Principles of Programming Languages Topic: Formal Languages I
CS 314, LS,LTM: L2, Formal Languages I
1
Review • This class will teach you – some common principles underlying most languages – some new ways of thinking about programs (new paradigms)
• A programming language must be – as easy as possible for people to • learn • read • write
– as easy as possible for a compiler to translate into efficient machine code or an interpreter to execute CS 314, LS,LTM: L2, Formal Languages I
2
Defining a Language • To define a computer language we need to say – How do we tell if a file of characters is a legal (grammatical) program in this language? ( syntax ) – How do we define what a program in this language means? (semantics )
• Semantics: several approaches but in practice we just use English to explain the meaning • Syntax: defined by a formal grammar
CS 314, LS,LTM: L2, Formal Languages I
3
Grammars S R NP VP NP R Name | Det Noun VP R Verb | Verb NP Name R john | mary Det R a | the Det R some | every Noun R boy | girl Verb R runs | likes
CS 314, LS,LTM: L2, Formal Languages I
S NP Det
VP
Noun
Verb
NP Name
the
boy
likes
Mary
4
Grammars A grammar, G, is a quadruple , where • T is a set of terminal symbols (e.g., john, mary, a, the, some, every, boy, girl, runs, likes); • N is a set of nonterminal symbols (e.g., S, NP, VP, Name, Det, Noun, Verb); • P is a set of productions, or rewrite rules (e.g., Det R a the); • S is a special start symbol. The language of G, L(G), is the set of all terminal sequences that can be produced by applying the rewrite rules, repeatedly, starting with S .
CS 314, LS,LTM: L2, Formal Languages I
5
Grammars • For Programming Languages: Stmt R Identifier := Digit Identifier R Letter | Identifier Letter | Identifier Digit Letter R a | b | c | ... | x | y | z Digit R 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
• Backus-Naur Form (BNF): ::= := ::= | | ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 CS 314, LS,LTM: L2, Formal Languages I
6
Derivation in a Grammar Q: Generate x2:=0 in this grammar (call it G)? R R R R R R
:= := := x := x 2 := x 2 := 0
Yes! This is a leftmost or canonical derivation in G.
CS 314, LS,LTM: L2, Formal Languages I
7
Parsing in a Grammar Q: Recognize x2:=0 as a terminal sequence in L(G)? x 2 := 0 R R R R R R
2 := 0 2 := 0 := 0 := 0 :=
Yes! This is a parse of the sentence x2:=0 in G.
CS 314, LS,LTM: L2, Formal Languages I
8
Parse Trees
:=
2
0
x Each internal node is a nonterminal; its children are drawn from the right-hand side of one of the productions for that nonterminal. CS 314, LS,LTM: L2, Formal Languages I
9
Grammars are not Unique • Consider a grammar G’: ::= := ::= | ::= | ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
• The grammar G’ generates the same language as G, but it has different parse trees.
CS 314, LS,LTM: L2, Formal Languages I
10
Grammars are not Unique
:=
2
x Parse Tree for G
CS 314, LS,LTM: L2, Formal Languages I
0
:=
0
x
2
Parse Tree for G’
11
Types of Grammars • Context Free Grammars: – Every production has a single nonterminal on the lefthand side: A R … – Disallowed: X A R X a
• Regular Grammars: – Productions take the form: A R c , or are all either left-linear: A R B a , or right-linear: A R a B – Disallowed: S R a S b – Cannot generate the language { an bn | n = 1,2,3, ... }
CS 314, LS,LTM: L2, Formal Languages I
12
Types of Grammars • Context Free Grammars (CFGs) are used to specify the overall structure of a programming language: – if/then/else, ... – brackets: ( ), { }, begin/end, ...
• Regular Grammars (RGs) are used to specify the structure of tokens: – identifiers, numbers, keywords, ...
• Note: The recognition problem for CFGs and RGs requires a different computational model (more on this later). CS 314, LS,LTM: L2, Formal Languages I
13
Ambiguity S R NP VP NP R Name | Det Noun | NP PP PP R Prep NP VP R Verb | Verb NP NP Name R john | mary Name Det R a | the | some | every Prep R on | with | under | ... john Noun R man | hill | telescope | ... Verb R saw | runs | likes | ... CS 314, LS,LTM: L2, Formal Languages I
S VP Verb
NP
saw
....
14
Ambiguity NP NP NP Det
...
a
Noun
man
CS 314, LS,LTM: L2, Formal Languages I
PP PP
Prep
on
Prep NP
NP Det
Det
Noun
a
hill
with
Noun
a telescope 15
Ambiguity NP NP Det
Noun
PP Prep
NP NP Det
... a
man
CS 314, LS,LTM: L2, Formal Languages I
on
a
PP Noun
hill
Prep
with
NP Det
Noun
a
telescope 16
Dangling Else Here is a simplified grammar for Pascal: ::= | | ... ::= if then | if then else ::= := ::= = 0 ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
How are compound “if” statements parsed using this grammar? CS 314, LS,LTM: L2, Formal Languages I
17
if x = 0 then if y = 0 then z := 1 else w := 2 Parse Tree 1
if then
= 0 x
if then else
= 0 y
:= z
CS 314, LS,LTM: L2, Formal Languages I
1
:= w
2
18
if x = 0 then if y = 0 then z := 1 else w := 2 Parse Tree 2
if then
= 0 x
else
if then
= 0 y
:= z
CS 314, LS,LTM: L2, Formal Languages I
:=
1
w
2
Q: which tree is correct?
19
How to Fix the Dangling Else? • Algol60: use block structure if x = 0 then begin if y = 0 then z := 1 end else w := 2
• Algol68: use statement begin/end markers if x = 0 then if y = 0 then z := 1 fi else w := 2 fi
• Pascal: change the grammar of “if” statement to disallow the second parse tree, i.e., always associate an “else” with the closest “if”.
CS 314, LS,LTM: L2, Formal Languages I
20
How to Fix the Dangling Else? Here is a revised grammar for Pascal: ::= | ::= if then else | | ... ::= if then | if then else ::= := ::= = 0 ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 CS 314, LS,LTM: L2, Formal Languages I
21
if x = 0 then if y = 0 then z := 1 else w := 2 In the new grammar there is only one parse tree!
if then
= 0 x
if then else
= 0 y
:= z
CS 314, LS,LTM: L2, Formal Languages I
1
:= w
2
22
Arithmetic Expressions Here is a grammar for arithmetic expressions: ::= + | - | * | / | | ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Using this grammar, how would we parse: x + 3 * y ?
CS 314, LS,LTM: L2, Formal Languages I
23
Two Parse Trees
+
*
x
y *
+
3
y
x
3
CS 314, LS,LTM: L2, Formal Languages I
24
Precedence Modify the grammar to add precedence: ::= + | - | ::= * | / | ::= | | ( ) ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Using this grammar, how would we parse: x + 3 * y ? Using this grammar, how would we parse: 7 - 4 - 2 ? CS 314, LS,LTM: L2, Formal Languages I
25
Only One Parse Tree + x * 3
CS 314, LS,LTM: L2, Formal Languages I
y
But there are two parse trees for the second example: - 2 7 - 4 7 - 4 - 2
26
Associativity Modify the grammar to add associativity: ::= + | - | ::= * | / | ::= | | ( ) ::= a | b | c | ... | x | y | z ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Using this grammar, how would we parse: 7 - 4 - 2 ? CS 314, LS,LTM: L2, Formal Languages I
27
Only One Parse Tree - 2 -
4
7 CS 314, LS,LTM: L2, Formal Languages I
28
Concrete vs. Abstract Syntax
+
+ x * 3
CS 314, LS,LTM: L2, Formal Languages I
x
*
3
y
Abstract Syntax
y
29
Extended BNF (EBNF) Write nonterminals as in BNF. (Variant: Write them with initial capital letters, or using a different font.) Use additional metasymbols, as shortcuts: – {…} means repeat the enclosed text zero or more times – […] means the enclosed text is optional – (…) is used for grouping, usually with the alternation symbol, e.g., (… | ...).
If { }, [ ], or ( ) are used as terminal symbols in the language being defined, then they must be quoted. (Variant: They must be underlined.) CS 314, LS,LTM: L2, Formal Languages I
30
Extended BNF (EBNF) Examples: ::= { ( + | - ) } ::= { ( * | / ) } ::= | | ‘(’ ‘)’ ::= if then [ else ] ::= { ( | ) }
CS 314, LS,LTM: L2, Formal Languages I
31
