How to use infix notation in Android properly (Kotlin language)? - infix-notation

I have read Kotlin docs as well as the wikipedia links (https://en.wikipedia.org/wiki/Infix_notation#:~:text=Infix%20notation%20is%20the%20notation,plus%20sign%20in%202%20%2B%202), but unfortunately I am still unable to make use of this notation in my code.
Could someone please let me know, how can I use it?

Before jumping on to the code, let’s look at the rules of it’s usage (From the docs: https://kotlinlang.org/docs/functions.html#infix-notation):
Infix notation must be used with member functions or extension functions
They must have a single parameter
The parameter must not accept variable number of arguments and must have no default value.
Keeping these pointers in mind, you can achieve the following:
fun Int.add(x: Int) = this.plus(x) //This is a simple Extension function without infix notation
infix fun Int.subtract(x: Int) = this.minus(x) //Added infix notation here
fun main() {
val sum = 10.add(20)
println(sum) //prints 30
val sub = 100 subtract 30 //Notice that there is no dot(.) and parenthesis
println(sub) //prints 70
}
This is how we can use infix notations and get rid of the dots(.) and parenthesis and they will work the same.
This increases code readability.

Related

What is the point of op_Quotation if it cannot be used?

According the F# specification for operator overloading
<# #> op_Quotation
<## ##> op_QuotationUntyped
is given as with many other operators. Unless I'm missing something I don't believe that I can use this for custom types, so why is it listed?
I think you are right that there is no way of actually using those as custom operators. I suspect those are treated as operators in case this was useful, at some point in the future of the language, for some clever new feature.
The documentation really merely explains how the names of the operators get encoded. For non-special operator names, F# encodes those in a systematic way. For the ones listed in the page, it has a special nicer name. Consider this type:
type X() =
static member (<^><>) (a:int,b:int) = a + b
static member (<# #>) (a:int,b:int) = a + b
If you look at the names of those members:
[ for m in typeof<X>.GetMembers() -> m.Name ]
You see that the first operator got compiled as op_LessHatGreaterLessGreater, while the second one as op_Quotation. So this is where the name memntioned in the table comes in - it is probably good this is documented somewhere, but I think you're right, that this is not particularly useful!

Why is "do" allowed inside a function?

I noticed that the following code compiles and works in VS 2013:
let f() =
do Console.WriteLine(41)
42
But when looking at the F# 3.0 specification I can't find any mention of do being used this way. As far as I can tell, do can have the following uses:
As a part of loop (e.g. while expr do expr done), that's not the case here.
Inside computation expressions, e.g.:
seq {
for i in 1..2 do
do Console.WriteLine(i)
yield i * 2
}
That's not the case here either, f doesn't contain any computation expressions.
Though what confuses me here is that according to the specification, do should be followed by in. That in should be optional due to lightweight syntax, but adding it here causes a compile error (“Unexpected token 'in' or incomplete expression”).
Statement inside a module or class. This is also not the case here, the do is inside a function, not inside a module or a class.
I also noticed that with #light "off", the code doesn't compile (“Unexpected keyword 'do' in binding”), but I didn't find anything that would explain this in the section on lightweight syntax either.
Based on all this, I would assume that using do inside a function this way should not compile, but it does. Did I miss something in the specification? Or is this actually a bug in the compiler or in the specification?
From the documentation on MSDN:
A do binding is used to execute code without defining a function or value.
Even though the spec doesn't contain a comprehensive list of the places it is allowed, it is merely an expression asserted to be of type unit. Some examples:
if ((do ()); true) then ()
let x: unit = do ()
It is generally omitted. Each of the preceding examples are valid without do. Therefore, do serves only to assert that an expression is of type unit.
Going through the F# 3.0 specification expression syntax has do expr as a choice of class-function-or-value-defn (types) [Ch 8, A.2.5] and module-function-or-value-defn (modules) [Ch 10, A.2.1.1].
I don't actually see in the spec where function-defn can have more than one expression, as long all but the last one evaluate to unit -- or that all but the last expression is ignored in determining the functions return value.
So, it seems this is an oversight in the documentation.

F# operator overloading strange behavoir

Let's say that for some strange reason I want to have this function:
let (~-) (str:string) = 42
So I can do something like this and get 42 as result:
-"test"
val it : int = 42
Which is fine. But now when I do:
let a = 100
-a
I get:
error FS0001: This expression was expected to have type
string
but here has type
int
Any idea why is this happening?
When you define operators using let, the new definition hides all previous definition of the operator. So in your example, you are hiding the default implementation of the unary minus (which works for numbers) and replacing it with a new operator that only works on strings.
It is not easy to re-define overloaded operators on built-in types. If you need that, it is probably better idea to avoid using operators (just use a function). However, if you want to provide an overloaded operator for a custom type, you can do this by adding operator as a static member:
type MinusString(s:string) =
member x.Value = s
/// Provide unary minus for MinusString values
static member (~-) (ms:MinusString) =
MinusString("-" + ms.Value)
-(MinusString "hi") // Returns "-hi"
If you really want to redefine built-in operator like unary minus and make it work on string, then there is actually a way to do this using a trick described in earlier SO answers. However, I would only use this if you have a good reason.
Simply, you overwrote the minus operator with one that takes a string and returns an int, then tried to apply it to an int, which it can't do anymore.

Questions about the definition of lazy

On line 5633 in prim-types.fs (v1.9.7.8) there is the following type abbreviation:
type 'T ``lazy`` = Lazy<'T>
I have a few questions about it.
What do the double backticks mean?
Is this definition equivalent to type lazy<'T> = Lazy<'T>? (If not, how is it different?)
The double back ticks are a way of allowing an F# keyword to be used as an identifier. Another example would be
let ``let`` = 42
To answer the second half of your question, generic types in F# can be specified using either the O'Caml-style syntax where the generic parameter precedes the type (e.g 'a list, int array, etc.), or the .NET-style with angle brackets (e.g. list<'a>, array<int>, etc.), so the two definitions are indeed basically equivalent (except that your version as written is syntactically invalid because lazy is a keyword). For multi-parameter generic types, the O'Caml style is deprecated and will generate a warning (e.g. let (m:(int,string) Map) = Map.empty should be rewritten as let (m:Map<int,string>) = Map.empty).

Point-free style with objects/records in F#

I'm getting stymied by the way "dot notation" works with objects and records when trying to program in a point-free functional style (which I think is a great, concise way to use a functional language that curries by default).
Is there an operator or function I'm missing that lets me do something like:
(.) object method instead of object.method?
(From what I was reading about the new ? operator, I think it works like this. Except it requires definition and gets into the whole dynamic binding thing, which I don't think I need.)
In other words, can I apply a method to its object as an argument like I would apply a normal function to its argument?
Short answer: no.
Longer answer: you can of course create let-bound functions in a module that call a method on a given type... For example in the code
let l = [1;2;3]
let h1 = l.Head
let h2 = List.hd l
there is a sense in which "List.hd" is the version of what you want for ".Head on a list". Or locally, you can always do e.g.
let AnotherWay = (fun (l:list<_>) -> l.Head)
let h3 = AnotherWay l
But there is nothing general, since there is no good way to 'name' an arbitrary instance method on a given type; 'AnotherWay' shows a way to "make a function out of the 'Head' property on a 'list<_>' object", but you need such boilerplate for every instance method you want to treat as a first-class function value.
I have suggested creating a language construct to generalize this:
With regards to language design
suggestions, what if
SomeType..Foo optArgs // note *two* dots
meant
fun (x : SomeType) -> x.Foo optArgs
?
In which case you could write
list<_>..Head
as a way to 'functionize' this instance property, but if we ever do anything in that arena in F#, it would be post-VS2010.
If I understand your question correctly, the answer is: no you can't. Dot (.) is not an operator in F#, it is built into the language, so can't be used as function.

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