What is the correct grammar for a standard member expression?
E.g. the ast from the code:
test.test.function()
would be
MemberExpression("test", MemberExpression("test", MethodCall("function")))
And likewise for a variable:
test.test.test.variable
MemberExpression("test", MemberExpression("test", MemberExpression("test", Variable("variable"))))
Depends on the language, surely :-) But it's pretty straight-up in most grammars (see below).
One comment, though. As indicated by the grammars below, member access (like function calls and, usually, subscripting) acts like a postfix operator; the symbol after the dot (or arrow, in C-like languages) is a symbol representing a member name. It is not an expression; the only expression in the member lookup is on the left-hand side of the operator. So a.b.c should correspond to an AST node something like:
MemberLookup(MemberLookup(Variable("a"), "b"), "c")
and a.b.func(2, c) should be turned into:
MethodCall(MemberLookup(Variable("a"), "b"),
"func",
List(Number(2), Variable("c")))
or, perhaps,
Apply(MemberLookup(MemberLookup(Variable("a"), "b"), "func"),
List(Number(2), Variable("c"))
(The difference has to do with the implicit self/this argument; there are various strategies for handling this. Contrast Java, Python and Lua for three completely different strategies.)
Anyway, here's a couple of simple grammar fragments:
C
Here's an excerpt from the C grammar (as found in Appendix A of the C standard:
postfix-expression:
primary-expression
postfix-expression '[' expression ']'
postfix-expression '(' argument-expression-listopt ')'
postfix-expression '.' identifier
postfix-expression '->' identifier
postfix-expression '++'
postfix-expression '--'
I included more than just the member access functions, because it shows that .identifier and ->identifier are handled just like any other postfix operator, which is a useful insight. The same production also includes two postfix bracketed operators, subscripting ([...]) and function call ((...)), which seem relevant here. But I left out compound literals (which I would have put into primary-expression).
Python
The comparable excerpt from the Python 3.9 docs:
primary:
primary '.' NAME
primary '(' [arguments] ')'
primary '[' slices ']'
primary genexp
atom
Related
typename ::= typename DOT ID.
typename ::= ID.
lvalue ::= lvalue DOT lvalue2.
lvalue ::= lvalue2.
lvalue2 ::= ID LSQB expr RSQB. // LSQB & RSQB: left & right square bracket. i.e. [ ].
lvalue2 ::= ID.
typename is a rule for the names of types. It matches the following code:
ClassA
package_a.ClassA
while lvalue is a rule for left values. It matches the following code:
varA
varB.C
varD.E[i].F
Now the 2 rules conflicts with each other. Maybe it is because lvalue can also match package_a.ClassA?
How can I solve this?
You can't resolve this issue grammatically because your syntax is ambiguous. a.b = 3 is valid if a.b is a member of a, and invalid if a.b is a type, but the semantics of a.b cannot be determined by the syntax.
You could resolve this in a fairly messy way if you have some way to figure that out in the lexer (which will probably involve some kind of lexical feedback, since the lexer would presumably need access to the symbol table in order to provide that information). Then the lexer could use two different token types for IDs, based on whether or not they are type names.
But probably the best option is to either abandon the idea of distinguishing grammatically between lvalues and rvalues, and or to assume that all selection operations (a.b) produce lvalues, and then to validate the use of an expression as an lvalue in the semantic action or some subsequent semantic analysis.
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.
I was having some trouble with Bison creating an operator as such:
<- = identity postfix operator with a low precedence to force evaluation of what's on the left first, e.g. 1+2<-*3 (equivalent (1+2)*3) as well as -> which is a prefix operator which does the same thing but to the right.
I was not able to get the syntax to work properly and tested with Python using - not False, which resulted in a syntax error (in Python, - has a greater precedence than not). However, this is not a problem in C or C++, where - and !/not have the same precedence.
Of course, the difference in precedence has nothing to do with the relationship between the 2 operators, only a relationship with other operators that result in the relative precedences between them.
Why is chaining prefix or postfix operators with different precedences a problem when parsing and how can implement the <- and -> operators while still having higher-precedence operators like !, ++, NOT, etc.?
Obligatory Bison (this pattern is repeated for all operators, where copy has greater precedence than post_unary):
post_unary:
copy
| post_unary "++"
| post_unary "--"
| post_unary '!'
;
Chaining operators in this category, e.g. x ! -- ! works fine syntactically.
Ok, let me suggest a possible erroneous grammar based on your sketch:
low_postfix:
mid_infix
| low_postfix "<-"
mid_infix:
high_postfix
| mid_infix '+' high_postfix
high_postfix:
term
| high_postfix "++"
term:
ID
'(' expr ')'
It should be clear just looking at those productions that var <- ++ is not part of the language. The only things that can be used as an operand to ++ are terms and other applications of ++. var <- is neither of these things.
On the other hand, var ++ <- is fine, because the operand to <- can be a mid_infix which can be a high_postfix which is an application of the ++ operator.
If the intention were to allow both of those postfix sequences, then that grammar is incorrect.
A version of that cascade is present in the Python grammar (albeit using prefix operators) which is why not - False is OK, but - not False is a syntax error. I'm reluctant to call that a bug because it may have been intentional. (Really, neither of those expressions makes much sense.) We could disagree about the value of such an intention but not on SO, which prefers to avoid opinionated discussions.
Note that what we might call "strict precedence" in this grammar and the Python grammar is by no means restricted to combinations of unary operators. Here's another one which you have likely never tried:
$ python3 -c 'print(41 + not False)'
File "<string>", line 1
print(41 + not False)
^
SyntaxError: invalid syntax
So, how can we fix that?
On some level, it would be nice to be able to just write an unambiguous grammar which conveyed our intention. And it is certainly possible to write an unambiguous grammar, which would convey the intention to bison. But it's at least an open question as to whether it would convey anything to a human reader, because the massive clutter of multiple rules necessary in order to keep track of what is and is not an acceptable grouping would be pretty daunting.
On the other hand, it's dead simple to do with bison/yacc precedence declarations. We just list the operators in order, and the parser generator resolves all the ambiguities accordingly. [See Note 1 below]
Here's a similar grammar to the above, with precedence declarations. (I left the actions in place in case you want to play with it, although it's by no means a Reproducible Example; the infrastructure it relies upon is much bigger than the grammar itself, and of little use to anyone other than me. So you'll have to define the three functions and fill in some of the bison type declarations. Or just delete the AST functions and use your own.)
%left ','
%precedence "<-"
%precedence "->"
%left '+'
%left '*'
%precedence NEG
%right "++" '('
%%
expr: expr ',' expr { $$ = make_binop(OP_LIST, $1, $3); }
| "<-" expr { $$ = make_unop(OP_LARR, $2); }
| expr "->" { $$ = make_unop(OP_RARR, $1); }
| expr '+' expr { $$ = make_binop(OP_ADD, $1, $3); }
| expr '*' expr { $$ = make_binop(OP_MUL, $1, $3); }
| '-' expr %prec NEG { $$ = make_unop(OP_NEG, $2); }
| expr '(' expr ')' %prec '(' { $$ = make_binop(OP_CALL, $1, $3); }
| "++" expr { $$ = make_unop(OP_PREINC, $2); }
| expr "++" { $$ = make_unop(OP_POSTINC, $1); }
| VALUE { $$ = make_ident($1); }
| '(' expr ')' { $$ = $2; }
A couple of notes:
I used %prec NEG on the unary minus production in order to separate that production from the subtraction production. I also used a %prec declaration to modify the precedence of the call production (the default would be ')'), although in this particular case that's unnecessary. It is necessary to put '(' into the precedence list, though. ( is the lookahead symbol which is used in precedence comparisons.
For many unary operators, I used bison %precedence declaration in the precedence list, rather than %right or %left. Really, there is no such thing as associativity with unary operators, so I think that it's more self-documenting to use %precedence, which doesn't resolve conflicts involving reductions and shifts in the same precedence level. However, even though there is no such thing as associativity between unary operators, the nature of the precedence resolution algorithm is that you can put prefix operators and postfix operators in the same precedence level and choose whether the postfix or prefix operators have priority by using %right or %left, respectively. %right is almost always correct. I did that with ++, because I was a bit lazy by the time I got to that point.
This does "work" (I think). It certainly resolves all the conflicts; bison happily produces a parser without warnings. And the tests that I tried worked at least as I expected them to:
? a++->
=> [-> [++/post a]]
? a->++
=> [++/post [-> a]]
? 3*f(a)+2
=> [+ [* 3 [CALL f a]] 2]
? 3*f(a)->+2
=> [+ [-> [* 3 [CALL f a]]] 2]
? 2+<-f(a)*3
=> [+ 2 [<- [* [CALL f a] 3]]]
? 2+<-f(a)*3->
=> [+ 2 [<- [-> [* [CALL f a] 3]]]]
But there are some expressions where the operator precedence, while "correct", might not be easily explained to a novice user. For example, although the arrow operators look somewhat like parentheses, they don't group that way. Furthermore, the decision as to which of the two operators has higher precedence seems to me to be totally arbitrary (and indeed I might have done it differently from what you expected). Consider:
? <-2*f(a)->+3
=> [<- [+ [-> [* 2 [CALL f a]]] 3]]
? <-2+f(a)->*3
=> [<- [* [-> [+ 2 [CALL f a]]] 3]]
? 2+<-f(a)->*3
=> [+ 2 [<- [* [-> [CALL f a]] 3]]]
There's also something a bit odd about how the arrow operators override normal operator precedence, so that you can't just drop them into a formula without changing its meaning:
? 2+f(a)*3
=> [+ 2 [* [CALL f a] 3]]
? 2+f(a)->*3
=> [* [-> [+ 2 [CALL f a]]] 3]
If that's your intention, fine. It's your language.
Note that there are operator precedence problems which are not quite so easy to solve by just listing operators in precedence order. Sometimes it would be convenient for a binary operator to have different binding power on the left- and right-hand sides.
A classic (but perhaps controversial) case is the assignment operator, if it is an operator. Assignment must associate to the right (because parsing a = b = 0 as (a = b) = 0 would be ridiculous), and the usual expectation is that it greedily accepts as much to the right as possible. If assignment had consistent precedence, then it would also accept as much to the left as possible, which seems a bit strange, at least to me. If a = 2 + b = 7 is meaningful, my intuitions say that its meaning should be a = (2 + (b = 7)) [Note 2]. That would require differential precedence, which is a bit complicated but not unheard of. C solves this problem by restricting the left-hand side of the assignment operators to (syntactic) lvalues, which cannot be binary operator expressions. But in C++, it really does mean a = ((2 + b) = 7), which is semantically valid if 2 + b has been overloaded by a function which returns a reference.
Notes
Precedence declarations do not really add any power to the parser generator. The languages it can produce a parser for are exactly the same languages; it produces the same sort of parsing machine (a pushdown automaton); and it is at least theoretically possible to take that pushdown automaton and reverse engineer a grammar out of it. (In practice, the grammars produced by this process are usually monstrous. But they exist.)
All that the precedence declarations do is resolve parsing conflicts (typically in an ambiguous grammar) according to some user-supplied rules. So it's worth asking why it's so much simpler with precedence declarations than by writing an unambiguous grammar.
The simple hand-waving answer is that precedence rules only apply when there is a conflict. If the parser is in a state where only one action is possible, that's the action which remains, regardless of what the precedence rules might say. In a simple expression grammar, an infix operator followed by a prefix operator is not at all ambiguous: the prefix operator must be shifted, because there is no reduce action for a partial sequence ending with an infix operator.
But when we're writing a grammar, we have to specify explicitly what constructs are possible at each point in the grammar, which we usually do by defining a bunch of non-terminals, each corresponding to some parsing state. An unambiguous grammar for expressions already has split the expression non-terminal into a cascading series of non-terminals, one for each operator precedence value. But unary operators do not have the same binding power on both sides (since, as noted above, one side of the unary operator cannot take an operand). That means that a binary operator could well be able to accept a unary operator for one of its operands, and not be able to accept the same unary operator for its other operand. Which in turn means that we need to split all of our non-terminals again, corresponding to whether the non-terminal appears on the left or the right side of a binary operator.
That's a lot of work, and it's really easy to make a mistake. If you're lucky, the mistake will result in a parsing conflict; but equally it could result in the grammar not being able to recognise a particular construct which you would never think of trying, but which some irate language user feels is an absolute necessity. (Like 41 + not False)
It's possible that my intuitions have been permanently marked by learning APL at a very early age. In APL, all operators associate to the right, basically without any precedence differences.
My graduate student and I are working on a training compiler, which we will use to teach students at the subject "Compilers and Interpreters".
The input program language is a limited subset of the Java language and the compiler implementation language is Java.
The grammar of the input language syntax is LL(1), because it is easier to be understood and implemented by students. We have the following general problem in the parser implementation. How to differentiate identifier from function call during the parsing?
For example we may have:
b = sum(10,5) //sum is a function call
or
b = a //a is an identifier
In both cases after the = symbol we have an identifier.
Is it possible to differentiate what kind of construct (a function call or an identifier) we have after the equality symbol =?
May be it is not possible in LL(1) parser, as we can look only 1 symbol ahead? If this is true, how do you recommend to define the function call in the grammar? Maybe some additional symbol in front of the function call is necessary, e.g. b = #sum(10,5)?
Do You think this symbol would be confusing for students? What kind of symbol for the function call would be proper?
You indeed can't have separate rules for function calls and variables in an LL(1) grammar because that would require additional lookahead. The common solution to this is to combine them into one rule that matches an identifier, optionally followed by an argument list:
primary_expression ::= ID ( "(" expression_list ")" )?
| ...
In a language where a function can be an arbitrary expression, not just an identifier, you'll want to treat it just like any other postfix operator:
postfix_expression ::= primary_expression postfix_operator*
postfix_operator ::= "++"
| "--"
| "[" expression "]"
| "(" expression_list ")"
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).