Let's say that we have the following expression (band 'x (bor 'y 'z)), where band and bor boolean structs that have arg1 and arg2.
If I want to change the variables 'x and 'y to 'a and 'b by deep recursion on the expression, how can I do that?
There is a special form for functionally updating just some fields in a struct which is very nice to use where you have a lot of fields:
(struct person (name age occupation) #:transparent)
(define p (person "Per" 19 "Painter"))
(define (change-occupation p new-occupation)
(struct-copy person p [occupation new-occupation]))
(change-occupation p "Programmer") ; ==> (person "Per" 19 "Programmer")
Of course this is just a fancy way of writing:
(define (change-occupation p new-occupation)
(person (person-name p)
(person-age p)
new-occupation))
Now I don't know the names of your two structs but you may need to make a generic accessor unless one is subtype of the other:
(define (change-first obj new-value)
(if (band? obj)
(band new-value (band-second obj))
(bor new-value (bor-arg2 obj))))
Or you can just have similar case analysis in your procedure.
Related
Hello(my english isn't very well I hope you will understand) , I have a misson to make a compiler, already made the language in the lex and yacc but i'm pretty stuck, our teacher asked from us to build AST tree from the language and print it with pre-order. he gave us example for the text:
function void foo(int x, y, z; real f) {
if (x>y) {
x = x + f;
}
else {
y = x + y + z;
x = f * 2;
z = f;
}
}
the output of the AST tree with pre-order should be:
(CODE
(FUNCTION
foo
(ARGS
(INT x y z)
(REAL f)
)
(TYPE VOID)
(BODY
(IF-ELSE
(> x y)
(BLOCK
(= x
(+ x f)
)
)
(BLOCK
(= y
(+
(+ x y)
z
)
)
(
(= x
(* f 2)
)
(= z f)
)
)
)
)
my question is how should I build the tree? I mean which token will go to left which will go to right so I can get the same output ?
like
makeTree($1,$2,$3);
node,left,right
Help please :)
Stephen Johnson wrote a technical paper to accompany yacc about 42 years ago. Have you read it, and do you understand it?
If yes, a syntax rule like:
expr : expr '+' expr { $$ = node( '+', $1, $3 ); }
node is effectively creating an abstract syntax tree node; and each reduction performed by yacc is the opportunity to build this tree from the bottom up. That is the most important thing to know about yacc; it builds from the bottom up, and you need to construct your data structures likewise.
When the parse is complete ( for whatever version of complete your grammar yields ), the resultant value ($$) is the root of your syntax tree.
followup
You might want to devise a node data structure something like this:
typedef struct Node Node;
typedef struct NodeList NodeList;
struct NodeList {
int Num;
Node *List;
};
struct Node {
int Type;
union {
unsigned u; unsigned long ul; char *s, ... ;
Variable *var;
Node *Expression;
NodeList *List;
} Operands[3];
};
With this you could devise a node of type '+', which defined 2 Operands, corresponding to the Left and Right sides of the '+' opcode.
You could also have a node of type IF, which had three operands:
a conditional Expression
a List of Nodes to perform if the conditional was true
a List of Nodes to perform if the conditional was false.
You could have a node of type Func, which had three operands:
A Type of return value.
A List of arguments.
A List of Nodes comprising the body of the function
I would give more examples but formatting lists with this UI is as much fun as kicking a whale down a beach.
I am trying to create a dictionary or hash-table where keys are string and values are integers with following code:
(define dict #())
(dict-set! dict "bash" 1)
(displayln dict)
(dict-set! dict "racket" 1)
(displayln dict)
However, it gives following error:
dict-set!: contract violation
expected: (dict-implements/c dict-set!)
given: '#()
in: the d argument of
(->i
((d (dict-implements/c dict-set!))
(k (d) (dict-key-contract d))
(value (d) (dict-value-contract d)))
(_r void?))
contract from: <collects>/racket/dict.rkt
Where is the problem and how can it be solved?
The problem is that the literal #() is an empty vector.
To make a mutable hash table, use (make-hash).
(define dict (make-hash))
So let me start by saying this was part of a past homework I couldn't solve but as I am preparing for a test I would like to know how to do this. I have these implementations of map_tree and fold_tree provided by the instructor:
let rec map_tree (f:'a -> 'b) (t:'a tree) : 'b tree =
match t with
| Leaf x -> Leaf (f x)
| Node (x,lt,rt) -> Node (f x,(map_tree f lt),(map_tree f rt))
let fold_tree (f1:'a->'b) (f2:'a->'b->'b->'b) (t:'a tree) : 'b =
let rec aux t =
match t with
| Leaf x -> f1 x
| Node (x,lt,rt) -> f2 x (aux lt) (aux rt)
in aux t
I need to implement a function that verifies a tree is a BST using the above functions, so far this is what I've accomplished and I'm getting the error:
Error: This expression has type bool but an expression was expected of type
'a tree
This is my code:
let rec smaller_than t n : bool =
begin match t with
| Leaf x -> true
| Node(x,lt,rt) -> (x<n) && (smaller_than lt x) && (smaller_than rt x)
end
let rec greater_equal_than t n : bool =
begin match t with
| Leaf x -> true
| Node(x,lt,rt) -> (x>=n) && (greater_equal_than lt x) && (greater_equal_than rt x)
end
let check_bst t =
fold_tree (fun x -> true) (fun x lt rt -> ((check_bst lt) && (check_bst rt)&&(smaller_than lt x)&&(greater_equal_than rt x))) t;;
Any suggestions? I seem to have trouble understanding exactly how higher order functions work in OCAML
What is the specification of a BST? It's a binary tree where:
all the elements in the left subtree (which is also a BST) are strictly smaller than the value stored at the node
and all the ones in the right subtree (which is also a BST) are bigger or equal than the value stored at the node
A fold is an induction principle: you have to explain how to deal with the base cases (the Leaf case here) and how to combine the results for the subcases in the step cases (the Node case here).
A Leaf is always a BST so the base case is going to be pretty simple. However, in the Node case, we need to make sure that the values are in the right subtrees. To be able to perform this check, we are going to need extra information. The idea is to have a fold computing:
whether the given tree is a BST
and the interval in which all of its values live
Let's introduce type synonyms to structure our thoughts:
type is_bst = bool
type 'a interval = 'a * 'a
As predicted, the base case is easy:
let leaf_bst (a : 'a) : is_bst * 'a interval = (true, (a, a))
In the Node case, we have the value a stored at the node and the results computed recursively for the left (lih as in left induction hypothesis) and right subtrees respectively. The tree thus built is a BST if and only if the two subtrees are (b1 && b2) and their values respect the properties described earlier. The interval in which this new tree's values live is now the larger (lb1, ub2).
let node_bst (a : 'a) (lih : is_bst * 'a interval) (rih : is_bst * 'a interval) =
let (b1, (lb1, ub1)) = lih in
let (b2, (lb2, ub2)) = rih in
(b1 && b2 && ub1 < a && a <= lb2, (lb1, ub2))
Finally, the function checking whether a tree is a BST is defined by projecting out the boolean out of the result of calling fold_tree leaf_bst node_bst on it.
let bst (t : 'a tree) : bool =
fst (fold_tree leaf_bst node_bst t)
I investigated Coq a little, with its dependent types. I have only the foggiest idea about it all, but now I have in mind that I want a bit vector as a bool list, in which the length of the vector is part of the type.
(This question is the possible predecessor of another question. In the next question, if I ask it, I'll ask whether I can recover what I lose, when I use typedef as below.)
(For this question, the question is at the bottom)
Here are the requirements for the type I want:
It has to use bool list, so that I can directly or indirectly do pattern matching and recursion on the list, and
the length of the vector has to be specified in the type.
Here is what I have:
typedef bitvec_4 = "{bl::bool list. length bl = 4}"
by(auto, metis Ex_list_of_length)
It's important that the length of the list be part of the type, because I want to use the type with a definition, where all lists are known to be of the same size, like with this simple example:
definition two_bv4_to_bv4 :: "bitvec_4 => bitvec_4 => bitvec_4" where
"two_bv4_to_bv4 x y = x"
What I don't want, in a theorem, is to have to specify the length of the lists. Type classes would eventually come into play somehow, but I want, as I say, the length to be specified in the type definition.
Definition and type signatures. Where do I let n = 4!!? (a tech joke of minimal humor-value)
Now, I try to generalize with a typedef like this, in which the length is a variable:
typedef bitvec_n = "{(bl::bool list, n::nat). length bl = n}"
by(auto)
That's no good. In a definition like this next one, my type doesn't guarantee that all lists are of the same length:
definition two_bvn_to_bvn :: "bitvec_n => bitvec_n => bitvec_n" where
"two_bvn_to_bvn x y = x"
The question? (I think so)
I've experimented a little with types like bitvec_4 above. If I don't run into big roadblocks, I might try to make big use of them.
I could define types like the above for powers of 2, up to, say, 1024 bits, along with type classes that reflect their common properties.
But, is there a better way to do this? It has to be somewhat straightforward, I think, with the use of bool list.
Update (got the answer for what it was actually about)
Based on Manuel's answer, I include here a self-contained theory.
It's mostly a duplication of Manuel's source, but at the end, my functions swap_bl and swap_2bv, along with the final use of value, show the end result of what I was trying to accomplish. My comments emphasize the problems that were on my mind, and possibly, my end application shows why I haven't looked to HOL/Word as a solution.
For a typedef type, to do pattern matching indirectly, similar to that with swap_bl and 2 bitvec, I was using the Abs and Rep functions together as inverses. One problem, as Manuel pointed out, is that I can feed an Abs function a bool list of the wrong length, and it won't give an error. Another big problem is the abstraction violations due to the use of the Abs function.
Those problems, and wanting to know if I could recover the use of value for my typedef type, would have been parts of my next question, but all that's been answered here.
theory i141210ac__testing_out_manuels_answer
imports Main "~~/src/HOL/Library/Numeral_Type"
begin
(*Basic type definition.*)
typedef ('n::finite) bitvec = "{bs :: bool list. length bs = CARD('n)}"
morphisms bitvec_to_list Abs_bitvec
by (simp add: Ex_list_of_length)
setup_lifting type_definition_bitvec
lift_definition nth :: "('n::finite) bitvec => nat => bool" (infixl "$" 90)
is List.nth .
(*Can't use 'value' yet for 'nth', or I get an abstraction violation.*)
term "(Abs_bitvec [True,False] :: 2 bitvec) $ 1"
(*Truncate or fill the list: needed to set things up for 'value'.*)
definition set_length :: "nat => bool list => bool list" where
"set_length n xs = (if length xs < n
then xs # replicate (n - length xs) False
else take n xs)"
lemma length_set_length [simp]: "length (set_length n xs) = n"
unfolding set_length_def by auto
definition list_to_bitvec :: "bool list => ('n::finite) bitvec" where
"list_to_bitvec xs = Abs_bitvec (set_length CARD('n) xs)"
(*Finishing the magic needed for 'value'.*)
lemma list_to_bitvec_code [code abstract]:
"bitvec_to_list (list_to_bitvec xs :: ('n::finite) bitvec)
= set_length CARD('n) xs"
unfolding list_to_bitvec_def by(simp add: Abs_bitvec_inverse)
(*Inverses for lists of length 2: no abstraction violations.*)
value "list_to_bitvec (bitvec_to_list x) :: 2 bitvec"
value "bitvec_to_list (list_to_bitvec x :: 2 bitvec)"
(*The magic now kicks in for 'value' and 'nth'. Understanding is optional.*)
value "(list_to_bitvec [True,False] :: 2 bitvec) $ 1" (*OUTPUT: False.*)
(*For my use, the primary context of all this is pattern matching on lists.
I can't pattern match on a 'typedef' type directly with 'fun', because
it's not a 'datatype'. I do it indirectly.*)
fun swap_bl :: "bool list => bool list" where
"swap_bl [a,b] = [b,a]"
|"swap_bl _ = undefined"
definition swap_2bv :: "2 bitvec => 2 bitvec" where
"swap_2bv bv = list_to_bitvec (swap_bl (bitvec_to_list bv))"
value "swap_2bv (list_to_bitvec [a,b] :: 2 bitvec)" (*
OUTPUT: "Abs_bitvec [b, a]" :: "2 bitvec" *)
(*Whether that's all a good idea, that's about the future, but it appears the
hard work, recovering the use of 'value', along with generalizing length,
has been done by Manuel, and the authors of Numeral_Type and its imports.*)
end
Isabelle does not support dependent types, but there are ways to still do what you want to do. For instance, there is already a stack of type classes and type syntax for type-level natural numbers.
theory Scratch
imports Main "~~/src/HOL/Library/Numeral_Type"
begin
lemma "(UNIV :: 4 set) = {0,1,2,3}"
by (subst UNIV_enum) eval
As you can see, the type 4 is a type that contains the numbers from 0 to 3. Incidentally, this can also be used for computations in modular arithmetic:
lemma "((2 + 3) :: 4) = 1" by simp
lemma "((2 * 3) :: 4) = 2" by simp
You can use these numeral types to parametrise your bit vectors with a length:
typedef ('n::finite) bitvec = "{bs :: bool list. length bs = CARD('n)}"
morphisms bitvec_to_list Abs_bitvec
by (simp add: Ex_list_of_length)
setup_lifting type_definition_bitvec
You can access the n-th element of a bit vector by lifting the nth function from Boolean lists to bit vectors, which works automatically:
lift_definition nth :: "('n::finite) bitvec ⇒ nat ⇒ bool" (infixl "$" 90) is List.nth .
Converting boolean lists to bit vectors is a bit tricky, because the list you get in might not have the correct length; the expression list_to_bitvec [True] :: 2 bitvec would typecheck, but is obviously problematic. You could solve this either by returning undefined or, perhaps more appropriate in this instance, filling up the list with False or truncating it to get the right length:
definition set_length :: "nat ⇒ bool list ⇒ bool list" where
"set_length n xs = (if length xs < n then xs # replicate (n - length xs) False else take n xs)"
lemma length_set_length[simp]: "length (set_length n xs) = n"
unfolding set_length_def by auto
Now we can define a function that converts a list of Booleans to a bit vector:
definition list_to_bitvec :: "bool list ⇒ ('n::finite) bitvec" where
"list_to_bitvec xs = Abs_bitvec (set_length CARD('n) xs)"
However, we are not allowed to use Abs_bitvec in code equations; if you tried to evaluate, say, list_to_bitvec [True] :: 1 bitvec, you would get an abstraction violation. We have to give an explicit code abstract equation in terms of the morphism list_to_bitvec:
lemma list_to_bitvec_code[code abstract]:
"bitvec_to_list (list_to_bitvec xs :: ('n::finite) bitvec) = set_length CARD('n) xs"
unfolding list_to_bitvec_def by (simp add: Abs_bitvec_inverse)
And now we are basically done and can do e.g. this:
definition myvec :: "4 bitvec" where "myvec = list_to_bitvec [True, False, True]"
value myvec
(* Output: "Abs_bitvec [True, False, True, False]" :: "4 bitvec" *)
value "myvec $ 2"
(* Output: "True" :: "bool" *)
Note that you always have to annotate the result of list_to_bitvec with its length; Isabelle can not infer the length.
You may also want to have a look at the Word theory in ~~/src/HOL/Word/; it implements machine words of fixed length with all kinds of bit operations like NOT, AND, OR, etc.:
value "42 AND 23 :: 32 word"
(* Output: "2" :: "32 word" *)
value "293 :: 8 word"
(* Output: "37" :: "8 word" *)
value "test_bit (42 :: 8 word) 1"
(* Output: "True" :: "bool" *)
value "set_bit (42 :: 8 word) 2 True"
(* Output: "46" :: "8 word" *)
value "(BITS i. i < 4) :: 8 word"
(* Output: "15" :: "8 word" *)
Another related type are the vectors in src/HOL/Library/Multivariate_Analysis/Finite_Cartesian_Product.
Can the following polymorphic functions
let id x = x;;
let compose f g x = f (g x);;
let rec fix f = f (fix f);; (*laziness aside*)
be written for types/type constructors or modules/functors? I tried
type 'x id = Id of 'x;;
type 'f 'g 'x compose = Compose of ('f ('g 'x));;
type 'f fix = Fix of ('f (Fix 'f));;
for types but it doesn't work.
Here's a Haskell version for types:
data Id x = Id x
data Compose f g x = Compose (f (g x))
data Fix f = Fix (f (Fix f))
-- examples:
l = Compose [Just 'a'] :: Compose [] Maybe Char
type Natural = Fix Maybe -- natural numbers are fixpoint of Maybe
n = Fix (Just (Fix (Just (Fix Nothing)))) :: Natural -- n is 2
-- up to isomorphism composition of identity and f is f:
iso :: Compose Id f x -> f x
iso (Compose (Id a)) = a
Haskell allows type variables of higher kind. ML dialects, including Caml, allow type variables of kind "*" only. Translated into plain English,
In Haskell, a type variable g can correspond to a "type constructor" like Maybe or IO or lists. So the g x in your Haskell example would be OK (jargon: "well-kinded") if for example g is Maybe and x is Integer.
In ML, a type variable 'g can correspond only to a "ground type" like int or string, never to a type constructor like option or list. It is therefore never correct to try to apply a type variable to another type.
As far as I'm aware, there's no deep reason for this limitation in ML. The most likely explanation is historical contingency. When Milner originally came up with his ideas about polymorphism, he worked with very simple type variables standing only for monotypes of kind *. Early versions of Haskell did the same, and then at some point Mark Jones discovered that inferring the kinds of type variables is actually quite easy. Haskell was quickly revised to allow type variables of higher kind, but ML has never caught up.
The people at INRIA have made a lot of other changes to ML, and I'm a bit surprised they've never made this one. When I'm programming in ML, I might enjoy having higher-kinded type variables. But they aren't there, and I don't know any way to encode the kind of examples you are talking about except by using functors.
You can do something similar in OCaml, using modules in place of types, and functors (higher-order modules) in place of higher-order types. But it looks much uglier and it doesn't have type-inference ability, so you have to manually specify a lot of stuff.
module type Type = sig
type t
end
module Char = struct
type t = char
end
module List (X:Type) = struct
type t = X.t list
end
module Maybe (X:Type) = struct
type t = X.t option
end
(* In the following, I decided to omit the redundant
single constructors "Id of ...", "Compose of ...", since
they don't help in OCaml since we can't use inference *)
module Id (X:Type) = X
module Compose
(F:functor(Z:Type)->Type)
(G:functor(Y:Type)->Type)
(X:Type) = F(G(X))
let l : Compose(List)(Maybe)(Char).t = [Some 'a']
module Example2 (F:functor(Y:Type)->Type) (X:Type) = struct
(* unlike types, "free" module variables are not allowed,
so we have to put it inside another functor in order
to scope F and X *)
let iso (a:Compose(Id)(F)(X).t) : F(X).t = a
end
Well... I'm not an expert of higher-order-types nor Haskell programming.
But this seems to be ok for F# (which is OCaml), could you work with these:
type 'x id = Id of 'x;;
type 'f fix = Fix of ('f fix -> 'f);;
type ('f,'g,'x) compose = Compose of ('f ->'g -> 'x);;
The last one I wrapped to tuple as I didn't come up with anything better...
You can do it but you need to make a bit of a trick:
newtype Fix f = In{out:: f (Fix f)}
You can define Cata afterwards:
Cata :: (Functor f) => (f a -> a) -> Fix f -> a
Cata f = f.(fmap (cata f)).out
That will define a generic catamorphism for all functors, which you can use to build your own stuff. Example:
data ListFix a b = Nil | Cons a b
data List a = Fix (ListFix a)
instance functor (ListFix a) where
fmap f Nil = Nil
fmap f (Cons a lst) = Cons a (f lst)