Gold Parsing System - What can it be used for in programming? - parsing

Question

I have read the GOLD Homepage ( http://www.devincook.com/goldparser/ ) docs, FAQ and Wikipedia to find out what practical application there could possibly be for GOLD. I was thinking along the lines of having a programming language (easily) available to my systems such as ABAP on SAP or X++ on Axapta - but it doesn't look feasible to me, at least not easily - even if you use GOLD.
The final use of the parsed result produced by GOLD escapes me - what do you do with the result of the parse?
EDIT: A practical example (description) would be great.

Answer 1

Parsing really consists of two phases. The first is "lexing", which convert the raw strings of character in to something that the program can more readily understand (commonly called tokens).
Simple example, lex would convert:
if (a + b > 2) then
In to:
IF_TOKEN LEFT_PAREN IDENTIFIER(a) PLUS_SIGN IDENTIFIER(b) GREATER_THAN NUMBER(2) RIGHT_PAREN THEN_TOKEN
The parse takes that stream of tokens, and attempts to make yet more sense out of them. In this case, it would try and match up those tokens to an IF_STATEMENT. To the parse, the IF _STATEMENT may well look like this:
IF ( BOOLEAN_EXPRESSION ) THEN
Where the result of the lexing phase is a token stream, the result of the parsing phase is a Parse Tree.
So, a parser could convert the above in to:
if_statement
|
v
boolean_expression.operator = GREATER_THAN
| |
| v
V numeric_constant.string="2"
expression.operator = PLUS_SIGN
| |
| v
v identifier.string = "b"
identifier.string = "a"
Here you see we have an IF_STATEMENT. An IF_STATEMENT has a single argument, which is a BOOLEAN_EXPRESSION. This was explained in some manner to the parser. When the parser is converting the token stream, it "knows" what a IF looks like, and know what a BOOLEAN_EXPRESSION looks like, so it can make the proper assignments when it sees the code.
For example, if you have just:
if (a + b) then
The parser could know that it's not a boolean expression (because the + is arithmetic, not a boolean operator) and the parse could throw an error at this point.
Next, we see that a BOOLEAN_EXPRESSION has 3 components, the operator (GREATER_THAN), and two sides, the left side and the right side.
On the left side, it points to yet another expression, the "a + b", while on the right is points to a NUMERIC_CONSTANT, in this case the string "2". Again, the parser "knows" this is a NUMERIC constant because we told it about strings of numbers. If it wasn't numbers, it would be an IDENTIFIER (like "a" and "b" are).
Note, that if we had something like:
if (a + b > "XYZ") then
That "parses" just fine (expression on the left, string constant on the right). We don't know from looking at this whether this is a valid expression or not. We don't know if "a" or "b" reference Strings or Numbers at this point. So, this is something the parser can't decided for us, can't flag as an error, as it simply doesn't know. That will happen when we evaluate (either execute or try to compile in to code) the IF statement.
If we did:
if [a > b ) then
The parser can readily see that syntax error as a problem, and will throw an error. That string of tokens doesn't look like anything it knows about.
So, the point being that when you get a complete parse tree, you have some assurance that at first cut the "code looks good". Now during execution, other errors may well come up.
To evaluate the parse tree, you just walk the tree. You'll have some code associated with the major nodes of the parse tree during the compile or evaluation part. Let's assuming that we have an interpreter.
public void execute_if_statment(ParseTreeNode node) {
// We already know we have a IF_STATEMENT node
Value value = evaluate_expression(node.getBooleanExpression());
if (value.getBooleanResult() == true) {
// we do the "then" part of the code
}
}
public Value evaluate_expression(ParseTreeNode node) {
Value result = null;
if (node.isConstant()) {
result = evaluate_constant(node);
return result;
}
if (node.isIdentifier()) {
result = lookupIdentifier(node);
return result;
}
Value leftSide = evaluate_expression(node.getLeftSide());
Value rightSide = evaluate_expression(node.getRightSide());
if (node.getOperator() == '+') {
if (!leftSide.isNumber() || !rightSide.isNumber()) {
throw new RuntimeError("Must have numbers for adding");
}
int l = leftSide.getIntValue();
int r = rightSide.getIntValue();
int sum = l + r;
return new Value(sum);
}
if (node.getOperator() == '>') {
if (leftSide.getType() != rightSide.getType()) {
throw new RuntimeError("You can only compare values of the same type");
}
if (leftSide.isNumber()) {
int l = leftSide.getIntValue();
int r = rightSide.getIntValue();
boolean greater = l > r;
return new Value(greater);
} else {
// do string compare instead
}
}
}
So, you can see that we have a recursive evaluator here. You see how we're checking the run time types, and performing the basic evaluations.
What will happen is the execute_if_statement will evaluate it's main expression. Even tho we wanted only BOOLEAN_EXPRESION in the parse, all expressions are mostly the same for our purposes. So, execute_if_statement calls evaluate_expression.
In our system, all expressions have an operator and a left and right side. Each side of an expression is ALSO an expression, so you can see how we immediately try and evaluate those as well to get their real value. The one note is that if the expression consists of a CONSTANT, then we simply return the constants value, if it's an identifier, we look it up as a variable (and that would be a good place to throw a "I can't find the variable 'a'" message), otherwise we're back to the left side/right side thing.
I hope you can see how a simple evaluator can work once you have a token stream from a parser. Note how during evaluation, the major elements of the language are in place, otherwise we'd have got a syntax error and never got to this phase. We can simply expect to "know" that when we have a, for example, PLUS operator, we're going to have 2 expressions, the left and right side. Or when we execute an IF statement, that we already have a boolean expression to evaluate. The parse is what does that heavy lifting for us.
Getting started with a new language can be a challenge, but you'll find once you get rolling, the rest become pretty straightforward and it's almost "magic" that it all works in the end.
Note, pardon the formatting, but underscores are messing things up -- I hope it's still clear.

Answer 2

I would recommend antlr.org for information and the 'free' tool I would use for any parser use.

Answer 3

GOLD can be used for any kind of application where you have to apply context-free grammars to input.
elaboration:
Essentially, CFGs apply to all programming languages. So if you wanted to develop a scripting language for your company, you'd need to write a parser- or get a parsing program. Alternatively, if you wanted to have a semi-natural language for input for non-programmers in the company, you could use a parser to read that input and spit out more "machine-readable" data. Essentially, a context-free grammar allows you to describe far more inputs than a regular expression. The GOLD system apparently makes the parsing problem somewhat easier than lex/yacc(the UNIX standard programs for parsing).