes One very useful method is tree, in the Parser. Test class n n n Another is assert. Stack. Top, which is just n n tree just takes Objects and builds a tree from them This method lets me build parse trees for use in assert. Equals tests private void assert. Stack. Top(Tree bt) { assert. Equals(bt, parser. stack. peek()); } Examples: n n Tree condition = tree("=", "2"); assert. Stack. Top(tree("if", condition, "list"));

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