CSC 4181 Compiler Construction Context-Free Grammars Using grammars - PDF document

CSC 4181 Compiler Construction Context-Free Grammars Using grammars in parsers CFG 1 1 Parsing Process  Call the scanner to get tokens  Build a parse tree from the stream of tokens  A parse tree shows the syntactic structure of the source program.  Add information about identifiers in the symbol table  Report error, when found, and recover from thee error CFG 2 2 1

Context-Free Grammar  a tuple ( V, T, P, S ) where  V is a finite set of nonterminals, containing S �  T is a finite set of terminals,  P is a set of production rules in the form of  β where  is in V and β is in ( V U T )* , and  S is the start symbol.  Any string in (V U T)* is called a sentential form � CFG 3 3 Examples E  E O E S  SS E  � S  (S)S (E) E  id S  () O  � S  �  + O  � - O  � * � O  � / CFG 4 4 2

Backus-Naur Form (BNF)  Nonterminals are in < > .  Terminals are any other symbols.  ::= means  �  | means or.  Examples: < E> ::= < E> < O> < E> | (< E> ) | ID < O> ::= + | - | * | / < S> ::= < S> < S> | (< S> )< S> | () |  CFG 5 5 Derivation  A sequence of replacement of a substring in a sentential form. Definition ( V , T , P , S ) be a CFG,  ,  ,  be  Let G � � � strings in ( V U T ) * and A is in V �  A  �  G �  if A �  is in P �   � G denotes a derivation in zero step or more. CFG 6 6 3

Examples S  SS | (S)S | () |  E  E O E | (E) | id O  + | - | * | / S  SS E  (S)SS  E O E  (S)S(S)S  (E) O E  (S)S(())S  (E O E) O E  ((S)S)S(())S  ((E O E) O E) O E  ((S)())S(())S  ((id O E)) O E) O E  ((())())S(())S  ((id + E)) O E) O E  ((())()) (())S  ((id + id)) O E) O E  ((())())(())  ((id + id)) * id) + id CFG 7 7 Leftmost Derivation Rightmost Derivation  Each step of the derivation  Each step of the derivation is a replacement of the is a replacement of the leftmost nonterminals in a rightmost nonterminals in sentential form. a sentential form. E E  E O E  E O E  ( E ) O E  E O id  ( E O E) O E  E * id  (id O E) O E  ( E ) * id  (id + E ) O E  (E O E ) * id  (id + id) O E  (E O id) * id  (id + id) * E  ( E + id) * id  (id + id) * id  (id + id) * id CFG 8 8 4

Right/Left Recursive  A grammar is a left  A grammar is a right recursive if its recursive if its production rules can production rules can generate a derivation generate a derivation of the form A  * A X. of the form A  * X A.  Examples:  Examples:  E  E O id | (E) | id  E  id O E | (E) | id  E  F + id | (E) | id  E  id + F | (E) | id F  E * id | id F  id * E | id  E  F + id  E  id + F  E * id + id  id + id * E CFG 9 9 Parse Tree  A labeled tree in which  the interior nodes are labeled by nonterminals  leaf nodes are labeled by terminals  the children of an interior node represent a replacement of the associated nonterminal in a derivation  corresponding to a derivation E � F id id E � id CFG 10 10 5

Parse Trees and Derivations E � E  E + E (1) E �  id + E E � (2) �  id + E * E E � E � (3) id �  id + id * E (4) id id  id + id * id Preorder numbering (5) E  E + E E � (1)  E + E * E E � (2) E � �  E + E * id (3) E � E � id �  E + id * id (4) id id  id + id * id (5) Reverse or postorder numbering CFG 11 11 Grammar: Example <St> ::=  | s; | s; <St> | { <St> } <St> List of statements: <St>  No statement  { <St> } <St>  One statement:  { { <St> } <St> } <St>  { { s; <St> } <St> } <St>  s;  { { s; { <St> } <St> } <St> } <St>  More than one  { { s; { s; <St> } <St> } <St> } <St> statement:  { { s; { s; s; } <St> } <St> } <St>  s; s; s;  { { s; { s; s; } s; <St> } <St> } <St>  A statement can be a  { { s; { s; s; } s; { <St> } <St> } <St> } <St> block of statements.  { { s; { s; s; } s; { } <St> } <St> } <St>  { { s; { s; s; } s; { } } <St>} <St>  { s; s; s;}  { { s; { s; s;} s; { } } s; } <St> Is the following correct?  { { s; { s; s;} s; { } } s; } { { s; { s; s;} s; { } } s; } CFG 12 12 6

Abstract Syntax Tree  Representation of actual source tokens  Interior nodes represent operators.  Leaf nodes represent operands. CFG 13 13 Abstract Syntax Tree for Expression E � � E E � id1 E E � id1 id2 id3 id2 id3 CFG 14 14 7

Abstract Syntax Tree for If Statement st ifStatement if elsePart st � � exp if true st return else st true return CFG 15 15 Ambiguous Grammar  A grammar is ambiguous if it can generate two different parse trees for one string.  Ambiguous grammars can cause inconsistency in parsing. CFG 16 16 8

Example: Ambiguous Grammar E  � E + E E  � E - E E  � E * E E  E / E E  id E E + E * E E E E E E E id3 * id1 + id2 id3 id1 id2 CFG 17 17 Ambiguity in Expressions  Which operation is to be done first?  solved by precedence  An operator with higher precedence is done before one with lower precedence.  An operator with higher precedence is placed in a rule (logically) further from the start symbol.  solved by associativity  If an operator is right-associative (or left-associative), an operand in between 2 operators is associated to the operator to the right (left).  Right-associated : W + (X + (Y + Z))  Left-associated : ((W + X) + Y) + Z CFG 18 18 9

Precedence E  � E E + E E E  E - E + E * E E E E  E * E E E E E id3 E  E / E * + id1 E  id id1 id2 id2 id3 E  � E + E E E  E - E E + E E  F F F F  � F * F  F / F F * F id1 F �  id F � id2 id3 CFG 19 19 Precedence (cont’d) E  � E + E | E - E | F E F  F * F | F / F | X F X  � ( E ) | id F * F X F * F X X ( E ) (id1 + id2) * id3 * id4 id4 id3 E + E F F X X id1 id2 CFG 20 20 10

Associativity E  Left-associative operators F E  � E + F | E - F | F X F / F  � F * X | F / X | X X  � * X id4 ( E ) | id F X id3 ( E ) (id1 + id2) * id3 / id4 + E F = (((id1 + id2) * id3) / id4) F X X id2 id1 CFG 21 21 Ambiguity in Dangling Else St  � IfSt | ... IfSt  � if ( E ) St | if ( E ) St else St E  � { if (0) 0 | 1 | … { if (1) St } else St } { if (0) { if (1) St else St } } IfSt IfSt � � if E St else St if � E � St 0 IfSt IfSt 0 � � if E St if � E � St else St 1 1 CFG 22 22 11

Disambiguating Rules for Dangling Else St  MatchedSt | UnmatchedSt UnmatchedSt  if (E) St | St if (E) MatchedSt else UnmatchedSt MatchedSt  � UnmatchedSt if (E) MatchedSt else MatchedSt| ... E  if � E � St 0 | 1 MatchedSt  if (0) if (1) St else St = if (0) if � E � MatchedSt else MatchedSt if (1) St else St CFG 23 23 Extended Backus-Naur Form (EBNF)  Kleene’s Star/ Kleene’s Closure  Seq ::= St { ; St}  Seq ::= { St ;} St  Optional Part  IfSt ::= if ( E ) St [else St]  E ::= F [+ E ] | F [- E] CFG 24 24 12

Syntax Diagrams  Graphical representation of EBNF rules IfSt  nonterminals: id  terminals:  sequences and choices:  Examples � � E  X ::= (E) | id id St  Seq ::= { St ;} St � St F �  E ::= F [+ E ] E CFG 25 25 13