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I am trying to write a simple parser for a small calculator project, that should be able to parse e.g. the following inputs:

5 + 3
5 + f(4)
5 + f(x)
x = 5
f(x) = 3*x

so basically, I want to be able to parse expressions (that may contain variables and function calls), variable assignments, and function definitions using the = operator.

The problem is that in function definitions, only identifiers must be allowed in the definition, e.g.

f(5+1) = 3*x

should not be legal in the grammar. Thus, I need to define two distinct cases for functions, where the first (bFunc, containing only ids) could occur on both sides of =, and the second (func, can also contain expressions) must not occur on the left side.

This is the grammar so far:

stmt -> expr.
stmt -> fdef.
stmt -> vdef.

fdef -> bFunc = expr.
vdef -> id = expr.

bFunc -> id ( idList ).

func -> id ( list ).
func -> id ( ).

list -> expr.
list -> list , expr.

idList -> id.
idList -> idList , id.

expr -> term.
expr -> expr + term.
expr -> expr - term.

term -> atom.
term -> term / atom.
term -> term * atom.

atom -> id.
atom -> num.
atom -> func.
atom -> ( expr ).

However, I was not able to figure out how to specify a valid LALR(1) grammar because of this problem. The tool http://mdaines.github.io/grammophone/#/ reports reduce-reduce conflicts. The problem is that the parser would not know whether id should be reduced to idList or atom.

Surely it would be possible to simply use func in both cases and catch invalid left sides later in the program. But my question is now, is it even possible to write a LALR(1) grammar for this problem? How does one decide whether it is possible or not? And may my problem be the reason why programming languages use keywords like def or function for function definitions?

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  • $\begingroup$ @rici actually not, thank you $\endgroup$
    – David
    Feb 15 at 0:07
  • $\begingroup$ Does this mean you now know how to solve your own problem? If so, perhaps I can encourage you to write an answer to your explanation, to provide the solution and explain why it works? $\endgroup$
    – D.W.
    Feb 15 at 6:27
  • $\begingroup$ unfortunately not, the problem still remains. I think it should be possible, because once the = appears one can decide whether the f(..) on the left is valid or not. So my guess is that the grammar is lalr(k), and apparently any lalr(k) can be transformed into lalr(1). However I have not yet figured out how to do it. I'm new to this topic and I'm grateful for any hint. $\endgroup$
    – David
    Feb 15 at 8:17
  • $\begingroup$ Sorry, I missed that. (Although $atom \to bFunc$ is still wrong IMHO.) I'll add an answer in a bit. $\endgroup$
    – rici
    Feb 15 at 16:40

2 Answers 2

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As modified by your edit, your grammar is unambiguous. Unfortunately, it is not deterministic; no limited lookahead is sufficient to decide whether when the parser reaches a comma: $$\bf{ID}\;\bf{(}\;\bf{ID}\;\bullet\;\bf{,}\;\cdots$$ it should predict $\it{bFunc}$ by reducing $\bf{ID}$ to $\it{idList}$ or predict $\it{func}$ by reducing it to $\it{atom}$ (and thus eventually to $\it{expr}$ and $\it{list}$).

One reasonably straightforward solution is to differentiate between $\it{expr}$ and $\bf{ID}$ by creating the non-terminal $\it{expr_{!ID}}$ which matches $\tt{expr}\setminus\{\bf{ID}\}$ and using it in $\it{list}$. (Note that $\it{list}$ is not a simple list of $\it{expr_{!ID}}$; rather, it is a list in which at least one element is a $\it{expr_{!ID}}$. See below for sample grammar).

That makes $\it{list}$ and $\it{idList}$ disjoint and the grammar is deterministic. However, the distinction is not strictly related to the semantics of the language. It creates a curious hybrid parse tree node in which some of the nodes in the argument list's parse subtree are $\it{idList}$ and the rest are $\it{list}$. (Look at the parse tree for f(a,b,c,d+3), for example). Indeed, a complete function call might be parsed as a $\it{bFunc}$, which will need to be converted to $\it{bFunc}$ when it turns out that it is not followed by an $\bf{=}$.

A practical parser will need to repair the parse tree for $\it{func}$ by converting the $\it{idList}$ nodes to $\it{list}$. That could be done in a reduction action, or it could be done in a post-parse tree walk, but it will almost certainly need to be done, since the semantic distinction is real: the use of an $\bf{ID}$ in a parameter list is a binding, while the use of the same $\bf{ID}$ in a function call is a use.

So at the high level, what we end up with is: $$\begin{align} \tt{bFunc}&\to\;ID\;\tt{(}\;\tt{idList}\;\tt{)}\\ \tt{func}&\to\;ID\;\tt{(}\;\tt{list}\;\tt{)}\\ \tt{func}&\to\;ID\;\tt{(}\;\tt{)}\\ \tt{func}&\to\;ID\;\tt{(}\;\tt{idList}\;\tt{)}\\ \\ \tt{list}&\to\;\tt{expr_{!ID}}\\ \tt{list}&\to\;\tt{list}\;,\;\tt{expr}\\ \tt{list}&\to\;\tt{idList}\;,\;\tt{expr_{!ID}}\\ \\ \tt{idList}&\to\;\tt{ID}\\ \tt{idList}&\to\;\tt{idList}\;,\;\tt{ID}\\ \end{align}$$

We also need to define the non-terminal $\it{expr_{!ID}}$ which doesn't match $\bf{ID}$ (and its pars at the other chained precedence levels). The straightforward solution is to remove the production $\text{atom}\;\to\;ID$, thus removing the token $\bf{ID}$ from the direct unit-production chain. Then we add it back in the non-terminals in which it is permitted (i.e. the ones which are not restricted to $\it{!ID}$). So the expression grammar now looks like this: $$\begin{align} \tt{expr}&\to\;\tt{expr_{!ID}}\\ \tt{expr}&\to\;\tt{ID}\\ \tt{term}&\to\;\tt{term_{!ID}}\\ \tt{term}&\to\;\tt{ID}\\ \tt{atom}&\to\;\tt{atom_{!ID}}\\ \tt{atom}&\to\;\tt{ID}\\ \\ \tt{expr_{!ID}}&\to\;\tt{expr}\;\bf{+}\;\tt{term}\\ \tt{expr_{!ID}}&\to\;\tt{expr}\;\bf{-}\;\tt{term}\\ \tt{expr_{!ID}}&\to\;\tt{term_{!ID}}\\ \\ \tt{term_{!ID}}&\to\;\tt{term}\;\bf{/}\;\tt{atom}\\ \tt{term_{!ID}}&\to\;\tt{term}\;\bf{*}\;\tt{atom}\\ \tt{term_{!ID}}&\to\;\tt{atom_{!ID}}\\ \\ \tt{atom_{!ID}}&\to\;\tt{NUM}\\ \tt{atom_{!ID}}&\to\;\tt{func}\\ \tt{atom_{!ID}}&\to\;\bf{(}\;\tt{expr}\;\bf{)}\\ \end{align}$$

which is only a little longer than the equivalent lines in your original.

It's interesting to note that the syntax formalism used by ECMAScript (and some other semi-related technologies) uses a macro-enhanced form of BNF in which non-terminals can be given boolean parameters, like the $\tt{!ID}$ subscript I used above. With that feature, the grammar could be written even more compactly (although the macro expansion would be a bit bigger).

As another interesting note, your grammar will work perfectly without the definition of $\it{expr_{!ID}}$ if processed with Bison or almost any other Yacc-derivative parser generator. (You do need to add the extra productions in the list definitions to convert $\it{idList}$ to $\it{list}$, as indicated above.) The parser generator will signal four reduce-reduce conflicts, since there is an ambiguity. But as long as the list productions come before the expression productions, Yacc/Bison's automatic conflict resolution mechanism will produce the correct resolution. That's probably not the optimal solution, but it is certainly the shortest.

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  • $\begingroup$ wow, thank you for this solution and detailed explanation. There is much more to writing grammars than I would have guessed. I ended up with a version that has a shift/reduce conflict, but I'll try to integrate your solution. As a sidenote, there are still reduce/reduce conflicts if the ',' is not used in the list definition (according to grammophone). $\endgroup$
    – David
    Feb 17 at 21:02
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    $\begingroup$ @David: Indeed there are conflicts without the comma in the list; that was a typo and I'll fix it. The commas are needed, obviously; otherwise, you cannot distinguish the single expression a(1) from the list of two expressions a and (1). $\endgroup$
    – rici
    Feb 17 at 21:43
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The problem occurs with the parsing prefixes having the form f(id, id, ..., so that your grammar cannot determine by any means what do they denote: either the function definition or an expression starting with the function call.

Thinking in terms of PDAs, it is natural to parse such prefixes uniformly until the parse path is revealed. The parse path is always revealed deterministically: if the prefix ends with ) =, then you parse a definition, otherwise you parse an expression. I have used some simplified form of your grammar to check out the idea, and it seems that there are no SLR(1)-conflicts in the resulting grammar. But yeah, the complete grammar can be quite longer, and maybe some good grammar transforming techs (such as introducing new nonterminals for the suffixes of the expressions) would be on hand.

The core idea is introducing fstrange nonterminal with the meaning "for now, we do not know what is it, but keep the both possibilities in mind".

stmt -> num + expr.
stmt -> num.
stmt -> id.
stmt -> id + expr.
stmt -> id ( fstrange.
stmt -> id = expr.

fstrange -> id , fstrange.
fstrange -> id ) = expr.
fstrange -> id ).
fstrange -> num ).
fstrange -> func ).
fstrange -> num ) + expr.
fstrange -> func ) + expr.
fstrange -> + expr ) + expr.
fstrange -> + expr ).
fstrange -> id ) + expr.
fstrange -> + expr , list ).
fstrange -> num , list ).
fstrange -> func , list ).
fstrange -> + expr , list ) + expr.
fstrange -> num , list ) + expr.
fstrange -> func , list ) + expr.


func -> id ( list ).
func -> id ( ).

list -> expr.
list -> list , expr.


expr -> atom.
expr -> expr + atom.

atom -> id.
atom -> num.
atom -> func.

Regarding your second question: your language seems to be deterministic, so it is always possible to construct an SLR(1)-grammar for it (since SLR(1)=LR(k) in terms of the languages, not grammars!). The matter is the grammar size. I do not claim that the example for the simplified grammar given above is perfect, but still you can notice it is much less natural and less intuitive than the one given by you.

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  • $\begingroup$ Thank you @Tonita for your reply. I understand your approach of keeping a common prefix, however I think it will be difficult to retrieve the actual syntax tree from this grammar. I do hope that this is not how grammars are written in practice :-) $\endgroup$
    – David
    Feb 17 at 21:10

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