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COP3402-Assignment 2 Lexical Analyzer Solved

You have been selected to write a compiler for the PL/0 language. In this assignment you have to implement a lexical analyzer for the programming language PL/0. Your program must be capable to read in a source program written in PL/0, identify some errors, and produce, as output, the source program, the source program lexeme table, and a list of lexemes. For an example of input and output refer to Appendix A.  As follows we show you the grammar for the programming language PL/0 using the Extended Backus-Naur Form (EBNF).

 

Based on Wirth’s definition for EBNF we have the following rule:

[ ] means an optional item,

{ } means repeat 0 or more times.

Terminal symbols are enclosed in quote marks.

A period is used to indicate the end of the definition of a syntactic class.

 

EBNF of  PL/0:

 

program ::= block "." .

block ::= const-declaration  var-declaration proc-declaration statement.     

constdeclaration ::= [ “const” ident "=" number {"," ident "=" number} “;"].        

var-declaration  ::= [ "var" ident {"," ident} “;"].

proc-declaration::= {"procedure" ident ";" block ";" } statement .

statement   ::= [ ident ":=" expression

| "call" ident

                        | "{" statement { ";" statement } "}"

                        | "if" condition "then" statement [“else" statement]

                        | "while" condition "do" statement

                        | “read” ident

                        | “write” ident

                        | e ] .

 

condition ::= "odd" expression

                        | expression  rel-op  expression.

 

rel-op ::= "="|“<"|"<"|"<="|""|"=“.

expression ::= [ "+"|"-"] term { ("+"|"-") term}.

term ::= factor {("*"|"/") factor}.

factor ::= ident | number | "(" expression ")“.

number ::= digit {digit}.

ident ::= letter {letter | digit}.

digit ;;= "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9“.

letter ::= "a" | "b" | … | "y" | "z" | "A" | "B" | ... | "Y" | "Z".

 

Example of a program written in PL/0:

var x, w;

{

   read w;

   x:= 4;

   if w x then

            w:= w + 1

   else

            w:= x;

   write w;

}.

 

Lexical Conventions for PL/0:

A numerical value is assigned to each token (internal representation) as follows:

nulsym = 1, identsym = 2, numbersym = 3, plussym = 4, minussym = 5, multsym = 6,  slashsym = 7, oddsym = 8,  eqlsym = 9, neqsym = 10, lessym = 11, leqsym = 12,

gtrsym = 13, geqsym = 14, lparentsym = 15, rparentsym = 16, commasym = 17, semicolonsym = 18, periodsym = 19, becomessym = 20, lbracesym = 21, rbracesym = 22, ifsym = 23, thensym = 24, whilesym = 25, dosym = 26, callsym = 27, constsym = 28, varsym = 29, procsym = 30, writesym = 31, readsym = 32, elsesym = 33.

 

Reserved Words: const, var, procedure, call, if, then, else, while, do, read, write, odd.

Special Symbols: ‘+’, ‘-‘, ‘*’, ‘/’, ‘(‘, ‘)’, ‘=’, ’,’ , ‘.’, ‘ <’, ‘’,  ‘;’ , ’:’ .

Identifiers: identsym = letter (letter | digit)*

Numbers: numbersym = (digit)+

Invisible Characters: tab, white spaces, newline

Comments denoted by: /* . . .   */

Refer to Appendix B for a declaration of the token symbols that may be useful.

Constraints:

Input:

Identifiers can be a maximum of 11 characters in length.
Numbers can be a maximum of 5 digits in length.
Comments should be ignored and not tokenized.
Invisible Characters should be ignored and not tokenized.
Important Note: Input files may NOT be grammatically valid PL/0 code.

 

Output:

The token separator in the output's Lexeme List (Refer to Appendix A) can be either a space or a bar ('|').
In your output's Lexeme List, identifiers must show the token and the variable name separated by a space or bar.
In your output's Lexeme List, numbers must show the token and the value separated by a space or bar. The value must be transformed into ASCII Representation (as discussed in class)
Be consistent in output. Choose either bars or spaces and stick with them.
The token representation of the Lexeme List will be used in the Parser (Project 3). So, PLAN FOR IT!
 

Detect the Following Lexical Errors:

 

Variable does not start with letter.
Number too long.
Name too long.
Invalid symbols.
 

Hint: You could create a transition diagram (DFS) to recognize each lexeme on the source program and once accepted, generate the token otherwise emit an error message.

 


Appendix A:

 

If the input is:

var x, y;

{

            y := 3;

            x := y + 56;

}.

 

The output will be:

Source Program:

var x, y;

{

            y := 3;

            x := y + 56;

}.

 

Lexeme Table:

lexeme            token type      

var                   29                   

x                      2

,                       17                   

y                      2

;                       18

{                      21

y                      2

:=                     20

3                      3

;                       18

x                      2

:=                     20                   

y                      2

+                      4

56                    3

;                       18

}                      22

.                       19

 

Lexeme List:

29  2 x  17  2 y  18  21 2 y 20 3 3 18  2 x  20  2 y  4  3  56  18  22  19

 

Appendix B:

 

Declaration of Token Types:

typedef enum {

nulsym = 1, identsym, numbersym, plussym, minussym,

multsym,  slashsym, oddsym, eqsym, neqsym, lessym, leqsym,

gtrsym, geqsym, lparentsym, rparentsym, commasym, semicolonsym,

periodsym, becomessym, lbracesym, rbracesym, ifsym, thensym,

whilesym, dosym, callsym, constsym, varsym, procsym, writesym,

readsym , elsesym } token_type;

 

 

Example of Token Representation:

“29  2  x  17  2  y  18  21  2  x  20  2  y  4  3  56  18  22  19”

 

Is Equivalent:

varsym identsym  x  commasym  identsym  y  semicolonsym  lbracesym  identsym  x

becomessym  identsym  y  plussym  numbersym  56  semicolonsym rbracesym  periodsym

 

 

 

Appendix C:

 

Example of a PL/0 program:

const m = 7, n = 85; 

var  i,x,y,z,q,r; 

procedure mult;

  var a, b; 

  {

     a := x;  b := y; z := 0;  

     while b 0 do   

     {

        if odd x then z := z+a;      

           a := 2*a;

           b := b/2;    

     }  

  };

 

{

  x := m;

  y := n;

  call mult;

}.

 

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