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1.5.1 Using GLR on Unambiguous Grammars

In the simplest cases, you can use the GLR algorithm to parse grammars that are unambiguous but fail to be LR(1). Such grammars typically require more than one symbol of lookahead.

Consider a problem that arises in the declaration of enumerated and subrange types in the programming language Pascal. Here are some examples:

     type subrange = lo .. hi;
     type enum = (a, b, c);

The original language standard allows only numeric literals and constant identifiers for the subrange bounds (‘lo’ and ‘hi’), but Extended Pascal (ISO/IEC 10206) and many other Pascal implementations allow arbitrary expressions there. This gives rise to the following situation, containing a superfluous pair of parentheses:

     type subrange = (a) .. b;

Compare this to the following declaration of an enumerated type with only one value:

     type enum = (a);

(These declarations are contrived, but they are syntactically valid, and more-complicated cases can come up in practical programs.)

These two declarations look identical until the ‘..’ token. With normal LR(1) one-token lookahead it is not possible to decide between the two forms when the identifier ‘a’ is parsed. It is, however, desirable for a parser to decide this, since in the latter case ‘a’ must become a new identifier to represent the enumeration value, while in the former case ‘a’ must be evaluated with its current meaning, which may be a constant or even a function call.

You could parse ‘(a)’ as an “unspecified identifier in parentheses”, to be resolved later, but this typically requires substantial contortions in both semantic actions and large parts of the grammar, where the parentheses are nested in the recursive rules for expressions.

You might think of using the lexer to distinguish between the two forms by returning different tokens for currently defined and undefined identifiers. But if these declarations occur in a local scope, and ‘a’ is defined in an outer scope, then both forms are possible—either locally redefining ‘a’, or using the value of ‘a’ from the outer scope. So this approach cannot work.

A simple solution to this problem is to declare the parser to use the GLR algorithm. When the GLR parser reaches the critical state, it merely splits into two branches and pursues both syntax rules simultaneously. Sooner or later, one of them runs into a parsing error. If there is a ‘..’ token before the next ‘;’, the rule for enumerated types fails since it cannot accept ‘..’ anywhere; otherwise, the subrange type rule fails since it requires a ‘..’ token. So one of the branches fails silently, and the other one continues normally, performing all the intermediate actions that were postponed during the split.

If the input is syntactically incorrect, both branches fail and the parser reports a syntax error as usual.

The effect of all this is that the parser seems to “guess” the correct branch to take, or in other words, it seems to use more lookahead than the underlying LR(1) algorithm actually allows for. In this example, LR(2) would suffice, but also some cases that are not LR(k) for any k can be handled this way.

In general, a GLR parser can take quadratic or cubic worst-case time, and the current Bison parser even takes exponential time and space for some grammars. In practice, this rarely happens, and for many grammars it is possible to prove that it cannot happen. The present example contains only one conflict between two rules, and the type-declaration context containing the conflict cannot be nested. So the number of branches that can exist at any time is limited by the constant 2, and the parsing time is still linear.

Here is a Bison grammar corresponding to the example above. It parses a vastly simplified form of Pascal type declarations.

     %token TYPE DOTDOT ID
     
     %left '+' '-'
     %left '*' '/'
     
     %%
     type_decl: TYPE ID '=' type ';' ;
     
     type:
       '(' id_list ')'
     | expr DOTDOT expr
     ;
     
     id_list:
       ID
     | id_list ',' ID
     ;
     
     expr:
       '(' expr ')'
     | expr '+' expr
     | expr '-' expr
     | expr '*' expr
     | expr '/' expr
     | ID
     ;

When used as a normal LR(1) grammar, Bison correctly complains about one reduce/reduce conflict. In the conflicting situation the parser chooses one of the alternatives, arbitrarily the one declared first. Therefore the following correct input is not recognized:

     type t = (a) .. b;

The parser can be turned into a GLR parser, while also telling Bison to be silent about the one known reduce/reduce conflict, by adding these two declarations to the Bison grammar file (before the first ‘%%’):

     %glr-parser
     %expect-rr 1

No change in the grammar itself is required. Now the parser recognizes all valid declarations, according to the limited syntax above, transparently. In fact, the user does not even notice when the parser splits.

So here we have a case where we can use the benefits of GLR, almost without disadvantages. Even in simple cases like this, however, there are at least two potential problems to beware. First, always analyze the conflicts reported by Bison to make sure that GLR splitting is only done where it is intended. A GLR parser splitting inadvertently may cause problems less obvious than an LR parser statically choosing the wrong alternative in a conflict. Second, consider interactions with the lexer (see Semantic Tokens) with great care. Since a split parser consumes tokens without performing any actions during the split, the lexer cannot obtain information via parser actions. Some cases of lexer interactions can be eliminated by using GLR to shift the complications from the lexer to the parser. You must check the remaining cases for correctness.

In our example, it would be safe for the lexer to return tokens based on their current meanings in some symbol table, because no new symbols are defined in the middle of a type declaration. Though it is possible for a parser to define the enumeration constants as they are parsed, before the type declaration is completed, it actually makes no difference since they cannot be used within the same enumerated type declaration.