Week 4

1. Week 4 • Questions / Concerns • Comments about Lab1 • What’s due: • Lab1 check off this week (see schedule) • Homework #3 due Wednesday (Define grammar for your language) • Homework #4 due Thursday (Grammar modifications) • Coming up: • Lab2a & Lab2b posted. • Test#1 next week • Grammar Modifications • Recursive Descent Parser

2. Lab1 • Data structure for Symbol Table • List (dynamic) • Dynamic Array • std::vector<symbol> • Dynamically allocate more when needed but it’s done in binary (2, 4, 8, 16, etc.) • symbol * mySymbolArray • Dynamically allocate more space when needed (how many more at a time?) • Map • Maps string (name) to more info about the name • Sorted • Binary search tree (red/black tree). Tree is always balanced. • Unordered map • STL’s hashtable

3. Preprocessor / Symbol Table • Given the following code snippet: #define MAX 5 void main() { int x = 5; int y = 6; …… #if x == 5 //do something #endif const int MIN = 0; Add (MAX 5) to the symbol table main, x,y are not added to the symbol table in the preprocessor Why? There is no preprocessor symbol called x in the symbol table This can be added to the symbol in preprocessor because it’s just a constant that’s not going to change

4. Lab1 check-off Schedule • Wednesday: I will be in and out most of the day but can check-off labs whenever I am on. • Thursday: I will be available in the morning for lab check-off and again in late afternoon.

5. Structure of Compilers skeletal source program preprocessor Modified Source Program Syntax Analysis (Parser) Lexical Analyzer (scanner) Tokens Syntactic Structure Semantic Analysis Intermediate Representation Optimizer Symbol Table Code Generator Target machine code

6. Grammar Example S -> E E -> E + E E -> E * E E -> id • This grammar is ambiguous.

7. Revised & Expanded Grammar Example S -> id = E ; E -> E + T | E – T | T T -> T * F | T / F | F F -> ( E ) | id S ; E id (i) = E T + T F T * F F id(c) id(a) id(b) i = a + b * c;

8. But… S -> id = E ; E -> E + T |E – T | T T -> T * F |T / F | F F -> ( E ) | id This grammar doesn’t work for top-down because of left recursion

9. In-Class Exercise #6 S -> id = E ; E -> E + T |E – T | T T -> T * F |T / F | F F -> ( E ) | id • Remove left-recursion from this grammar

10. Recursive Descent Parser (RDP) • Is a top down parser • Start with grammar modifications • MUST remove all left recursion from the grammar. • Try to remove all unit productions. • Try to left factor so the grammar is one-token look ahead. • Input to the parser is a list of tokens. • Output: • Yes/No: Did the input parse? • Parse structure: representing all the statements.

11. Grammar HW#2 • Let’s look at the grammar for HW#2 • Identify left recursion • Identify opportunities for one-token look ahead • Identify unit productions

12. HW#2 Grammar COMPOUND_STAT -> begin OPTIONAL_STAT end STATEMENT -> VARIABLE :=EXPRESSION | COMPOUND_STAT | PROCEDURE_CALL | if EXPRESSION then STATEMENT else STATEMENT | while EXPRESSION do STATEMENT VARIABLE -> id PROCEDURE_CALL -> id | id ( EXPR_LIST )

13. Recursive Descent Parser Each token is a pair (Value, Type) Lab1 RDP List of Tokens ( symbol void keyword main ID ) symbol { symbol int keyword

14. Recursive Descent Parser Each token is a pair (Value, Type) Lab1 RDP List of Tokens void keywordmain ID ( symbol ) symbol Tokens can be in a separate file

15. Recursive Descent Parser • There are 2 types of rules in the grammar: • With  productions • Without  productions • Example: • Field -> `[´ Exp `]´ `=´ Exp | • Name `=´ Exp | Exp • Funcname2 -> ‘.’ Name Funcname2 | 

16. Recursive Descent Parser • Load tokens into a buffer. • Need ability to pull out tokens but also put them back if you can’t use them in a rule. (Backtracking) main ID void ( ) current

17. Recursive Descent Parser • For every rule in the grammar, generate a bool function. • This answers the yes/no question first. • Non- rules: • If tokens match the rule, return true. • If tokens do not match the rule, return false. •  rules: • This means that this rule doesn’t match ANY tokens. It’s not wrong, it’s just that it doesn’t match anything at this point in the matching. • One way to handle it is to just return True and let other rules handle the next token.

18. Bool Function S -> a S b | c • bool S() • { if currentToken == a • if (S()) • if nextToken == b • return True; • else • return False; • else return False; • else if currentToken == c • return True; • else • return False; • }

19. Bool Function S -> a S b |  • bool S() • { if currentToken == a • if (S()) • if nextToken == b • return True; • else • return False; • else return False; • else • return True; //for  • }

20. Bool Function Parser is a pushdown automata But where is the “stack”? S -> a S b| c • bool S() • { if currentToken == a • if (S()) • if nextToken == b • return True; • else • return False; • else return False; • else if currentToken == c • return True; • else • return False; • } Call stack

21. Bool function Field -> `[´ Exp `]´ `=´ Exp | Name `=´ Exp | Exp bool Field() { if currentToken == ‘[‘ if (Exp()) if nextToken == ‘]’ if nextToken == ‘=‘ if (Exp()) return True; else if currentToken == Name if nextToken == ‘=‘ if (Exp()) return True; else if (Exp()) return True; } Return false for all other conditions May need to put tokens back before checking this rule

22. Bool function Funcname2 -> ‘.’ Name Funcname2 |  • bool Funcname2() • { if currentToken == ‘.‘ • if (nextToken == Name) • if (Funcname2()) • return True; • else • return True; //for .. Don’t remove //any tokens • }