Chapter Three(1)

COMPILER CONSTRUCTION Principles and Practice Kenneth C. Louden

3. Context-Free Grammars and Parsing PART ONE

Contents PART ONE 3.1 The Parsing Process [More] 3.2 Context-Free Grammars [More] 3.3 Parse Trees and Abstract [More] 3.4 Ambiguity [More] PART TWO 3.5 Extended Notations: EBNF and Syntax Diagrams 3.6 Formal Properties of Context-Free Languages 3.7 Syntax of the TINY Language

Introduction Parsing is the task of Syntax Analysis Determining the syntax, or structure, of a program. The syntax is defined by the grammar rules of a Context-Free Grammar The rules of a context-free grammar are recursive The basic data structure of Syntax Analysis is parse tree or syntax tree The syntactic structure of a language must also be recursive

Function of a Parser Takes the sequence of tokens produced by the scanner as its input and produces the syntax tree as its output. Parser Sequence of tokens Syntax-Tree

Issues of the Parsing The sequence of tokens is not an explicit input parameter The parser calls a scanner procedure getToken to fetch the next token from the input as it is needed during the parsing process. The parsing step of the compiler reduces to a call to the parser as follows: SyntaxTree = parse( )

Issues of the Parsing The parser incorporate all the other phases of a compiler in a single-pass compiler No explicit syntax tree needs to be constructed The parser steps themselves will represent the syntax tree implicitly by a call Parse ( )

Issues of the Parsing In Multi-Pass, the further passes will use the syntax tree as their input The structure of the syntax tree is heavily dependent on the particular syntactic structure of the language This tree is usually defined as a dynamic data structure Each node consists of a record whose fields include the attributes needed for the remainder of the compilation process (i.e., not just those computed by the parser).

Issues of the Parsing What is more difficult for the parser than the scanner is the treatment of errors . Error in the scanner Generate an error token and consume the offending character.

Issues of the Parsing Error in the parser The parser must not only report an error message but it must recover from the error and continue parsing ( to find as many errors as possible ) A parser may perform error repair Error recovery is the reporting of meaningful error messages and the resumption of parsing as close to the actual error as possible Back

Basic Concept A context-free grammar is a specification for the syntactic structure of a programming language Similar to the specification of the lexical structure of a language using regular expressions Except involving recursive rules For example: exp  exp op exp | (exp) | number op  + | – | *

3.2.1 Comparison to Regular Expression Notation

Comparing an Example The context-free grammar: exp  exp op exp | (exp) | number op  + | – | * The regular expression: number = digit digit* digit = 0|1|2|3|4|5|6|7|8|9

Basic Regular Expression Rules Three operations : Choice, concatenation, repetition Equal sign represents the definition of a name for a regular expression; Name was written in italics to distinguish it from a sequence of actual characters.

Grammars Rules Vertical bar appears as meta-symbol for choice. Concatenation is used as a standard operation. No meta-symbol for repetition (like the * of regular expressions) Use the arrow symbol  instead of equality to express the definitions of names Names are written in italic( in a different font)

Grammars Rules Grammar rules use regular expressions as components The notation was developed by John Backus and adapted by Peter Naur for the Algol60 report Grammar rules in this form are usually said to be in Backus-Naur form, or BNF

3.2.2 Specification of Context-Free Grammar Rules

Symbols of Grammar Rules Grammar rules are defined over an alphabet , or set of symbols. The symbols are tokens representing strings of characters Using the regular expressions to represent the tokens A token is a fixed symbol, as in the reserved word while or the special symbols such as + or :=, write the string itself in the code font used in Chapter 2 Tokens such as identifiers and numbers, representing more than one string, use code font in italics, just as though the token is a name for a regular expression.

Symbols in TINY Represent the alphabet of tokens for the TINY language: {if. then, else, end, repeat, until, read, write, identifier, number, +, -, *, /, =, <, (, ), ; , := } Instead of the set of tokens (as defined in the TINY scanner) {IF,THEN,ELSE,END,REPEAT , UNTIL,READ,WRITE,ID,NUM, PLUS,MINUS,TIMES, OVER,EQ, LT, LPAREN,RPAREN, SEMI, ASSIGN }

Construction of a CFG rule Given an alphabet, a context-free grammar rule in BNF consists of a string of symbols. The first symbol is a name for a structure. The second symbol is the meta-symbol"  ". This symbol is followed by a string of symbols , each of which is either a symbol from the alphabet, a name for a structure, or the metasymbol "| ".

Construction of a CFG rule A grammar rule in BNF is interpreted as follows The rule defines the structure whose name is to the left of the arrow The structure is defined to consist of one of the choices on the right-hand side separated by the vertical bars The sequences of symbols and structure names within each choice defines the layout of the structure For example: exp  exp op exp | (exp) | number op  + | – | *

More about the Conventions The meta-symbols and conventions used here are in wide use but there is no universal standard for these conventions Common alternatives for the arrow metasymbol '  ' include "=" (the equal sign), ":" (the colon), and "::=" ("double-colon-equals") In normal text files, replacing the use of italics, by surrounding structure names with angle brackets <...> and by writing italicized token names in uppercase Example: <exp> ::= <exp> <op> <exp> | ( <exp> ) | NUMBER <op> ::= + | - | *

3.2.3 Derivations & Language Defined by a Grammar

How Grammar Determine a Language Context-free grammar rules determine the set of syntactically legal strings of token symbols for the structures defined by the rules. For example, the arithmetic expression (34-3)*42 Corresponds to the legal string of seven tokens (number - number ) * number While (34-3*42 is not a legal expression, There is a left parenthesis that is not matched by a right parenthesis and the second choice in the grammar rule for an exp requires that parentheses be generated in pairs

Derivations Grammar rules determine the legal strings of token symbols by means of derivations A derivation is a sequence of replacements of structure names by choices on the right-hand sides of grammar rules A derivation begins with a single structure name and ends with a string of token symbols At each step in a derivation, a single replacement is made using one choice from a grammar rule

Derivations The example exp  exp op exp | (exp) | number op  + | – | * A derivation (1) exp => exp op exp [ exp  exp op exp ] (2) => exp op number [ exp  number] (3) => exp * number [ op  * ] (4) => ( exp ) * number [ exp  ( exp ) ] (5) =>{ exp op exp ) * number [ exp  exp op exp} (6) => (exp op number) * number [ exp  number] (7) => (exp - number) * number [ op  - ] (8) => (number - number) * number [ exp  number] Derivation steps use a different arrow from the arrow meta-symbol in the grammar rules.

Language Defined by a Grammar The set of all strings of token symbols obtained by derivations from the exp symbol is the language defined by the grammar of expressions L(G) = { s | exp =>* s } G represents the expression grammar s represents an arbitrary string of token symbols (sometimes called a sentence) The symbols =>* stand for a derivation consisting of a sequence of replacements as described earlier. (The asterisk is used to indicate a sequence of steps, much as it indicates repetition in regular expressions.) Grammar rules are sometimes called productions Because they "produce" the strings in L(G) via derivations

Grammar for a Programming Language The grammar for a programming language often defines a structure called program The language of this structure is the set of all syntactically legal programs of the programming language. For example: a BNF for Pascal program  program-heading; program-block. program-heading  …. program-block  ….. The first rule says that a program consists of a program heading, followed by a semicolon, followed by a program block, followed by a period.

Symbols in rules Start symbol The most general structure is listed first in the grammar rules. Non-terminals Structure names are also called non-terminals, since they always must be replaced further on in a derivation. Terminals Symbols in the alphabet are called terminals, since they terminate a derivation. Terminals are usually tokens in compiler applications.

Examples Example 3.1: The grammar G with the single grammar rule E  (E) | a This grammar generates the language L(G) = { a,(a),((a)),(((a))),…} = { ( n a) n | n an integer >= 0 } Derivation for ((a)) E => (E) => ((E)) => ((a))

Examples Example 3.2: The grammar G with the single grammar rule E  (E) This grammar generates no strings at all , there is no way we can derive a string consisting only of terminals.

Examples Example 3.3: Consider the grammar G with the single grammar rule E  E + a | a This grammar generates all strings consisting of a's separated by +'s: L(G) ={a, a + a ,a + a + a, a + a + a + a,...} Derivation: E => E+a => E+a+a => E+a+a+a => … … finally replace the E on the left using the base E  a

Prove the Example 3.3 (1) Every string a + a + … + a is in L(G) by induction on the number of a's. The derivation E => a shows that a is in L(G); Assume now that s = a + a + … + a, with n–1 a's, is in L(G). Thus, there is a derivation E => * s: Now the derivation E => E + a => * s + a shows that the string s + a, with n + a's, is in L(G).

Prove the example 3.3 (2) Any strings from L(G) must be of the form a + a + …. + a. If the derivation has length 1, then it is of the form E => a, and so s is of the correct form. Now, assume the truth of the hypothesis for all strings with derivations of length n – 1; And let E =>* s be a derivation of length n > 1. This derivation must begin with the replacement of E by E + a, and so is of the form E => E + a =>* s' + a = s. Then, s' has a derivation of length n - 1, and so is of the form a + a + … + a. Hence, s itself must have this same form.

Example Example 3.4 Consider the following extremely simplified grammar of statements: Statement  if-stmt | other if-stmt  if ( exp ) statement | if ( exp ) statement else statement exp  0 | 1 Examples of strings in this language are other if (0) other if (1) other if (0) other else other if (1) other else other if (0) if (0) other if (0) if (1) other else other if (1) other else if (0) other else other

Recursion The grammar rule: A  A a | a or A  a A | a Generates the language {an | n an integer >=1 } (the set of all strings of one or more a's) The same language as that generated by the regular expression a+ The string aaaa can be generated by the first grammar rule with the derivation A => Aa => Aaa => Aaaa => aaaa

Recursion left recursive: The non-terminal A appears as the first symbol on the right-hand side of the rule defining A A  A a | a right recursive: The non-terminal A appears as the last symbol on the right-hand side of the rule defining A A  a A | a

Examples of Recursion Consider a rule of the form: A  A  |  where  and  represent arbitrary strings and  does not begin with A. This rule generates all strings of the form  ,  ,  ,  , ... (all strings beginning with a  , followed by 0 or more  '’s). This grammar rule is equivalent in its effect to the regular expression  *. Similarly, the right recursive grammar rule A   A |  ( where  does not end in A) generates all strings  ,  ,  ,  , ....

Examples of Recursion To generate the same language as the regular expression a* we must have a notation for a grammar rule that generates the empty string use the epsilon meta-symbol for the empty string empty   , called an  -production (an "epsilon production"). A grammar that generates a language containing the empty string must have at least one  -production. A grammar equivalent to the regular expression a* A  A a |  or A  a A |  Both grammars generate the language { an | n an integer >= 0} = L (a*).

Examples Example 3.5: A  (A) A |  generates the strings of all "balanced parentheses." For example, the string (( ) (( ))) ( ) generated by the following derivation (the  -production is used to make A disappear as needed): A => (A) A => (A)(A)A => (A)(A) =>(A)( ) => ((A)A)( ) =>( ( )A)() => (( ) (A)A ) () => (( )( A ))( ) => (( )((A)A))( ) => (( )(( )A))( ) => (( )(( )))( )

Examples Example 3.6: The statement grammar of Example 3.4 can be written in the following alternative way using an  -production: statement  if-stmt | other if-stmt  if ( exp ) statement else-part else-part  else statement |  exp  0 | 1 The  -production indicates that the structure else-part is optional.

Examples Example 3.7: Consider the following grammar G for a sequence of statements: stmt-sequence  stmt ; stmt-sequence | stmt stmt  s This grammar generates sequences of one or more statements separated by semicolons (statements have been abstracted into the single terminal s ): L ( G )= { s , s; s, s; s; s,... )

Examples If allow statement sequences to be empty, write the following grammar G' : stmt-sequence  stmt ; stmt-sequence |  stmt  s semicolon is a terminator rather than a separator : L ( G' )= {  , s; , s;s;, s;s;s;,... } If allow statement sequences to be empty, but retain the semicolon as a separator , write the grammar as follows: stmt-sequence  nonempty-stmt-sequence |  nonempty-stmt-sequence  stmt ； nonempty-stmt-sequence | stmt stmt  s L ( G )= {  , s , s; s, s; s; s,... ) Back

3.3 Parse trees and abstract syntax trees

Derivation V.S. Structure Derivations do not uniquely represent the structure of the strings There are many derivations for the same string. The string of tokens: ( number - number ) * number There exist two different derivations for above string

Derivation V.S. Structure (1) exp => exp op exp [ exp  exp op exp ] (2) => exp op number [ exp  number ] (3) => exp * number [ op  * ] (4) => ( exp ) * number [ exp  ( exp ) ] (5) =>( exp op exp ) * number [ exp  exp op exp} (6) => (exp op number) * number [ exp  number ] (7) => (exp - number) * number [ op  - ] (8) => (number - number) * number [ exp  number ]

Derivation V.S. Structure (1) exp => exp op exp [ exp  exp op exp ] (2) => ( exp) op exp [ exp  ( exp )] (3) => ( exp op exp) op exp [ exp  exp op exp ] (4) => ( number op exp) op exp [ exp  number ] (5) =>( number - exp) op exp [ op  - ] (6) => ( number - number) op exp [ exp  number ] (7) => ( number - number) * exp [ op  *] (8) =>( number - number) * number [ exp  number]

Parsing Tree A parse tree corresponding to a derivation is a labeled tree . The interior nodes are labeled by non-terminals, the leaf nodes are labeled by terminals; And the children of each internal node represent the replacement of the associated non-terminal in one step of the derivation. The example: exp => exp op exp => number op exp => number + exp => number + number

Parsing Tree The example: exp => exp op exp => number op exp => number + exp => number + number Corresponding to the parse tree: exp exp op exp number + number The above parse tree is corresponds to the three derivations:

Parsing Tree Left most derivation （ 1 ） exp => exp op exp （ 2 ） => number op exp （ 3 ） => number + exp （ 4 ） => number + number Right most derivation (1) exp => exp op exp (2) => exp op number (3) => exp + number (4) => number + number

Parsing Tree Neither leftmost nor rightmost derivation （ 1 ） exp => exp op exp （ 2 ） => exp + exp （ 3 ） => number + exp （ 4 ） => number + number Generally, a parse tree corresponds to many derivations represent the same basic structure for the parsed string of terminals. It is possible to distinguish particular derivations that are uniquely associated with the parse tree.

Parsing Tree A leftmost derivation: A derivation in which the leftmost non-terminal is replaced at each step in the derivation. Corresponds to the preorder numbering of the internal nodes of its associated parse tree. A rightmost derivation: A derivation in which the rightmost non-terminal is replaced at each step in the derivation. Corresponds to the postorder numbering of the internal nodes of its associated parse tree.

Parsing Tree The parse tree corresponds to the first derivation. 1 exp 2 exp 3 op 4 exp number + number

Example: The expression (34-3)*42 The parse tree for the above arithmetic expression 1 exp 4 exp 3 op 2 exp ( 5 exp ) * number 8 exp 7 op 6 exp number – number

Way Abstract Syntax-Tree The parse tree contains more information than is absolutely necessary for a compiler For the example: 3*4 exp exp op exp number * number (3) (4)

Why Abstract Syntax-Tree The principle of syntax-directed translation The meaning, or semantics, of the string 3+4 should be directly related to its syntactic structure as represented by the parse tree. In this case, the parse tree should imply that the value 3 and the value 4 are to be added. A much simpler way to represent this same information, namely, as the tree + 3 4

Tree for expression (34-3)*42 The expression (34-3)*42 whose parse tree can be represented more simply by the tree: * - 42 34 3 The parentheses tokens have actually disappeared still represents precisely the semantic content of subtracting 3 from 34, and then multiplying by 42.

Abstract Syntax Trees or Syntax Trees Syntax trees represent abstractions of the actual source code token sequences, The token sequences cannot be recovered from them (unlike parse trees). Nevertheless they contain all the information needed for translation , in a more efficient form than parse trees.

Abstract Syntax Trees or Syntax Trees A parse tree is a representation for the structure of ordinary called concrete syntax when comparing it to abstract syntax. Abstract syntax can be given a formal definition using a BNF-like notation , just like concrete syntax. The BNF-like rules for the abstract syntax of the simple arithmetic expression: exp  OpExp(op,exp,exp) | ConstExp(integer) op  Plus | Minus | Times

Abstract Syntax Trees or Syntax Trees Data type declaration. ： the C data type declarations. typedef enum {Plus,Minus,Times} OpKind; typedef enum {OpK.ConstK} ExpKind; typedef struct streenode { ExpKind kind; OpKind op; struct streenode *lchild,*rchild; int val; } STreeNode; typedef STreeNode *SyntaxTree;

Examples Example 3.8: The grammar for simplified if-statements statement  if-stmt | other if-stmt  if ( exp ) statement | if ( exp ) statement else s tatement exp  0 | 1

Examples The parse tree for the string: if (0) other else other statement if-stmt if ( exp ) statement else statement 0 other other

Examples Using the grammar of Example 3.6 statement  if-stmt | other if-stmt  if ( exp ) statement else-part else-part  else statement |  exp  0 | 1

Examples This same string has the following parse tree: if (0) other else other statement if-stmt if ( exp ) statement else-part 0 other else statement other

Examples A syntax tree for the previous string (using either the grammar of Example 3.4 or 3.6) would be: if (0) other else other if 0 other other

Examples A set of C declarations that would be appropriate for the structure of the statements and expressions in this example’ is as follows: typedef enum {ExpK, StmtK) NodeKind; typedef enum {Zero, One} ExpKind; typedef enum {IfK, OtherK) StmtKind; typedef struct streenode { NodeKind kind; ExpKind ekind; . StmtKind skind; struct streenode *test,*thenpart,*elsepart; } STreeNode; typedef STreeNode * SyntaxTree;

Examples Example 3.9: The grammar of a sequence of statements separated by semicolons from Example 3.7: stmt-sequence  stmt ; stmt-sequence| stmt stmt  s

Examples The string s; s; s has the following parse tree with respect to this grammar: stmt-sequence stmt ; stmt-sequence s stmt ; stmt-sequence s stmt s

Examples A possible syntax tree for this same string is: ; s ; s s Bind all the statement nodes in a sequence together with just one node, so that the previous syntax tree would become seq s s s

Problem & Solution The solution: use the standard leftmost-child right-sibling representation for a tree (presented in most data structures texts) to deal with arbitrary number of children The only physical link from the parent to its children is to the leftmost child . The children are then linked together from left to right in a standard linked list, which are called sibling links to distinguish them from parent-child links.

Problem & Solution The previous tree now becomes, in the leftmost-child right-sibling arrangement: seq s s s With this arrangement, we can also do away with the connecting seq node, and the syntax tree then becomes simply: s s s Back

What is Ambiguity Parse trees and syntax trees uniquely express the structure of syntax But it is possible for a grammar to permit a string to have more than one parse tree For example, the simple integer arithmetic grammar: exp  exp op exp| ( exp ) | number op  + | - | * The string: 34-3*42

What is Ambiguity exp exp op exp * number exp op exp number - number exp exp op exp _ exp op exp number number * number This string has two different parse trees.

What is Ambiguity exp => exp op exp => exp op exp op exp , => number op exp op exp => number - exp op exp => number - number op exp => number - number * exp => number - number * number exp => exp op exp => number op exp => number - exp => number - exp op exp => number - number op exp => number - number * exp => number - number * number Corresponding to the two leftmost derivations

What is Ambiguity The associated syntax trees are * _ 42 34 3 AND _ 34 * 3 42

An Ambiguous Grammar A grammar that generates a string with two distinct parse trees Such a grammar represents a serious problem for a parser Not specify precisely the syntactic structure of a program In some sense, an ambiguous grammar is like a non-deterministic automaton Two separate paths can accept the same string

An Ambiguous Grammar Ambiguity in grammars cannot be removed nearly as easily as non-determinism in finite automata No algorithm for doing so , unlike the situation in the case of automata Ambiguous grammars always fail the tests that we introduce later for the standard parsing algorithms A body of standard techniques have been developed to deal with typical ambiguities that come up in programming languages.

Two Basic Methods dealing with Ambiguity One is to state a rule that specifies in each ambiguous case which of the parse trees (or syntax trees) is the correct one, called a disambiguating rule. The advantage : it corrects the ambiguity without changing (and possibly complicating) the grammar. The disadvantage : the syntactic structure of the language is no longer given by the grammar alone.

Two Basic Methods dealing with Ambiguity Change the grammar into a form that forces the construction of the correct parse tree, thus removing the ambiguity. Of course, in either method we must first decide which of the trees in an ambiguous case is the correct one.

Remove The Ambiguity in Simple Expression Grammar Simply state a disambiguating rule that establishes the relative precedence of the three operations represented. The standard solution is to give addition and subtraction the same precedence, and to give multiplication a higher precedence. A further disambiguating rule is the associativity of each of the operations of addition, subtraction, and multiplication. S pecify that all three of these operations are left associative

Remove the Ambiguity in simple Expression Grammar Specify that an operation is nonassociative A sequence of more than one operator in an expression is not allowed. For instance, writing simple expression grammar in the following form: fully parenthesized expressions exp  factor op factor | factor factor  ( exp ) | number op  + |- | *

Remove the Ambiguity in simple Expression Grammar Strings such as 34-3-42 and even 34-3*42 are now illegal, and must instead be written with parentheses such as (34-3) -42 and 34- (3*42). Not only changed the grammar, also changed the language being recognized.

3.4.2 Precedence and Associativity

Group of Equal Precedence The precedence can be added to our simple expression grammar as follows: exp  exp addop exp | term addop  + | - term  term mulop term| factor mulop  * factor  ( exp ) | number Addition and subtraction will appear "higher" (that is, closer to the root) in the parse and syntax trees Receive lower precedence.

Precedence Cascade Grouping operators into different precedence levels. Cascade is a standard method in syntactic specification using BNF. Replacing the rule exp  exp addop exp | term by exp  exp addop term |term or exp  term addop exp |term A left recursive rule makes operators associate on the left A right recursive rule makes them associate on the right

Removal of Ambiguity Removal of ambiguity in the BNF rules for simple arithmetic expressions write the rules to make all the operations left associative exp  exp addop term |term addop  + | - term  term mulop factor | factor mulop  * factor  ( exp ) | number

New Parse Tree The parse tree for the expression 34-3*42 is

New Parse Tree The parse tree for the expression 34-3-42 The precedence cascades cause the parse trees to become much more complex The syntax trees, however, are not affected

3.4.3 The dangling else problem

An Ambiguity Grammar Consider the grammar from: statement  if-stmt | other if-stmt  if ( exp ) statement | if ( exp ) statement else statement exp  0 | 1 This grammar is ambiguous as a result of the optional else. Consider the string if (0) if (1) other else other

Dangling else problem Which tree is correct depends on associating the single else-part with the first or the second if-statement. The first associates the else-part with the first if-statement; The second associates it with the second if-statement. This ambiguity called dangling else problem This disambiguating rule is the most closely nested rule implies that the second parse tree above is the correct one.

An Example For example: if (x != 0) if (y = = 1/x) ok = TRUE; else z = 1/x; Note that, if we wanted we could associate the else-part with the first if-statement by using brackets {...} in C, as in if (x != 0) { if (y = = 1/x) ok = TRUE; } else z = 1/x;

A Solution to the dangling else ambiguity in the BNF statement  matched-stmt | unmatched-stmt matched-stmt  if ( exp ) matched-stmt else matched-stmt | other unmatched-stmt  if ( exp ) statement | if ( exp ) matched-stmt else unmatched-stmt exp  0 | 1 Permitting only a matched-stmt to come before an else in an if-statement forcing all else-parts to be matched as soon as possible.

More about dangling else The dangling else problem has its origins in the syntax of Algol60. It is possible to design the syntax in such a way that the dangling else problem does not appear. Require the presence of the else-part , and this method has been used in LISP and other functional languages (where a value must also be returned). Use a bracketing keyword for the if-statement languages that use this solution include Algol68 and Ada.

More About Dangling else For example, in Ada, the programmer writes if x /= 0 then if y = 1/x then ok := true; else z := 1/x; end if; end if; Associate the else-part with the second if-statement, the programmer writes if x /= 0 then if y = 1/x then ok := true; end if else z := 1/x; end if;

More about dangling else BNF in Ada (somewhat simplified) is if-stmt  if condition then statement-sequence end if | if condition then statement-sequence else statement-sequence end if

Why Inessential A grammar may be ambiguous and yet always produce unique abstract syntax trees. The grammar ambiguously as stmt-sequence  stmt-sequence ; stmt-sequence | stmt stmt  s Either a right recursive or left recursive grammar rule would still result in the same syntax tree structure, Such an ambiguity could be called an inessential ambiguity

Why Inessential Inessential ambiguity: the associated semantics do not depend on what disambiguating rule is used. Arithmetic addition or string concatenation, that represent associative operations (a binary operator • is associative if (a • b) • c = a • (b • c) for all values a, b, and c). In this case the syntax trees are still distinct, but represent the same semantic value, and we may not care which one we use. Nevertheless, a parsing algorithm will need to apply some disambiguating rule that the compiler writer may need to supply

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Chapter Three(1)