1. BNF
9-Apr-19

2. BNF BNF stands for either Backus-Naur Form or Backus Normal Form
BNF is a metalanguage used to describe the grammar of a programming language
BNF is formal and precise
BNF is a notation for context-free grammars
BNF is essential in compiler construction

3. BNF
< > indicate a nonterminal that needs to be further expanded, e.g. <variable>
Symbols not enclosed in < > are terminals; they represent themselves, e.g. if, while, (
The symbol ::= means is defined as
The symbol | means or; it separates alternatives, e.g. <addop> ::= + | -

4. BNF uses recursion
<integer> ::= <digit> | <integer> <digit> or <integer> ::= <digit> | <digit> <integer>
Recursion is all that is needed (at least, in a formal sense)
"Extended BNF" allows repetition as well as recursion
Repetition is usually better when using BNF to construct a compiler

5. BNF Examples I <digit> ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
<if statement> ::= if ( <condition> ) <statement> | if ( <condition> ) <statement> else <statement>

6. BNF Examples II
<unsigned integer> ::= <digit> | <unsigned integer> <digit>
<integer> ::= <unsigned integer> | + <unsigned integer> | - <unsigned integer>

7. BNF Examples III
<identifier> ::= <letter> | <identifier> <letter> | <identifier> <digit>
<block> ::= { <statement list> }
<statement list> ::= <statement> | <statement list> <statement>

8. BNF Examples IV
<statement> ::= <block> | <assignment statement> | <break statement> | <continue statement> | <do statement> | <for loop> | <goto statement> | <if statement> |

9. Extended BNF The following are pretty standard:
[ ] enclose an optional part of the rule
Example: <if statement> ::= if ( <condition> ) <statement> [ else <statement> ]
{ } mean the enclosed can be repeated any number of times (including zero)
Example: <parameter list> ::= ( ) | ( { <parameter> , } <parameter> )

10. Parser Lexical Parser Analyzer Parser works on a stream of tokens.
The smallest item is a token.
token
Parser
source
program
Lexical
Analyzer
parse tree
get next token

11. Parse Tree
Inner nodes of a parse tree are non-terminal symbols.
The leaves of a parse tree are terminal symbols.
A parse tree can be seen as a graphical representation of a derivation.
Example:
E  E + E | E – E | E * E | E / E | - E
E  ( E )
E  id
E  -E E -
 -(E) E - ( ) E + - ( )
 -(E+E) E + - ( ) id E id + - ( )
 -(id+E)
 -(id+id)

12. Two groups of parser We categorize the parsers into two groups:
Top-Down Parser
the parse tree is created top to bottom, starting from the root.
Bottom-Up Parser
the parse is created bottom to top; starting from the leaves
Both top-down and bottom-up parsers scan the input from left to right (one symbol at a time).
Efficient top-down and bottom-up parsers can be implemented only for sub-classes of context-free grammars.
LL for top-down parsing
LR for bottom-up parsing

13. Left-Most and Right-Most Derivations
Left-Most Derivation
E  -E  -(E)  -(E+E)  -(id+E)  -(id+id)
Right-Most Derivation
E  -E  -(E)  -(E+E)  -(E+id)  -(id+id) lm lm lm lm lm rm rm rm rm rm

14. Ambiguity

15. Ambiguity rules For the most parsers, the grammar must be unambiguous.
We should eliminate the ambiguity in the grammar during the design phase of the compiler.
An unambiguous grammar should be written to eliminate the ambiguity.
We have to prefer one of the parse trees of a sentence (generated by an ambiguous grammar) to disambiguate that grammar to restrict to this choice.

16. Ambiguity
A grammar produces more than one parse tree for a sentence is
called as an ambiguous grammar.
Example:
E  E + E | E * E
E  id E + id *
E  E+E
 id+E
 id+E*E
id+id*E
 id+id*id E id + *
E  E*E
E+E*E
 id+E*E
id+id*E
 id+id*id

17. Unambiguous grammar unique selection of the parse tree for the grammar(Sentence).

18. Which is Ambiguity? 1? 2? BNF : stmt  if expr then stmt |
if expr then stmt else stmt | otherstmts
if E1 then if E2 then S1 else S2
1? 2?

19. Ambiguity stmt  matchedstmt | unmatchedstmt
We prefer the second parse tree (else matches with closest if).
So, we have to disambiguate our grammar to reflect this choice.
The disambiguous grammar will be:
stmt  matchedstmt | unmatchedstmt
matchedstmt  if expr then matchedstmt else matchedstmt | otherstmts
unmatchedstmt  if expr then stmt |
if expr then matchedstmt else unmatchedstmt

20. Operator Precedence- Ambiguity
Ambiguous grammars (because of ambiguous operators) can be disambiguated according to the precedence and associatively rules.
E  E+E | E*E | E^E | id | (E)
disambiguate the grammar
precedence: ^ (right to left)
* (left to right)
+ (left to right) 

21. Left Recursion

22. Left Recursion
A grammar is left recursive if it has a non-terminal A such that there is a derivation.
A  A for some string 
Top-down parsing techniques cannot handle left-recursive grammars.
So, we have to convert our left-recursive grammar into an equivalent grammar which is not left-recursive.
The left-recursion may appear in a single step of the derivation (immediate left-recursion), or may appear in more than one step of the derivation. +

23. Immediate Left-Recursion
A  A  |  where  does not start with A
 eliminate immediate left recursion
A   A’
A’   A’ |  an equivalent grammar

24. In general
A  A 1 | ... | A m | 1 | ... | n where 1 ... n do not start with A
 eliminate immediate left recursion
A  1 A’ | ... | n A’
A’  1 A’ | ... | m A’ |  an equivalent grammar

25. Immediate Left-Recursion
E  E+T | T
T  T*F | F
F  id | (E)

26. After eliminate immediate left recursion
E  T E’
E’  +T E’ | 
T  F T’
T’  *F T’ | 
F  id | (E)

27. Creating a top-down parser
Top-down parsing can be viewed as the problem of constructing a parse tree for the input string, starting form the root and creating the nodes of the parse tree in preorder.
An example follows.

28. Top-down parsing
A top-down parsing program consists of a set of procedures, one for each non-terminal.
Execution begins with the procedure for the start symbol, which halts and announces success if its procedure body scans the entire input string.

29. Creating a top-down parser (Cont.)
Given the grammar :
E → TE’
E’ → +TE’ | λ
T → FT’
T’ → *FT’ | λ
F → (E) | id
The input: id + id * id