Lesson 1.4: Disambiguating Parses

The purpose of this lesson is to teach how to use K's builtin features for disambiguation to transform an ambiguous grammar into an unambiguous one that expresses the intended ASTs.

Priority blocks

In practice, very few formal languages outside the domain of natural language processing are ambiguous. The main reason for this is that parsing unambiguous languages is asymptotically faster than parsing ambiguous languages. Programming language designers instead usually use the notions of operator precedence and associativity to make expression grammars unambiguous. These mechanisms work by instructing the parser to reject certain ASTs in favor of others in case of ambiguities; it is often possible to remove all ambiguities in a grammar with these techniques.

While it is sometimes possible to explicitly rewrite the grammar to remove these parses, because K's grammar specification and AST generation are inextricably linked, this is generally discouraged. Instead, we use the approach of explicitly expressing the relative precedence of different operators in different situations in order to resolve the ambiguity.

For example, in C, && binds tighter in precedence than ||, meaning that the expression true && false || false has only one valid AST: (true && false) || false.

Consider, then, the third iteration on the grammar of this definition (lesson-04-a.k):

module LESSON-04-A syntax Boolean ::= "true" | "false" | "(" Boolean ")" [bracket] > "!" Boolean [function] > Boolean "&&" Boolean [function] > Boolean "^" Boolean [function] > Boolean "||" Boolean [function] endmodule

In this example, some of the | symbols separating productions in a single block have been replaced with >. This serves to describe the priority groups associated with this block of productions. The first priority group consists of the atoms of the language: true, false, and the bracket operator. In general, a priority group starts either at the ::= or > operator and extends until either the next > operator or the end of the production block. Thus, we can see that the second, third, fourth, and fifth priority groups in this grammar all consist of a single production.

The meaning of these priority groups becomes apparent when parsing programs: A symbol with a lesser priority, (i.e., one that binds looser), cannot appear as the direct child of a symbol with a greater priority (i.e., one that binds tighter. In this case, the > operator can be seen as a greater-than operator describing a transitive partial ordering on the productions in the production block, expressing their relative priority.

To see this more concretely, let's look again at the program true && false || false. As noted before, previously this program was ambiguous because the parser could either choose that && was the child of || or vice versa. However, because a symbol with lesser priority (i.e., ||) cannot appear as the direct child of a symbol with greater priority (i.e., &&), the parser will reject the parse where || is under the && operator. As a result, we are left with the unambiguous parse (true && false) || false. Similarly, true || false && false parses unambiguously as true || (false && false). Conversely, if the user explicitly wants the other parse, they can express this using brackets by explicitly writing true && (false || false). This still parses successfully because the || operator is no longer the direct child of the && operator, but is instead the direct child of the () operator, and the && operator is an indirect parent, which is not subject to the priority restriction.

Astute readers, however, will already have noticed what seems to be a contradiction: we have defined () as also having greater priority than ||. One would think that this should mean that || cannot appear as a direct child of (). This is a problem because priority groups are applied to every possible parse separately. That is to say, even if the term is unambiguous prior to this disambiguation rule, we still reject that parse if it violates the rule of priority.

In fact, however, we do not reject this program as a parse error. Why is that? Well, the rule for priority is slightly more complex than previously described. In actual fact, it applies only conditionally. Specifically, it applies in cases where the child is either the first or last production item in the parent's production. For example, in the production Bool "&&" Bool, the first Bool non-terminal is not preceded by any terminals, and the last Bool non-terminal is not followed by any terminals. As a result of this, we apply the priority rule to both children of &&. However, in the () operator, the sole non-terminal is both preceded by and followed by terminals. As a result, the priority rule is not applied when () is the parent. Because of this, the program we mentioned above successfully parses.


Parse the program true && false || false using kast, and confirm that the AST places || as the top level symbol. Then modify the definition so that you will get the alternative parse.


Even having broken the expression grammar into priority blocks, the resulting grammar is still ambiguous. We can see this if we try to parse the following program (assoc.bool):

true && false && false

Priority blocks will not help us here: the problem comes between two parses where both possible parses have a direct parent and child which is within a single priority block (in this case, && is in the same block as itself).

This is where the notion of associativity comes into play. Associativity applies the following additional rules to parses:

  • a left-associative symbol cannot appear as a direct rightmost child of a symbol with equal priority;
  • a right-associative symbol cannot appear as a direct leftmost child of a symbol with equal priority; and
  • a non-associative symbol cannot appear as a direct leftmost or rightmost child of a symbol with equal priority.

In C, binary operators are all left-associative, meaning that the expression true && false && false parses unambiguously as (true && false) && false, because && cannot appear as the rightmost child of itself.

Consider, then, the fourth iteration on the grammar of this definition (lesson-04-b.k):

module LESSON-04-B syntax Boolean ::= "true" | "false" | "(" Boolean ")" [bracket] > "!" Boolean [function] > left: Boolean "&&" Boolean [function] > left: Boolean "^" Boolean [function] > left: Boolean "||" Boolean [function] endmodule

Here each priority group, immediately after the ::= or > operator, can be followed by a symbol representing the associativity of that priority group: either left: for left associativity, right: for right associativity, or non-assoc: for non-associativity. In this example, each priority group we apply associativity to has only a single production, but we could equally well write a priority block with multiple productions and an associativity.

For example, consider the following, different grammar (lesson-04-c.k):

module LESSON-04-C syntax Boolean ::= "true" | "false" | "(" Boolean ")" [bracket] > "!" Boolean [function] > left: Boolean "&&" Boolean [function] | Boolean "^" Boolean [function] | Boolean "||" Boolean [function] endmodule

In this example, unlike the one above, &&, ^, and || have the same priority. However, viewed as a group, the entire group is left associative. This means that none of &&, ^, and || can appear as the right child of any of &&, ^, or ||. As a result of this, this grammar is also not ambiguous. However, it expresses a different grammar, and you are encouraged to think about what the differences are in practice.


Parse the program true && false && false yourself, and confirm that the AST places the rightmost && at the top of the expression. Then modify the definition to generate the alternative parse.

Explicit priority and associativity declarations

Previously we have only considered the case where all of the productions which you wish to express a priority or associativity relation over are co-located in the same block of productions. However, in practice this is not always feasible or desirable, especially as a definition grows in size across multiple modules.

As a result of this, K provides a second way of declaring priority and associativity relations.

Consider the following grammar, which we will name lesson-04-d.k and which will express the exact same grammar as lesson-04-b.k

module LESSON-04-D syntax Boolean ::= "true" [group(literal)] | "false" [group(literal)] | "(" Boolean ")" [group(atom), bracket] | "!" Boolean [group(not), function] | Boolean "&&" Boolean [group(and), function] | Boolean "^" Boolean [group(xor), function] | Boolean "||" Boolean [group(or), function] syntax priority literal atom > not > and > xor > or syntax left and syntax left xor syntax left or endmodule

This introduces a couple of new features of K. First, the group(_) attribute is used to conceptually group together sets of sentences under a common user-defined name. For example, literal in the syntax priority sentence is used to refer to all the productions marked with the group(literal) attribute, i.e., true and false. A production can belong to multiple groups using syntax such as group(myGrp1,myGrp2).

Once we understand this, it becomes relatively straightforward to understand the meaning of this grammar. Each syntax priority sentence defines a priority relation where > separates different priority groups. Each priority group is defined by a list of one or more group names, and consists of all productions which are members of at least one of those named groups.

In the same way, a syntax left, syntax right, or syntax non-assoc sentence defines an associativity relation among left-, right-, or non-associative groups. Specifically, this means that:

syntax left a b

is different to:

syntax left a
syntax left b

As a consequence of this, syntax [left|right|non-assoc] should not be used to group together labels with different priority.


Sometimes priority and associativity prove insufficient to disambiguate a grammar. In particular, sometimes it is desirable to be able to choose between two ambiguous parses directly while still not rejecting any parses if the term parsed is unambiguous. A good example of this is the famous "dangling else" problem in imperative C-like languages.

Consider the following definition (lesson-04-E.k):

module LESSON-04-E syntax Exp ::= "true" | "false" syntax Stmt ::= "if" "(" Exp ")" Stmt | "if" "(" Exp ")" Stmt "else" Stmt | "{" "}" endmodule

We can write the following program (dangling-else.if):

if (true) if (false) {} else {}

This is ambiguous because it is unclear whether the else clause is part of the outer if or the inner if. At first we might try to resolve this with priorities, saying that the if without an else cannot appear as a child of the if with an else. However, because the non-terminal in the parent symbol is both preceded and followed by a terminal, this will not work.

Instead, we can resolve the ambiguity directly by telling the parser to "prefer" or "avoid" certain productions when ambiguities arise. For example, when we parse this program, we see the following ambiguity as an error message:

[Error] Inner Parser: Parsing ambiguity.
1: syntax Stmt ::= "if" "(" Exp ")" Stmt

2: syntax Stmt ::= "if" "(" Exp ")" Stmt "else" Stmt


Roughly, we see that the ambiguity is between an if with an else or an if without an else. Since we want to pick the first parse, we can tell K to "avoid" the second parse with the avoid attribute. Consider the following modified definition (lesson-04-f.k):

module LESSON-04-F syntax Exp ::= "true" | "false" syntax Stmt ::= "if" "(" Exp ")" Stmt | "if" "(" Exp ")" Stmt "else" Stmt [avoid] | "{" "}" endmodule

Here we have added the avoid attribute to the else production. As a result, when an ambiguity occurs and one or more of the possible parses has that symbol at the top of the ambiguous part of the parse, we remove those parses from consideration and consider only those remaining. The prefer attribute behaves similarly, but instead removes all parses which do not have that attribute. In both cases, no action is taken if the parse is not ambiguous.


  1. Parse the program if (true) if (false) {} else {} using lesson-04-f.k and confirm that else clause is part of the innermost if statement. Then modify the definition so that you will get the alternative parse.

  2. Modify your solution from Lesson 1.3, Exercise 2 so that unary negation should bind tighter than multiplication and division, which should bind tighter than addition and subtraction, and each binary operator should be left associative. Write these priority and associativity declarations explicitly, and then try to write them inline.

  3. Write a simple grammar containing at least one ambiguity that cannot be resolved via priority or associativity, and then use the prefer attribute to resolve that ambiguity.

  4. Explain why the following grammar is not labeled ambiguous by the K parser when parsing abb, then make the parser realize the ambiguity.

module EXERCISE4 syntax Expr ::= "a" Expr "b" | "abb" | "b" endmodule

Next lesson

Once you have completed the above exercises, you can continue to Lesson 1.5: Modules, Imports, and Requires.