CS 353 Lab #9: Parsing a Simple Expression Language

The problem description

For this lab, you will write a parser that reads in an infix expression, including variables and assignment operators, and produces an internal abstract syntax tree (AST) which you will use to re-write the expression (i.e., to "pretty-print" it). The trees you build will use the classes you developed in the last lab (the interpreter). In the next lab you will use this parser to compile the same expression language into machine code for the PC-231 architecture. In working on this lab, you should benefit from the design work you did on the last lab and should take into account the fact that you will be re-using this code in a later lab: plan ahead so that your work will require minimal changes.

The specific expression language you will work with is described below in more detail using a context-free grammar written in extended BNF syntax. The grammar given is for abstract syntax, so it only describes the different kinds of expressions and the type and order of appearance of their constituent sub-expressions. Real expressions will use parentheses to disambiguate structure: we'll see some precedence rules below.

Your major problem in this lab is to decide how to parse the input. Several natural choices come to mind:

use a hand-coded parser based on recursive descent parsing (as discussed in lecture): this may require you to "massage" the grammar a bit;
use a parser generator such as YACC or javacc: again, a certain amount of modification to the grammar might be required;
use an informal, ad-hoc technique suitable for very simple grammars, e.g., the "shunting yard algorithm" (see the home page for a Wikipedia link). I have provided a summary description for this technique in a separate section below, and we have discussed it in lecture.

The expression grammar

We will use a fairly rich set of expressions as our input language in order to better approximate the kinds of language constructs which are used in more realistic programming languages. The grammar given below does not specify the precedence or associativity of the operations involved; precedence rules are given following the grammar. The lexical structure of variables, literals and operators is also given in an extended BNF style: if you use a parser generator or other such tool, you may want to split off lexical issues into a separate part suitable as input to a scanner generator.

<expr>  ::=  <int-lit>			(integer literals)

         |   <var>			(variables)
         |   READ			(input)
         |   WRITE <expr>		(output)
         |   <var> = <expr>		(assignment to variables)

         |   <var> <p-op>		(postfix operators)
         |   <p-op> <var> 		(prefix operators)
         |   <expr> <bin-op> <expr>	(infix operators)
         |   <expr> ? <expr> : <expr>	(conditional expression)

         |   ( <expr> )			(parenthesized expressions)

<int-lit>  ::=  <digit>+		(non-empty sequences of digits)
<digit>    ::=  0 | 1 | ... | 9


<var>      ::=  <alpha>+		(lowercase only, no digits)
<alpha>    ::=  a | b | ... | z

<bin-op>   ::=  +  |  -  |  *   	(arithmetic, but no division)
             |  <  |  >  |  ==		(relational)
             |  &  |  |			(logical)

<p-op>     ::=  ++ | --

Below are a few notes on the syntax (and informal semantics).

variable names are lowercase only and may not include digits; all "keywords" are uppercase only. Variables denote the usual readable, writable, named locations holding (in this case) integer values.
integer literals (sequences of digits only) mean exactly what you would expect, i.e., they denote integers. They are written in decimal and there is no provision for negative numbers (although they can be easily simulated with expressions).
the READ expression returns a value read from the user, dynamically; the WRITE expression writes the value of its sub-expression out, but has that same value as its value. This is useful for debugging, etc.
assignment expressions are similar to those used in Java (and C). Their constituents are a variable and an expression: the expression is evaluated and the value stored in the variable.
conditional expressions are as in Java and C: the first constituent is evaluated and treated as a boolean (zero is false, non-zero values true); depending on its value, one of the two other branches is evaluated and its value is returned as the result (note: the other branch is not evaluated!).
the prefix and postfix operators are just the increment (++) and decrement (--) operators familiar from Java. Note that these operators can only be applied to variables, not to arbitrary expressions.
the binary operators are just the usual multiplication, addition and subtraction operators.

Precedence and associativity

Prefix and postfix operators only apply to variables, so they are clearly performed immediately on the variable given (i.e., there is no potential conflict since expressions such as "3*x++" could never be interpreted as "(3*x)++" anyway).

Multiplication takes precedence over addition and subtraction; the latter two are of the same precedence. Relational operators are all of the same precedence (but lower than the arithmetic ones), and logical operators are lower than relational ones. Operators of the same precedence associate to the left, so that "2+3-4" is the same as "(2+3)-4". Assignment has the next highest precedence, so that it is always performed last of the "operators" (i.e., "x=3+2" is interpreted as "x=(3+2)" rather than "(x=3)+2"). Of course "x=y++" parses as "(x=(y++))". The WRITE form has a lower precedence than any infix operator (even considering assignment one of these), and the conditional expression has the lowest precedence of all, so that "x + WRITE x = y++ * 2" is parsed as "(x+(WRITE(x=((y++)*2))))" and "x==3 ? y++ : WRITE y" is parsed as "(x==3) ? (y++) : (WRITE y)".

Some sample expressions

Here is a set of sample expressions you might want to test your parser on, along with their fully parenthesized forms.

2+3*5
produces: (2+(3*5)).
2+3*7-1
produces: ((2+(3*7))-1).
2*x*x + 3*x=7 + 2
produces: (((2*x)*x)+(3*(x=(7+2))))
READ + READ * 2
produces: (READ + (READ * 2)).
x*x+++10-x=3
produces: (((x*(x++))+10)-(x=3)).

Basic plan of action

As described above, your job is to read in an expression (as a String), then parse it into a tree form. (You should already have designed your Expression classes in the last lab.) After reading in and parsing an expression, you should print it out again to verify that the parse worked correclty (you can use your fully-parenthesized toString method from the last lab for this).

Note that error-checking is an important part of the assignment: you should be able to reject any expressions which do not fit the syntax given in this lab write-up. The minimum requirement is to just reject the ill-formed expressions: of course, it's better if you can describe to the user just what has gone wrong, but (as you'll find out), giving good feedback can be quite difficult with syntax errors.

It may be convenient to convert the input string into tokens (or "chunks") before trying to apply a parsing strategy (note: this is a different use of the term "token" than Java uses in "StringTokenizer", although the latter may be helpful). You can either cull out tokens as a separate phase or just handle them dynamically as part of the parsers input routines. See the section Parsing strategy below for details on token handling.

You will probably want to work through several example expressions by hand, using the parsing strategy of your choice, in order to make sure you understand the structures and process involved.

In memoriam of hours wasted by Zac Bentley ☹


			
			original: 2*x+++10-x=3+x=READ
			
			
			paren'ed: (((2*(x++))+10)-(x=(3+(x=READ))))

Parsing strategy

As described above, there are several approaches you might take to parsing the input, including recursive descent, a parser generator or more ad-hoc methods. The following section describes one ad-hoc method which will work for the simple grammar here (it is based on operator-precedence parsing, but presented here more informally).

Even if you use one of the other strategies, you may want to read through this section in order to get a feel for some of the parsing issues this grammar presents.

The first stage in parsing, conceptually at least, is to break the input string up into tokens. Our expression language has been designed to make this phase very simple: in most cases, characters in the input can be categorized and handled in a general way (e.g., alphabetic characters and digits can be consumed one after the other and combined to produce a variable or literal). White space can be essentially ignored, since variable names and integer literals can never occur immediately adjacent to another instance of the same token class. The "+" and "-" characters should signal a one-character "look-ahead", since two adjacent occurrences should be taken as a postfix increment (++) or decrement (--) operator.

(There is some possible ambiguity here due to the possibility of binary operators such as "+" occurring next to prefix and postfix operators such as "++", as in an example like "x+++y". Let's agree that the tokenizer can "eagerly" grab duplicate occurrences of a character like "+" and turn them into a single "++" token, so that the previous example would be seen as "[x] [++] [+] [y]" in terms of tokens. Then the parser will have no choice but to accept this as "(x++)+y".)

To continue with larger syntactic structures, infix expressions can be parsed from a series of tokens using two stacks, and the basic algorithm can be modified to accommodate postfix operators and assignment as they appear in our language. One stack is used to hold parse trees under construction (we'll call this the argument stack), the other to hold operators (and left parentheses, for matching purposes; we'll call this the operator stack).

As we read in each new token (from left to right), we either push the token (or a related tree) onto one of the stacks, or we reduce the stacks by combining an operator with some arguments. Along the way, it will be helpful to maintain a search state which tells us whether we should see an argument or operator next (the search state is a refinement which helps us to reject certain ill-formed expressions: if you don't see how to use it at first, leave it out until later).

The basic rules for handling individual tokens can be summarized as follows:

upon seeing a variable or literal, push it onto the tree stack (possible after wrapping it up as a leaf node). The search state can be changed from "argument" to "operator". (If a prefix operator is on top of the operator stack, combine it immediatley; see below.)
upon seeing a prefix or postfix operator, check that there is a variable leaf node on the argument stack, then pop it off and push a postfix operation node (built from the variable and the postfix operator) back onto the argument stack. In this case the search state is still for an operator. If the top node on the stack not a variable, push the p-fix operator on the stack in the hopes that a variable will come next. In this case, the search state is for an argument.
upon seeing a binary operator, we must check the status of the operator stack: if it is empty, we simply push the new operator. If it has some operator on top, we compare the precedence of the two operators and push the new one if it has higher precedence (strictly higher precedence: this will force the left associativity we want). Otherwise we can reduce the two stacks before continuing with the same input operator. In any case, pushing an operator means we should now search for an argument, whereas if we continue with the same operator the search state should clearly stay as "operator".
upon seeing a left parenthesis, we simply push it on the operator stack; the search should have been for an argument and should remain so.
upon seeing a right parenthesis, we have two cases: if there is a left parenthesis on the operator stack, we can "cancel" the pair and continue searching for an operator. If the operator stack contains some other operator on top, we should reduce the stacks and continue with the right parenthesis as input again.
upon seeing the end of the input, we should have one of the following situations: if there is only one tree on the argument stack and the operator stack is empty, we should finish and return the single tree as the result. If there are more trees and/or operators, we should attempt to reduce the stacks until they reach the first form.

Exercise: what is the appropriate way to treat the precedence of the left parenthesis as it sits on the operator stack? Try a short example to see what it should be.

The rules for reduction are a bit simpler: if we see a binary operator on top of the operator stack, we should have two trees on top of the argument stack, both representing expressions. We pop the operator and two trees off, combining them into a single tree node, which is then pushed back on the argument stack. Note that the trees on the argument stack represent the right and left arguments, respectively: this is the opposite of what you might expect. There is a special case when the operator is an assignment: we should then check that the second tree on the argument stack is just a variable leaf node before building and pushing the combined tree.

Any other combinations (e.g., no operator on top of the operator stack, or not enough trees on the argument stack, or no variable to go with an assignment operator) should be considered an error.