I have a rule that looks like this:
a : (b | c) d;
b : 'B';
c : 'C';
d : 'D';
With this grammar ANTLR builds a flat parse tree. How can I rewrite the first rule (and leave the other two unchanged) so that whatever is matched is being returned under a root node called A?
If the first production rule was like this:
a : b d;
then it could have been rewritten as
a : b d -> ^(A b d)
and it would have solved my problem. However the first grammar rule yields more than one possibility for the resulting parse tree ^(A b d) or ^(A c d).
How do I express this when rewriting the rule?
You can use the ? operator in the rewrite as follows.
a : (b | c) d -> ^(A b? c? d);
Related
I'm currently looking at two closure calculation examples using the tool at
http://jsmachines.sourceforge.net/machines/lr1.html
Example 1
S -> A c
A -> b B
B -> A b
Here, in the initial state ends up with a closure of:
[S -> .A c, $]; [A -> .b B, c]}
Example 2
S -> A B
A -> a
B -> b
B -> ''
The calculated first step closure is:
{[S -> .A B, $]; [A -> .a, b/$]}
In example 1, why is the follow of b from rule 3 not included in the lookahead? In case 2, we follow B to figure out that $ is part of the lookahead so is there some special reason to not consider all rules in case 1?
When doing a closure with ". A α" we use FIRST(α) as the lookahead, and only include the containing (parent) lookahead if ε ∈ FIRST(α). In example 1, ε ∉ FIRST(c), so the lookahead is just c. In example 2, ε ∈ FIRST(B), so we add the containing lookahead ($ in this case) to the lookahead.
FOLLOW is never relevant.
I have a rule in my grammar such as
A -> B C D E {: ...some actions... :}
;
D -> /*empty*/ {: some actions using attributes of B and C :}
;
To implement the actions associated with production rule of D, I need to access the parser stack. How can I do that in CUP?
Rewrite your grammar:
A -> A1 E
A1 -> B C D
If the action for the first production requires B and C as well, then the semantic value of A1 will have to be more complicated in order to pass the semantic values through.
I wanted to implement this statement in agda ;
A dedekind cut is a pair (L, U) of mere predicates L : Q -> Set and R : Q -> Set which is
1) inhibited : exists (q : Q) . L(q) ^ exists (r : Q) . U(r)
I have tried in this way,
record cut : Set where
field
L : Q -> Set
R : Q -> Set
inhibited : exists (q : Q) . L(q) ^ exists (r : Q) . U(r)
but this is not working. I want to write this and i am struck please help. And also what is the difference between 1)data R : Set and record R : Set and 2) data R : Set and data R : Q -> Set
I don't know about defining the whole of the dedekind cut, though I did find your definition on page 369 of Homotopy Type Theory: Univalent Foundations of Mathematics.
Here is the syntax for defining what you asked about in your question in two forms, one using the standard library and one expanded out to show what it is doing.
open import Data.Product using (∃)
record Cut (Q : Set) : Set₁ where
field
L U : Q → Set -- The two predicates
L-inhabited : ∃ L
U-inhabited : ∃ U
If we manually expand the definition of ∃ (exists) we have:
record Cut′ (Q : Set) : Set₁ where
field
L U : Q → Set -- The two predicates
q r : Q -- Witnesses
Lq : L q -- Witness satisfies predicate
Ur : U r -- Witness satisfies predicate
Note that the record has type Set₁ due to the types of fields L and U.
Regarding your question about records and inductively defined data types, there are lots of differences. You might want to start with the wiki and ask more specific questions if you get stuck somewhere: Data, Records
Unfortunately, it is not possible for ANTLR to support direct-left recursion when the rule has parameters passed. The only viable option is to remove the left recursion. Is there a way to remove the left-recursion in the following grammar ?
a[int x]
: b a[$x] c
| a[$x - 1]
(
c a[$x - 1]
| b c
)
;
The problem is in the second alternative involving left recursion. Any kind of help would be much appreciated.
Without the parameters and easier formatting, it would look like this:
a
: b a c
| a (c a | b c)
;
When a's left recursive alternative is matched n times, it would just mean that (c a | b c) will be matched n times, pre-pended with the terminating b a c (the first alternative). That means that this rule will always start with b a c, followed by zero or more occurrences of (c a | b c):
a
: b a c (c a | b c)*
;
First some boring imports:
import Relation.Binary.PropositionalEquality as PE
import Relation.Binary.HeterogeneousEquality as HE
import Algebra
import Data.Nat
import Data.Nat.Properties
open PE
open HE using (_≅_)
open CommutativeSemiring commutativeSemiring using (+-commutativeMonoid)
open CommutativeMonoid +-commutativeMonoid using () renaming (comm to +-comm)
Now suppose that I have a type indexed by, say, the naturals.
postulate Foo : ℕ -> Set
And that I want to prove some equalities about functions operating on this type Foo. Because agda is not very smart, these will be heterogeneous equalities. A simple example would be
foo : (m n : ℕ) -> Foo (m + n) -> Foo (n + m)
foo m n x rewrite +-comm n m = x
bar : (m n : ℕ) (x : Foo (m + n)) -> foo m n x ≅ x
bar m n x = {! ?0 !}
The goal in bar is
Goal: (foo m n x | n + m | .Data.Nat.Properties.+-comm n m) ≅ x
————————————————————————————————————————————————————————————
x : Foo (m + n)
n : ℕ
m : ℕ
What are these |s doing in the goal? And how do I even begin to construct a term of this type?
In this case, I can work around the problem by manually doing the substitution with subst, but that gets really ugly and tedious for larger types and equations.
foo' : (m n : ℕ) -> Foo (m + n) -> Foo (n + m)
foo' m n x = PE.subst Foo (+-comm m n) x
bar' : (m n : ℕ) (x : Foo (m + n)) -> foo' m n x ≅ x
bar' m n x = HE.≡-subst-removable Foo (+-comm m n) x
Those pipes indicate that reduction is suspended pending the result of the expressions in question, and it typically boils down to the fact that you had a with block whose result you need to know to proceed. This is because the rewrite construct just expands to a with of the expression in question along with any auxiliary values that might be needed to make it work, followed by a match on refl.
In this case, it just means that you need to introduce the +-comm n m in a with block and pattern match on refl (and you'll probably need to bring n + m into scope too, as it suggests, so the equality has something to talk about). The Agda evaluation model is fairly straightforward, and if you pattern match on something (except for the faux pattern matches on records), it won't reduce until you pattern match on that same thing. You might even be able to get away with rewriting by the same expression in your proof, since it just does what I outlined for you.
More precisely, if you define:
f : ...
f with a | b | c
... | someDataConstructor | boundButNonConstructorVariable | someRecordConstructor = ...
and then you refer to f as an expression, you will get the pipes you observed for expression a only, because it matches on someDataConstructor, so at the very least to get f to reduce you will need to introduce a and then match on someDataConstructor from it. On the other hand, b and c, although they were introduced in the same with block, do no halt evaluation, because b doesn't pattern match, and c's someRecordConstructor is known statically to be the only possible constructor because it's a record type with eta.