I have written a lexer and parser to analyze linear algebra statements. Each statement consists of one or more expressions followed by one or more declarations. I am using menhir and OCaml to write the lexer and parser.
For example:
Ax = b, where A is invertible.
This should be read as A * x = b, (A, invertible)
In an expression all ids must be either an uppercase or lowercase symbol. I would like to overload the multiplication operator so that the user does not have to type in the '*' symbol.
However, since the lexer also needs to be able to read strings (such as "invertible" in this case), the "Ax" portion of the expression is sent over to the parser as a string. This causes a parser error since no strings should be encountered in the expression portion of the statement.
Here is the basic idea of the grammar
stmt :=
| expr "."
| decl "."
| expr "," decl "."
expr :=
| term
| unop expr
| expr binop expr
term :=
| <int> num
| <char> id
| "(" expr ")"
decl :=
| id "is" kinds
kinds :=
| <string> kind
| kind "and" kinds
Is there some way to separate the individual characters and tell the parser that they should be treated as multiplication? Is there a way to change the lexer so that it is smart enough to know that all character clusters before a comma are ids and all clusters after should be treated as strings?
It seems to me you have two problems:
You want your lexer to treat sequences of characters differently in different places.
You want multiplication to be indicated by adjacent expressions (no operator in between).
The first problem I would tackle in the lexer.
One question is why you say you need to use strings. This implies that there is a completely open-ended set of things you can say. It might be true, but if you can limit yourself to a smallish number, you can use keywords rather than strings. E.g., invertible would be a keyword.
If you really want to allow any string at all in such places, it's definitely still possible to hack a lexer so that it maintains a state describing what it has seen, and looks ahead to see what's coming. If you're not required to adhere to a pre-defined grammar, you could adjust your grammar to make this easier. (E.g., you could use commas for only one purpose.)
For the second problem, I'd say you need to add adjacency to your grammar. I.e., your grammar needs a rule that says something like term := term term. I suspect it's tricky to get this to work correctly, but it does work in OCaml (where adjacent expressions represent function application) and in awk (where adjacent expressions represent string concatenation).
Related
I'm making a language in ANTLR, where a sequence of digits is a number. A sequence of digits, letters and an underscore is, however, an identifier. So, for example:
These are numbers: 234, 0243, 0, 11
These are identifiers: foo, 2foo, foo2, 2y8
However, I also have operators, like multiplication, addition, division... They all work fine, except for one operator, which is the scientific operator e (or E). Unlike in most other languages, where the e for a scientific number is considered part of the number itself (like 2e3), in my language the e is considered to be an operator itself. So for example, (2+5)e4 is valid.
However, this brings an issue: since e is a letter, unless I separe my scientific number into spaces, ANTLR recognizes 2e3 as an identifier instead of the operation 2-e-3.
I'd like the language to always treat e as an operator, and not as part of an identifier, if what's on both sides of the e is something other than a letter or an underscore, or if it's alone. So, for example:
The following are treated as operations: 2e3, 2 e 3, 2E3, 2e 3, 2 e3, 23.4e78.9, 5e($my_var), .0e4, -7e3, 7e+9, 0e-4, (2)e(3), 2e(hello)
The following are treated as identifiers: hello, 2ey9, w3e6, 6ee7, 7ep, e9, e.
I have the following minimal reproduction example:
grammar ScientificExample;
file: expr* EOF;
expr
: UNARY expr
| L_PAREN expr R_PAREN
| expr SCIENTIFIC expr
| atom;
atom: NUMBER | IDENTIFIER;
WS : [ \r\t\n]+ -> skip ;
SCIENTIFIC: [eE];
NUMBER: ( [0-9]* '.' )? [0-9]+;
L_PAREN: '(';
R_PAREN: ')';
IDENTIFIER: [a-zA-Z0-9_]+;
UNARY: [+-];
As I understand it, ANTLR gives priority to whichever parsing rule produces the largest possible result, which might be why it's not giving priority to scientific being first. Any ideas how I might engineer this so priority is given to valid scientific expressions, as long as both members are an expression that isn't a simple IDENTIFIER atom, or the lack thereof?
I've thought of over-engineering the IDENTIFIER lexing rule, however since ANTLR doesn't really have lookahead and lookbehind expressions, I'm not completely sure how I'd achieve that.
I am working on a SQL grammar where pretty much anything can be an expression, even places where you might not realize it. Here are a few examples:
-- using an expression on the list indexing
SELECT ([1,2,3])[(select 1) : (select 1 union select 1 limit 1)];
Of course this is an extreme example, but my point being, many places in SQL you can use an arbitrarily nested expression (even when it would seem "Oh that is probably just going to allow a number or string constant).
Because of this, I currently have one long rule for expressions that may reference itself, the following being a pared down example:
grammar DBParser;
options { caseInsensitive=true; }
statement:select_statement EOF;
select_statement
: 'SELECT' expr
'WHERE' expr // The WHERE clause should only allow a BoolExpr
;
expr
: expr '=' expr # EqualsExpr
| expr 'OR' expr # BoolExpr
| ATOM # ConstExpr
;
ATOM: [0-9]+ | '\'' [a-z]+ '\'';
WHITESPACE: [ \t\r\n] -> skip;
With sample input SELECT 1 WHERE 'abc' OR 1=2. However, one place I do want to limit what expressions are allowed is in the WHERE (and HAVING) clause, where the expression must be a boolean expression, in other words WHERE 1=1 is valid, but WHERE 'abc' is invalid. In practical terms what this means is the top node of the expression must be a BoolExpr. Is this something that I should modify in my parser rules, or should I be doing this validation downstream, for example in the semantic phase of validation? Doing it this way would probably be quite a bit simpler (even if the lexer rules are a bit lax), as there would be so much indirection and probably indirect left-recursion involved that it would become incredibly convoluted. What would be a good approach here?
Your intuition is correct that breaking this out would probably create indirect left recursion. Also, is it possible that an IDENTIFIER could represent a boolean value?
This is the point of #user207421's comment. You can't fully capture types (i.e. whether an expression is boolean or not) in the parser.
The parser's job (in the Lexer & Parser sense), put fairly simply, is to convert your input stream of characters into the a parse tree that you can work with. As long as it gives a parse tree that is the only possible way to interest the input (whether it is semantically valid or not), it has served its purpose. Once you have a parse tree then during semantic validation, you can consider the expression passed as a parameter to your where clause and determine whether or not it has a boolean value (this may even require consulting a symbol table to determine the type of an identifier). Just like your semantic validation of an OR expression will need to determine that both the lhs and rhs are, themselves, boolean expressions.
Also consider that even if you could torture the parser into catching some of your type exceptions, the error messages you produce from semantic validation are almost guaranteed to be more useful than the generated syntax errors. The parser only catches syntax errors, and it should probably feel a bit "odd" to consider a non-boolean expression to be a "syntax error".
I have the following grammar rules defined to cover tuples of the form: (a), (a,), (a,b), (a,b,) and so on. However, antlr3 gives the warning:
"Decision can match input such as "COMMA" using multiple alternatives: 1, 2
I believe this means that my grammar is not LL(1). This caught me by surprise as, based on my extremely limited understanding of this topic, the parser would only need to look one token ahead from (COMMA)? to ')' in order to know which comma it was on.
Also based on the discussion I found here I am further confused: Amend JSON - based grammar to allow for trailing comma
And their source code here: https://github.com/doctrine/annotations/blob/1.13.x/lib/Doctrine/Common/Annotations/DocParser.php#L1307
Is this because of the kind of parser that antlr is trying to generate and not because my grammar isn't LL(1)? Any insight would be appreciated.
options {k=1; backtrack=no;}
tuple : '(' IDENT (COMMA IDENT)* (COMMA)? ')';
DIGIT : '0'..'9' ;
LOWER : 'a'..'z' ;
UPPER : 'A'..'Z' ;
IDENT : (LOWER | UPPER | '_') (LOWER | UPPER | '_' | DIGIT)* ;
edit: changed typo in tuple: ... from (IDENT)? to (COMMA)?
Note:
The question has been edited since this answer was written. In the original, the grammar had the line:
tuple : '(' IDENT (COMMA IDENT)* (IDENT)? ')';
and that's what this answer is referring to.
That grammar works without warnings, but it doesn't describe the language you intend to parse. It accepts, for example, (a, b c) but fails to accept (a, b,).
My best guess is that you actually used something like the grammars in the links you provide, in which the final optional element is a comma, not an identifier:
tuple : '(' IDENT (COMMA IDENT)* (COMMA)? ')';
That does give the warning you indicate, and it won't match (a,) (for example), because, as the warning says, the second alternative has been disabled.
LL(1) as a property of formal grammars only applies to grammars with fixed right-hand sides, as opposed to the "Extended" BNF used by many top-down parser generators, including Antlr, in which a right-hand side can be a set of possibilities. It's possible to expand EBNF using additional non-terminals for each subrule (although there is not necessarily a canonical expansion, and expansions might differ in their parsing category). But, informally, we could extend the concept of LL(k) by saying that in every EBNF right-hand side, at every point where there is more than one alternative, the parser must be able to predict the appropriate alternative looking only at the next k tokens.
You're right that the grammar you provide is LL(1) in that sense. When the parser has just seen IDENT, it has three clear alternatives, each marked by a different lookahead token:
COMMA ↠ predict another repetition of (COMMA IDENT).
IDENT ↠ predict (IDENT).
')' ↠ predict an empty (IDENT)?.
But in the correct grammar (with my modification above), IDENT is a syntax error and COMMA could be either another repetition of ( COMMA IDENT ), or it could be the COMMA in ( COMMA )?.
You could change k=1 to k=2, thereby allowing the parser to examine the next two tokens, and if you did so it would compile with no warnings. In effect, that grammar is LL(2).
You could make an LL(1) grammar by left-factoring the expansion of the EBNF, but it's not going to be as pretty (or as easy for a reader to understand). So if you have a parser generator which can cope with the grammar as written, you might as well not worry about it.
But, for what it's worth, here's a possible solution:
tuple : '(' idents ')' ;
idents : IDENT ( COMMA ( idents )? )? ;
Untested because I don't have a working Antlr3 installation, but it at least compiles the grammar without warnings. Sorry if there is a problem.
It would probably be better to use tuple : '(' (idents)? ')'; in order to allow empty tuples. Also, there's no obvious reason to insist on COMMA instead of just using ',', assuming that '(' and ')' work as expected on Antlr3.
So I have been reading a bit on lexers, parser, interpreters and even compiling.
For a language I'm trying to implement I settled on a Recrusive Descent Parser. Since the original grammar of the language had left-recursion, I had to slightly rewrite it.
Here's a simplified version of the grammar I had (note that it's not any standard format grammar, but somewhat pseudo, I guess, it's how I found it in the documentation):
expr:
-----
expr + expr
expr - expr
expr * expr
expr / expr
( expr )
integer
identifier
To get rid of the left-recursion, I turned it into this (note the addition of the NOT operator):
expr:
-----
expr_term {+ expr}
expr_term {- expr}
expr_term {* expr}
expr_term {/ expr}
expr_term:
----------
! expr_term
( expr )
integer
identifier
And then go through my tokens using the following sub-routines (simplified pseudo-code-ish):
public string Expression()
{
string term = ExpressionTerm();
if (term != null)
{
while (PeekToken() == OperatorToken)
{
term += ReadToken() + Expression();
}
}
return term;
}
public string ExpressionTerm()
{
//PeekToken and ReadToken accordingly, otherwise return null
}
This works! The result after calling Expression is always equal to the input it was given.
This makes me wonder: If I would create AST nodes rather than a string in these subroutines, and evaluate the AST using an infix evaluator (which also keeps in mind associativity and precedence of operators, etcetera), won't I get the same result?
And if I do, then why are there so many topics covering "fixing left recursion, keeping in mind associativity and what not" when it's actually "dead simple" to solve or even a non-problem as it seems? Or is it really the structure of the resulting AST people are concerned about (rather than what it evaluates to)? Could anyone shed a light, I might be getting it all wrong as well, haha!
The shape of the AST is important, since a+(b*3) is not usually the same as (a+b)*3 and one might reasonably expect the parser to indicate which of those a+b*3 means.
Normally, the AST will actually delete parentheses. (A parse tree wouldn't, but an AST is expected to abstract away syntactic noise.) So the AST for a+(b*3) should look something like:
Sum
|
+---+---+
| |
Var Prod
| |
a +---+---+
| |
Var Const
| |
b 3
If you language obeys usual mathematical notation conventions, so will the AST for a+b*3.
An "infix evaluator" -- or what I imagine you're referring to -- is just another parser. So, yes, if you are happy to parse later, you don't have to parse now.
By the way, showing that you can put tokens back together in the order that you read them doesn't actually demonstrate much about the parser functioning. You could do that much more simply by just echoing the tokenizer's output.
The standard and easiest way to deal with expressions, mathematical or other, is with a rule hierarchy that reflects the intended associations and operator precedence:
expre = sum
sum = addend '+' sum | addend
addend = term '*' addend | term
term = '(' expre ')' | '-' integer | '+' integer | integer
Such grammars let the parse or abstract trees be directly evaluatable. You can expand the rule hierarchy to include power and bitwise operators, or make it part of the hierarchy for logical expressions with and or and comparisons.
I've got a simple grammar. Actually, the grammar I'm using is more complex, but this is the smallest subset that illustrates my question.
Expr ::= Value Suffix
| "(" Expr ")" Suffix
Suffix ::= "->" Expr
| "<-" Expr
| Expr
| epsilon
Value matches identifiers, strings, numbers, et cetera. The Suffix rule is there to eliminate left-recursion. This matches expressions such as:
a -> b (c -> (d) (e))
That is, a graph where a goes to both b and the result of (c -> (d) (e)), and c goes to d and e. I'm trying to produce an abstract syntax tree for these expressions, but I'm running into difficulty because all of the operators can accept any number of operands on each side. I'd rather keep the logic for producing the AST within the recursive descent parsing methods, since it avoids having to duplicate the logic of extracting an expression. My current strategy is as follows:
If a Value appears, push it to the output.
If a From or To appears:
Output a separator.
Get the next Expr.
Create a Link node.
Pop the first set of operands from output into the Link until a separator appears.
Erase the separator discovered.
Pop the second set of operands into the Link until a separator.
Push the Link to the output.
If I run this through without obeying steps 2.3–2.7, I get a list of values and separators. For the expression quoted above, a -> b (c -> (d) (e)), the output should be:
A sep_1 B sep_2 C sep_3 D E
Applying the To rule would then yield:
A sep_1 B sep_2 (link from C to {D, E})
And subsequently:
(link from A to {B, (link from C to {D, E})})
The important thing to note is that sep_2, crucial to delimit the left-hand operands of the second ->, does not appear, so the parser believes that the expression was actually written:
a -> (b c -> (d) (e))
In order to solve this with my current strategy, I would need a way to produce a separator between adjacent expressions, but only if the current expression is a From or To expression enclosed in parentheses. If that's possible, then I'm just not seeing it and the answer ought to be simple. If there's a better way to go about this, however, then please let me know!
I haven't tried to analyze it in detail, but: "From or To expression enclosed in parentheses" starts to sound a lot like "context dependent", which recursive descent can't handle directly. To avoid context dependence you'll probably need a separate production for a From or To in parentheses vs. a From or To without the parens.
Edit: Though it may be too late to do any good, if my understanding of what you want to match is correct, I think I'd write it more like this:
Graph :=
| List Sep Graph
;
Sep := "->"
| "<-"
;
List :=
| Value List
;
Value := Number
| Identifier
| String
| '(' Graph ')'
;
It's hard to be certain, but I think this should at least be close to matching (only) the inputs you want, and should make it reasonably easy to generate an AST that reflects the input correctly.