ITEC 380 2022fall ibarland

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Expr Trees
eval/parse B basics; adding identifiers

Over the course of several homeworks, we'll implement a “tower” of languages (B0, B1, …, B6) each language incorporating more features than the previous. B0 is provided for you.

1. Download & get the provided B0 running (both .rkt and .java.
2. Each time we add new syntax to our language, we must update three methods:
   • parse!, which turns source-code into an internal representation,
   • expr->string which turns internal representation back into source-code,
   and
   • eval, which actually interprets the program, returning a value (a number for B0-B2, and a number-or-function in B3).
   I recommend implementing and testing the methods in that order. See the provided test-cases, for examples. There might be other helper functions to implement as well.
3. Your solution may be in Java or in Racket (or see me, if you want to use a different language). (The exception is B1, which you must implement in both Racket and Java.)
4. You do not need to worry about error-checking the programs we process — we will only concern ourselves with syntactically correct Bi programs.
   (As a matter of defensive programming, you might still want to add some assertions/sanity-checks in your code.)
5. Use the D2L discussion board for group questions and answers.
6. This project is based, in part, on the first chapters of Programming Languages and Interpretation, by Shriram Krishnamurthi. As a result, this homework assignment is covered by the Creative Commons Attribution-NonCommercial-ShareAlike License. Although we're using a different dialect of scheme/racket than that book, you might find it helpful to skim it.

1. Implement B1 in racket (25pts) and in Java (25pts), both.
   B1 is just like B0, but with two additions:

   Op ::= … | # Interpretation: “modulo” (see A.iii for details) Expr ::= … | IfGT IfGT ::= obi Expr Expr wan Expr Expr Interpretation: if third Expr is > than fourth, result is the first Expr else the second.

   Be sure to write test cases first; To ensure everybody makes test cases to cover the basics, I've spelled out these strongly-suggested B2-initial-tests.html.
1. Add the # operator.
   1. update parse! (Java: Expr.parse) to handle this new operator (after writing test cases).
      (Or, if you don't need to update this function, understand why.)
   2. update expr->string (Java: Expr.toString) to handle this new operator (after writing test cases).
      (Or, if you don't need to update this function, understand why.)

   3. update eval to handle these new operators.

      • The semantics of ^x y# is:

        the remainder after dividing x by y, where the result is always between 0 (inclusive) and y (exclusive)¹

        In particular, the result should never be positive if y<0. Notice that this is slightly different behavior than either Java's built-in % (which behaves differently on negative numbers), and from Racket's built-in modulo (which only accepts integers). In both racket and Java, you can calculate this as y * (x/y - ⌊x/y⌋), where ⌊r⌋ means the the floor of r.

      Note that you are already provided sufficient test cases for #s, in the comments of the B0 test-case files.

2. Add the ifGT expression.
   1. update parse! (Java: Expr.parse) to handle this new type of expression (after writing test cases).
   2. update expr->string (Java: Expr.toString) to handle this new type of expression (after writing test cases).

   3. update eval to handle this new type of expression. The semantics of obi Expr₃ Expr₄ wan Expr₁ Expr₂ is:

      first evaluate just Expr₃ and Expr₄ to get values v₃; and v₄; if v₃ is greater v₄ then evaluate Expr₁ and return its value; otherwise evaluate Expr₂ and return that value.

      For example, obi 7.5 44 wan 3.4 2.1 evaluates to 2.1.
      (Note how the non-returned Expr never gets evaluated: you are implementing short-circuit semantics for ifGTs!)

      You must make your own test cases for IfGTs; include at least two as-simple-as-possible tests, and two tests with more deeply nested Exprs. I suggest including one where the IfGT is not the top-level expression (e.g., a / expression which contains a IfGT as one of its operands).

2. (50pts) Implement B2 in either racket or Java (your choice).
   B2 adds identifiers to B1:

   Expr ::= … | Id | LetExpr LetExpr ::= hrrm Expr in Expr, Id is2

   where Id can be pretty much any string: it can be any token returned by our scanner. This simplistic³ definition lets us keep our code easy: if we get a token (via peek/UtilIan#hasNextSplittingBy), and it's not the start of any other expression, then it's an identifier! We will assume that (a) an Id is never one of our “reserved” words like obi so you do not need to bother checking for that, and (b) for now any nested LetExpr expressions will use different Ids. We'll handle shadowing in B3, later.)

   Update your three methods parse, toString (a.k.a. expr->string), eval.

   1. add Ids to your data-definition (after deciding what data-type will represent them); then:
      1. update expr->string (or, Expr.toString) to handle this new type of expression (after test cases)
      2. update parse! (or, Expr.parse) to handle this new type of expression (after test cases)
      3. update eval (or, Expr.eval) to handle this new type of expression:

         eval'ing an identifier simply throws an error.

         You don't need test cases for evaling Ids. Though if you want to be spiffy, you can use check-error in racket, or assertThrows in JUnit5. In JUnit4, the hack-ish approach is to put “ExpectedException.none().expect(RunTimeException.class)” on the line before the one that should trigger an error.

   2. Next, add LetExprs to your data-definition (after deciding what data-type will represent them); then:
      1. update expr->string (or, Expr.toString) to handle this new type of expression (after test cases)
      2. update parse! (or, Expr.parse) to handle this new type of expression (after test cases). Since our parser splits on punctuation, < will be read as one token, and - as second (both racket and Java).
      3. Think about how to update eval to handle this new type of expression. Now we get to the heart of the issue! Write test cases, after reading the rest of this bullet.

         In order to write eval, we need to define the semantics of hrrm Expr₀ in Expr₁, Id is:

         • Evaluate Expr₀; let's call the result v₀.
         • Then, substitute v₀ for all occurrences of Id inside the tree Expr₁; name the result of the substitution E′.
         • Evaluate E′, and return that result.

         For example: hrrm 3 in ^x 2/, x is ⇒ ^3 2/ ⇒ 5⁴. Be sure to write test cases for your substitution function before you write its code; include several trivial and easy tests, along with a couple of more complicated nestings and one deeply nested expression.

         Observe that when evaluating a (legal) B2 program, eval will never actually encounter an Id — that Id will have been substituted out before we ever recur down to it.

      4. Now that we realize that eval will need to do substitution in a tree, and that's a smaller, simpler, self-contained task — perfect for its own helper-function substitute. This function only does substition in a tree, and does not attempt to do any evaluating.
         Go and write substitute (after test cases for it), before implementing eval for LetExprs. (And when starting substitute, start from the template for Exprs.) For the test cases, think about exactly types you'll be wanting to sent to substitute. Your simplest test-cases won't even contain a LetExpr — substitute is a function whose purpose-statement stands on its own, entirely independent of LetExprs and eval!

         Hint: Substituting a variable with a value in an syntax-tree is essentially the same as replacing every occurrence of one name with another in an anc-tree. (The only difference is that an anc-tree had only two cond-branches, while Expr has around five, though the code for most of those are very similar.)

         (Note: you must do substitution in the parse tree; no credit given for string-substitution ⁶.)

      5. Finally, with the substitute helper written, we're ready: write eval for LetExprs.

         Hint: Your code will correspond almost word-for-word to the semantics given above.

  eval(parse!("hrrm 3 in ^x 2/, x is"))
= eval(parse!("^3 2/"))
= eval(parse!("5"))

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