Is it possible, that a function can return a function in Ada? I am trying to get currying to work.
type Integer_Func_Type is access function (Y : Integer) return Integer;
function Add (X : Integer) return Integer_Func_Type is
function Inner (Y : Integer) return Integer is
begin
return X + Y;
end Inner;
begin
return Inner'Access;
end;
At the end, I do not want to provide all arguments of a function one at a time. For example: if x is a ternary function and y is curry(x), then I can use following function calls: y(a,b,c), y(a,b)(c), y(a)(b,c), y(a)(b)(c).
EDIT
I implemented 'Jacob Sparre Andersen' suggestions. But it does not look like currying will be easy to implement. I must implement every possible variant of any type I want to use in advance. Is this correct?
with Ada.Text_IO;
with R;
procedure Hello is
Add_Two : R.Test2 := (X => 2);
begin
Ada.Text_IO.Put_Line(Add_Two.Add(3)'Img);
end Hello;
r.adb
package body R is
function Add(A : Test2; Y : Integer) return Integer is
begin
return A.X + Y;
end Add;
end R;
r.ads
package R is
type Test is abstract tagged null record;
function Add(A : Test; Y : Integer) return Integer is abstract;
type Test2 is new Test with
record
X : Integer;
end record;
overriding
function Add(A : Test2; Y : Integer) return Integer;
end R;
This is how to do it with generics:
with Ada.Text_IO;
procedure Test is
-- shorthand Ada 2012 syntax; can also use full body
function Add (X, Y : Integer) return Integer is (X + Y);
generic
type A_Type (<>) is limited private;
type B_Type (<>) is limited private;
type Return_Type (<>) is limited private;
with function Orig (A : A_Type; B : B_Type) return Return_Type;
A : A_Type;
function Curry_2_to_1 (B : B_Type) return Return_Type;
function Curry_2_to_1 (B : B_Type) return Return_Type is
(Orig (A, B));
function Curried_Add is new Curry_2_to_1
(Integer, Integer, Integer, Add, 3);
begin
Ada.Text_IO.Put_Line (Integer'Image (Curried_Add (39)));
end Test;
As you see, it is quite verbose. Also, you need to provide a currying implementation for every count X of parameters of the original function and every number Y of parameters of the generated function, so you'd have a lot of Curry_X_to_Y functions. This is necessary because Ada does not have variadic generics.
A lot of the verbosity also comes from Ada not doing type inference: You need to explicitly specifiy A_Type, B_Type and Return_Type even though theoretically, they could be inferred from the given original function (this is what some functional programming languages do).
Finally, you need a named instance from the currying function because Ada does not support anonymous instances of generic functions.
So, in principle, currying does work, but it is not anything as elegant as in a language like Haskell. If you only want currying for a specific type, the code gets significantly shorter, but you also lose flexibility.
You can't do quite what you're trying to do, since Inner stops to exist as soon as Add returns.
You could do something with the effect you describe using tagged types.
One abstract tagged type with a primitive operation matching your function type.
And then a derived tagged type with X as an attribute and an implementation of the function matching Inner.
Many of the answers seem to deal with ways to have subprograms that deal with variable numbers of parameters. One way to deal with this is with a sequence of values. For example,
type Integer_List is array (Positive range <>) of Integer;
function Add (List : Integer_List) return Integer;
can be considered a function that takes an arbitrary number of parameters of type Integer. This is simple if all your parameters have the same type. It's more complicated, but still possible, if you deal with a finite set of possible parameter types:
type Number_ID is (Int, Flt, Dur);
type Number (ID : Number_ID) is record
case ID is
when Int =>
Int_Value : Integer;
when Flt =>
Flt_Value : Float;
when Dur =>
Dur_Value : Duration;
end case;
end record;
type Number_List is array (Positive range <>) of Number;
function Add (List : Number_List) return Number;
If you have to be able to handle types not known in advance, this technique is not suitable.
Related
Hello I started to write in sml and I have some difficulty in understanding a particular function.
I have this function:
fun isInRow (r:int) ((x,y)) = x=r;
I would be happy to get explain to some points:
What the function accepts and what it returns.
What is the relationship between (r: int) ((x, y)).
Thanks very much !!!
The function isInRow has two arguments. The first is named r. The second is a pair (x, y). The type ascription (r: int) says that r must be an int.
This function is curried, which is a little unusual for SML. What this means roughly speaking is that it accepts arguments given separately rather than supplied as a pair.
So, the function accepts an int and a pair whose first element is an int. These are accepted as separate arguments. It returns a boolean value (the result of the comparison x = r).
A call to the function would look like this:
isInRow 3 (3, 4)
There is more to say about currying (which is kind of cool), but I hope this is enough to get you going.
In addition to what Jeffrey has said,
You don't need the extra set of parentheses:
fun isInRow (r:int) (x,y) = x=r;
You don't need to specify the type :int. If you instead write:
fun isInRow r (x,y) = x=r;
then the function's changes type from int → (int • 'a) → bool into ''a → (''a • 'b) → bool, meaning that r and x can have any type that can be compared for equality (not just int), and y can still be anything since it is still disregarded.
Polymorphic functions are one of the strengths of typed, functional languages like SML.
You could even refrain from giving y a name:
fun isInRow r (x,_) = x=r;
In Haskell, I know that if I define a function like this add x y = x + y
then I call like this add e1 e2. that call is equivalent to (add e1) e2
which means that applying add to one argument e1 yields a new function which is then applied to the second argument e2.
That's what I don't understand in Haskell. in other languages (like Dart), to do the task above, I would do this
add(x) {
return (y) => x + y;
}
I have to explicitly return a function. So does the part "yields a new function which is then applied to the second argument" automatically do underlying in Haskell? If so, what does that "hiding" function look like? Or I just missunderstand Haskell?
In Haskell, everything is a value,
add x y = x + y
is just a syntactic sugar of:
add = \x -> \y -> x + y
For more information: https://wiki.haskell.org/Currying :
In Haskell, all functions are considered curried: That is, all functions > in Haskell take just single arguments.
This is mostly hidden in notation, and so may not be apparent to a new
Haskeller. Let's take the function
div :: Int -> Int -> Int
which performs integer division. The expression div 11 2
unsurprisingly evaluates to 5. But there's more that's going on than
immediately meets the untrained eye. It's a two-part process. First,
div 11
is evaluated and returns a function of type
Int -> Int
Then that resulting function is applied to the value 2, and yields 5.
You'll notice that the notation for types reflects this: you can read
Int -> Int -> Int
incorrectly as "takes two Ints and returns an Int", but what it's
really saying is "takes an Int and returns something of the type Int
-> Int" -- that is, it returns a function that takes an Int and returns an Int. (One can write the type as Int x Int -> Int if you
really mean the former -- but since all functions in Haskell are
curried, that's not legal Haskell. Alternatively, using tuples, you
can write (Int, Int) -> Int, but keep in mind that the tuple
constructor (,) itself can be curried.)
This question relates to the scala course from coursera so I want to please ask you to not give me the plain solution that I can copy-paste as this would break the coursera honor code.
This relates to the second assignment.
def Set = Int => Boolean
As it can be seen, Set is a function which returns weather or not the given int is or not part of the set. This is plain and simple so far. However the task asks me to create a union
def union(f: Set, s: Set): Set = ???
This union should return a set that satisfies the condition of both sets.
How could I do something like this:
I thought that such a thing could be done by adding the functions together however the following code:
f + s
Will not compile properly as expected
My question to is:
How would I be able to create a function from 2 other functions?
x => if x == 0 true else false //first
x => if x == 1 true else false //second
And what should equal:
x => if x==0 || x == 1 true else false
I'm not asking for a solution but rather how would I go around building something like this?
As I think you already understand, these Sets are functions that test whether a value meets the criteria for each Set.
The union of such a Set must also be a function that returns a Boolean (as shown by the type signature)
def union(f: Set, s: Set): Set
which (because Set is a type alias) is equivalent to:
def union(f: Int => Boolean, s: Int => Boolean): Int => Boolean
In plain English, union of two sets A and B means: "is the item in A or B".
Your task is to write a function that carries out that plain English specification.
You cannot "add" two functions together (at least, not in a way that is applicable to this question), but you can combine their results.
The Set has form of Set = Int => Boolean. Given the Int function will return true if the value is in a Set.
Well if we want to create a singleton set, we will return new function, which will compare any value passed to it, with the one passed to the function that created it.
The union of two sets, is one set plus the other. It means the element you're looking for must be either in one or the other set. But how do we get the new set, well we return a new function that does just that - checks if an element is either in one set or another.
Remember that in Scala functions can return functions, which may be evaluated later. I think that's the key.
The Set is defined as a function from Int to Boolean, "summing" two Sets won't return a Set object, the union means that one element should be either in one or in the other set but always expressed as a function.
I hope this is not too much, but given an element it should satisfy either f or s.
First of all, it's type Set =. Not def. Set is a type alias not a function definition.
Now, your question. You need a function which, when given two Int =>Boolean combines them with OR and returns a Int => Boolean.
First, how would you do this for two Boolean arguments?
def or(a: Boolean, b: Boolean) = a || b
So now we're half way there. What we have:
A pair of Int => Boolean functions.
A function that takes two Booleans and return a Boolean.
So all we need to do is apply each Set to an Int to get a Boolean and OR the result. The confusion is probably here.
The easiest way to curry a function is to do it explicitly
def union(f: Set, s: Set): Set = {
def doUnion(x: Int) = //apply x to f and s, return OR
doUnion
}
But we can, in Scala, so this inline by declaring an anonymous function
def union(f: Set, s: Set): Set = x => //apply x to f and s, return OR
Let's say I have 1000 functions defined as follows
void func dummy1(int a);
void func dummy2(int a, int aa);
void func dummy3(int a, int aa, int aaa);
.
.
.
void func dummy1000(int a, int aa, int aaa, ...);
I want to write a function that takes an integer, n (n < 1000) and calls nth dummy function (in case of 10, dummy10) with exactly n arguments(arguments can be any integer, let's say 0) as required. I know this can be achieved by writing a switch case statement with 1000 cases which is not plausible.
In my opinion, this cannot be achieved without recompilation at run time so languages like java, c, c++ will never let such a thing happen.
Hopefully, there is a way to do this. If so I am curious.
Note: This is not something that I will ever use, I asked question just because of my curiosity.
In modern functional languages, you can make a list of functions which take a list as an argument. This will arguably solve your problem, but it is also arguably cheating, as it is not quite the statically-typed implementation your question seems to imply. However, it is pretty much what dynamic languages such as Python, Ruby, or Perl do when using "manual" argument handling...
Anyway, the following is in Haskell: it supplies the nth function (from its first argument fs) a list of n copies of the second argument (x), and returns the result. Of course, you will need to put together the list of functions somehow, but unlike a switch statement this list will be reusable as a first-class argument.
selectApplyFunction :: [ [Int] -> a ] -> Int -> Int -> a
selectApplyFunction fs x n = (fs !! (n-1)) (replicate n x)
dummy1 [a] = 5 * a
dummy2 [a, b] = (a + 3) * b
dummy3 [a, b, c] = (a*b*c) / (a*b + b*c + c*a)
...
myFunctionList = [ dummy1, dummy2, dummy3, ... ]
-- (myfunction n) provides n copies of the number 42 to the n'th function
myFunction = selectApplyFunction myFunctionList 42
-- call the 666'th function with 666 copies of 42
result = myFunction 666
Of course, you will get an exception if n is greater than the number of functions, or if the function can't handle the list it is given. Note, too, that it is poor Haskell style -- mainly because of the way it abuses lists to (abusively) solve your problem...
No, you are incorrect. Most modern languages support some form of Reflection that will allow you to call a function by name and pass params to it.
You can create an array of functions in most of modern languages.
In pseudo code,
var dummy = new Array();
dummy[1] = function(int a);
dummy[2] = function(int a, int aa);
...
var result = dummy[whateveryoucall](1,2,3,...,whateveryoucall);
In functional languages you could do something like this, in strongly typed ones, like Haskell, the functions must have the same type, though:
funs = [reverse, tail, init] -- 3 functions of type [a]->[a]
run fn arg = (funs !! fn) $ args -- applies function at index fn to args
In object oriented languages, you can use function objects and reflection together to achieve exactly what you want. The problem of the variable number of arguments is solved by passing appropriate POJOs (recalling C stucts) to the function object.
interface Functor<A,B> {
public B compute(A input);
}
class SumInput {
private int x, y;
// getters and setters
}
class Sum implements Functor<SumInput, Integer> {
#Override
public Integer compute(SumInput input) {
return input.getX() + input.getY();
}
}
Now imagine you have a large number of these "functors". You gather them in a configuration file (maybe an XML file with metadata about each functor, usage scenarios, instructions, etc...) and return the list to the user.
The user picks one of them. By using reflection, you can see what is the required input and the expected output. The user fills in the input, and by using reflection you instantiate the functor class (newInstance()), call the compute() function and get the output.
When you add a new functor, you just have to change the list of the functors in the config file.
"Write a function lv: cfg -> (blabel -> ide set), which computes the live variables analysis on the given control flow graph."
Having cfg and blabel defined and ide set as a list of string, how can I create a function with that signature?
You're presumably familiar with the let syntax to define a function:
let f x = x + 1 in …
You can use this syntax anywhere, including in a function body. Now if you happen to use the inner function's name as the return value of the outer function, the outer function will be returning a function.
let outer_function x =
let inner_function y = x + y in
inner_function
The let syntax is in fact syntactic sugar for fun or function. In particular, if you define inner_function just to use the name once, you might as well use the fun notation and not bother giving the inner function a name.
let outer_function x =
fun y -> x + y
Furthermore, if all the outer function does when you pass it an argument is to build and return an inner function, then consider its behavior when you pass that function two arguments. First the outer function builds an inner function, using its first (and only) argument; then that inner function is applied to the second argument, so its body is executed. This is equivalent to having just one function that takes two arguments. This
observation is known as currying.
let outer_function x y = x + y
Note that the type of this function is int -> int -> int; this is the same type as int -> (int -> int) (the arrow type operator is right-associative).
Currying doesn't apply when the outer function does some work before building the inner function. In that case, the work is performed after receiving the first argument.
let outer_function x =
print_endline "outer";
fun y -> print_endline "inner"; x + y
So the structure of your code is likely to look like this:
let lv (c : cfg) : blabel -> ide set =
let c' = do_some_precomputation c in
fun (bl : blabel) -> (… : ide set)