Assignment 2:  Syntax Error Recovery

Here is an LL(1) grammar for the extended calculator language: 

P	→	SL  $$
SL	→	S ; SL  |  ε
S	→	int id := E  |  real id := E  |  id := E  |  read TP id  |  write E  |  if C then SL end  |  while C do SL end
TP	→	int  |  real  |  ε
C	→	E  RO  E
E	→	T  TT
TT	→	AO  T  TT  |  ε
T	→	F  FT
FT	→	MO  F  FT  |  ε
F	→	( E )  |  id  |  i_num  |  r_num  |  trunc ( E )  |  float ( E )
RO	→	==  |  <>  |  <  |  >  |  <=  |  >=
AO	→	+  |  -
MO	→	*  |  /

Integer and real numbers are differentiated by the presence or absence of a decimal point:

i_num  =  d⁺
r_num  =  ( d⁺ . d^* | d^* . d⁺ ) ( e ( + | - | ε ) d⁺ | ε )

As explained in lecture, $$ is a special token created by the scanner when it detects the end of the input; it is not a part of the actual program text.  The new nonterminal RO is meant to suggest a “relational operator.”  Identifiers are intended to be declared before use, either with an int or real statement that provides an initial value or with a read statement that indicates a type.  The scope of each declaration extends from the declaration itself through the end of the current statement list.  As an additional extension, integers and real numbers are not intended to be mixed in expressions unless explicitly converted with trunc and float.  Note, however, that we will not be checking these semantic requirements in the current assignment (unless you choose to pursue extra credit). 

As it turns out, if we assume that integers are unbounded, our extensions make the calculator language Turing complete (if still quite impractical).  As an illustration, here is a program that calculates the first n primes:

    read int n;
    int cp := 2;
    while n > 0 do
        int found := 0;
        int cf1 := 2;
        int cf1s := cf1 * cf1;
        while cf1s <= cp do
            int cf2 := 2;
            int pr := cf1 * cf2;
            while pr <= cp do
                if pr == cp then
                    found := 1;
                end;
                cf2 := cf2 + 1;
                pr := cf1 * cf2;
            end;
            cf1 := cf1 + 1;
            cf1s := cf1 * cf1;
        end;
        if found == 0 then
            write cp;
            n := n - 1;
        end;
        cp := cp + 1;
    end; 

Your AST for the primes-printing program should look like this:

    [ (int "n")
      (read "n")
      (int "cp")
      (:= "cp" "2")
      (while (> "n" "0")
        [ (int "found")
          (:= "found" "0")
          (int "cf1")
          (:= "cf1" "2")
          (int "cf1s")
          (:= "cf1s" (* "cf1" "cf1"))
          (while (<= "cf1s" "cp")
            [ (int "cf2")
              (:= "cf2" "2")
              (int "pr")
              (:= "pr" (* "cf1" "cf2"))
              (while (<= "pr" "cp")
                [ (if (== "pr" "cp")
                    [ (:= "found" "1")
                    ]
                  )
                  (:= "cf2" (+ "cf2" "1"))
                  (:= "pr" (* "cf1" "cf2"))
                ]
              )
              (:= "cf1" (+ "cf1" "1"))
              (:= "cf1s" (* "cf1" "cf1"))
            ]
          )
          (if (== "found" "0")
            [ (write "cp")
              (:= "n" (- "n" "1"))
            ]
          )
          (:= "cp" (+ "cp" "1"))
        ]
      )
    ] 

The executable /u/cs254/bin/ast_gen on the csug machines contains an AST generator for the extended calculator grammar, which you can use to check your code.  It reads a program from standard input and prints the corresponding AST on standard output.  It was used to generate the tree above.  (For those new to the Linux command line, if you paste characters into the terminal window as standard input, you have to hit control-D to indicate end-of-file before the generator will do anything.)  Note that the generator does not perform syntax error recovery, and is therefore not a complete solution for this project.  For a syntactically correct program, however, it will display (a pretty printed version of) the output we expect you to produce.  Please ensure that the non-whitespace characters in your output match what the generator gives you; your “correctness” score will depend on this. 

Suggestions

You do not have to build the syntax tree as an explicit data structure in your program in order to generate the right output.  You are welcome to build it if you want to, though, and extra credit options 2, 3, and 5 (realized as separate, post-parsing traversals of the tree) will be easier if you do. 

We’ve given you a trivial Makefile.  You should add to it a target test that causes make to pipe sample calculator programs (of your choosing) into your parser.  This will make it easier for the TAs to reproduce your tests.  Extra credit will be given to students who provide particularly well designed test mechanisms in their submission. 

Note that your code will employ both insertions and deletions: when match sees a token other than the one it expects, it will insert the expected token and continue (presumably after printing an error message).  When a recursive descent routine sees a token that is not in any of its PREDICT sets, it will delete tokens until it finds something in either its FIRST set or its FOLLOW set.

Extra Work for CSC 454

Students in 454 must implement immediate error detection:  epsilon productions should be predicted only when the upcoming token is in the context-specific FOLLOW set. 

Division of labor and writeup

As in most assignments this semester, you may work alone or in teams of two.  If you would like to work on a team but are in need of a partner, consider posting a note to the Blackboard discussion board.