I'm trying to convince the type checker that the type specified in the where clause matches one of the types of the top level function.
For example:
interface SomeInterface (e: Type -> Type) where
label : e l -> l
f : (SomeInterface e) => e l -> c
f x = ?hole_f
where
g : e l
g = ?hole_g
The type of ?hole_g should be e l, but the typechecker is unable to match the l type in g : e l to the l type in the toplevel f : (SomeInterface e) => e l -> c type. So according to the typechecker the type of ?hole_g is e l1
One way to fix this is to include l as a separate type variable in the toplevel function f, like this:
f : (SomeInterface e) => e l -> l -> c
f x _ = ?hole_f
where
g : e l
g = ?hole_g
Notice that the l in the toplevel function is completely neglected due to the _. But the typechecker is now somehow convinced that the type variable of ?hole_g is e l, like the toplevel type.
My two questions are:
1) Why is the typechecker unable to "link" the types of function g to f?
2) How can i fix this without including an extra type variable?
f : (SomeInterface e) => e l -> c
f x {l} = ?hole_f_1
where
g : e l
g = ?hole_g
Does this do what you want? I believe since e l is a type, you can still get the implicit information about the l part.
Related
I'm writing a function to test propositional equality of Nat, and it typechecks in Idris 1.
sameNat : (n : Nat) -> (m : Nat) -> Maybe (n = m)
sameNat Z Z = Just Refl
sameNat (S n) (S m) = case sameNat n m of
Just e => Just (cong e)
Nothing => Nothing
sameNat _ _ = Nothing
But it doesn't typecheck in Idris 2 (0.4.0) and I got this error.
Error: While processing right hand side of sameNat. When
unifying n = m and Nat m e -> :: ?x ?xs n m e.
Mismatch between: n = m and Nat m e -> :: ?x ?xs n m e.
It typechecks when I write a specific version of cong and use it.
cong' : n = m -> S n = S m
cong' Refl = Refl
Why doesn't this typecheck and how can I make it typecheck?
The type signature of cong changed:
Idris 1:
cong : (a = b) -> f a = f b
Idris 2:
Prelude.cong : (0 f : (t -> u)) -> a = b -> f a = f b
Select two prime no's. Suppose P = 53 and Q = 59.
Now First part of the Public key : n = P*Q = 3127.
We also need a small exponent say e :
But e Must be
An integer.
Not be a factor of n.
1 < e < Φ(n) [Φ(n) is discussed below],
Let us now consider it to be equal to 3.
Our Public Key is made of n and e
Generating Private Key :
We need to calculate Φ(n) :
Such that Φ(n) = (P-1)(Q-1)
so, Φ(n) = 3016
Now calculate Private Key, d :
d = (k*Φ(n) + 1) / e for some integer k
For k = 2, value of d is 2011.
Lets say I have the following
data Expr : Type -> Type where
Lift : a -> Expr a
Add : Num a => Expr a -> Expr a -> Expr a
Cnst : Expr a -> Expr b -> Expr a
data Context : Type -> Type where
Root : Context ()
L : Expr w -> Context x -> Expr y -> Expr z -> Context w
R : Expr w -> Context x -> Expr y -> Expr z -> Context w
M : Expr w -> Context x -> Expr y -> Expr z -> Context w
data Zipper : Type -> Type -> Type where
Z : Expr f -> Context g -> Zipper f g
E : String -> Context g -> Zipper String ()
I would like to write a function to rebuild a Zipper if I'm going up one level in the expression tree. Something like:
total
ZipUp : Zipper focus parent -> Type
ZipUp (Z e (R (Cnst {a} e1 e2) {x} c t u)) = Zipper a x
ZipUp (Z {f} e (L (Cnst l r) {x} c t u)) = Zipper f x
ZipUp (Z e (R (Add {a} e1 e2) {x} c t u)) = Zipper a x
ZipUp (Z {f} e (L (Add e1 e2) {x} c t u)) = Zipper f x
ZipUp _ = Zipper String ()
the problem comes when I want to write the function to return a ZipUp
up : (x : Zipper focus parent) -> ZipUp x
up (Z e (R (Cnst l r) c x y)) = Z (Cnst l e) c
up (Z e (L (Cnst l r) c x y)) = Z (Cnst e r) c
up (Z e (R (Add l r) c x y)) = Z (Add l e) c
up (Z e (L (Add l r) c x y)) = Z (Add e r) c
up (Z e (R (Lift x) c l r)) = E "Some error" c
up (Z e (L (Lift x) c l r)) = E "Some error" c
up (E s c) = E s c
This fails to typecheck on the Add case, because it can't infer that focus (type of e) matches with parent (expected type)
So my question has two parts then.
What can I do to express this relationship?
(that if the Zipper is an R constructed, e has the same type as r, or that e has the same type as l in the L constructed case. I've tried using {e = l} and other variants of this, but to no avail.)
The the code typechecks and runs if I comment out the last for lines of up to finish with:
up : (x : Zipper focus parent) -> ZipUp x
up (Z e (R (Cnst l r) c x y)) = Z (Cnst l e) c
up (Z e (L (Cnst l r) c x y)) = Z (Cnst e r) c
but the actual manipulation of the types should be the same in the Add case, yet this fails to typecheck, why is that?
Thanks for taking the time to read and/or answer!
This fails to typecheck on the Add case, because it can't infer that focus (type of e) matches with parent (expected type)
Because this is not always the case, e.g.
*main> :t Z (Lift "a") (R (Add (Lift 1) (Lift 2)) Root (Lift 4) (Lift 8))
Z (Lift "a") (R (Add (Lift 1) (Lift 2)) Root (Lift 4) (Lift 8)) : Zipper String Integer
And Add (Lift 1) (Lift "a") doesn't work because of the Num a constraint.
What can I do to express this relationship?
If you want to express the relationship within up: You have e : Expr f and can say that the same f should be used in the Add case:
up (Z {f} e (R (Add l r {a=f}) c x y)) = Z (Add l e) c
up (Z {f} e (L (Add l r {a=f}) c x y)) = Z (Add e r) c
up (Z e (R (Add l r) c x y)) = ?whattodo_1
up (Z e (L (Add l r) c x y)) = ?whattodo_2
Now you have a non total function because of the cases where a != f. I don't quite what you want to do, so I can't offer an option. :-)
If you want to express the relationship in Zipper you could do (very roughly) something like:
data Context : Bool -> Type -> Type where
Root : Context False ()
L : Expr w -> Context b x -> Expr y -> Expr z -> Context False w
R : Expr w -> Context b x -> Expr y -> Expr z -> Context True w
M : Expr w -> Context b x -> Expr y -> Expr z -> Context False w
data Zipper : Type -> Type -> Type where
Z : Expr f -> Context False g -> Zipper f g
ZR : Expr f -> Context True f -> Zipper f f
E : String -> Context b g -> Zipper String ()
Now you construct a proof when building up a zipper that f = g in a R-Context. Again, I cannot be specific, but I hope it helps in some way.
I want replace 1 word in text file (file format is not .txt)
file Unicode is (UTF16)
few text example:
I D = " f f 0 3 4 a 9 2 - d d 9 f - 4 3 7 4 - a 8 a d - f 5 5 4 0 0 2 a 4 1 9 b " I S S U E _ D A T E = " 2 0 1 7 - 0 2 - 1 6 T 1 7 : 2 9 : 1 8 . 9 7 0 2 2 9 4 Z " S E Q U E N C E = " 0 " M A N A G I N G _ A P P L I C A T I O N _ T O K E N = " " > < L I C E N S E P U B L I C _ I D = " 3 A A - U J F - 8 K P " U S E R N A M E = " N d a G 6 Z T w u v I X Z B i t h 8 g o d d Q x E r x 0 + O g M c t 0 2 3 f X K O E w = " P A S S W O R D = " F 9 b n 6 b v w l f I 5 Z A 2 t h M h 9 d d s x Q L w = " T Y P E = " T R I A L " F L A G S = " 4 " D I S P L A Y _ N A M E =
I want change T R I A L to other word
It's not too hard to modify your text file. Use the IO class to assign it to a text file, then use String.Replace(oldValue As String, newValue As String) to change your string. Then use IO again to save the string to the file. This should work so long as your file isn't open and being used in another program - regardless of file extensions.
An example, to help you, could be something such as this:
Dim myFileContents as String = IO.File.ReadAllText("Path\To\My\File\File.extension")
myFileContents = myFileContents.Replace("T R I A L", "Some other word")
IO.File.WriteAllText("Path\To\My\File\File.extension", myFileContents)
Modify the contents to suit your situation - however, this is only a basic implementation. Additionally, it is important to note that String.Replace() will change all occurrences of your word to the new word.
If the page is equal to 1, I want to display the value "T O T A L" and if not the value "T O T A L F O R W A R D E D" should display instead.
Since the variable Globals!PageNumber cannot be called outside Header and Footer, is there any other method?
The formula is supposed to be easy but does not work. It should be like this:
=iif(Globals!PageNumber > 1, "T O T A L F O R W A R D E D", "T O T A L")