I saw in TypeScript 2.2 the option for "overloading" function by defined Interface call signatures and I spent quite a time to understand how to use it.
so after working on it and "cracking" it I thought it will be worth to post it here.
the issue I started with was, for example:
interface Func1 {
(num1: number, num2: number): number;
(str1: number, str2: string): string;
}
function F1(num1: number, num2: number): number {
return num1 + num2;
}
const f1: Func1 = F1;
console.log(f1(1, 2));
but the compiler did not pass it because the Func1 can't accept the F1 function.
I want to make overloading and I don't know what to do.
see answer below.
after digging it I found we can do quite of overloading. Overloading in TypeScript is not different from any other JavaScript code. it must be implemented in one function and not in multiple ones, like in C++. you need to make your function accept all options.
for example:
interface Func1 {
(v1: number, v2: number): number;
(v1: string, v2: string): string;
}
const f1: Func1 = (v1, v2): any => {
return v1 + v2;
};
console.log(f1("1", "2"));
console.log(f1(1, 2));
the function f1 suppose to have types for all options of overloading interface call signature, if you set strong type it will block you because of the types are not castable. like in this example number not castable to string.
you also can create multiple functions which use of the interface call signature overloading and pass those to your f1 and if the type match it will pass. just make sure your function can handle all the interface call signature.
it important to check types in implemented function when using classes or if the function parameters are not simple, like above example, using typeof and control all options.
hope it helped.
Related
I can define a generic function:
package hello
func IsZero[T int64|float64](value T) bool {
return value == 0
}
Then if I try to alias that function in another package, it fails:
package world
import "hello"
var IsZero = hello.IsZero
The above doesn't compile with:
cannot use generic function hello.IsZero without instantiation
Instead this works:
var IsZero = hello.IsZero[int64]
Is it possible to do this, using some other syntax?
That's not an alias. And you already have your answer, actually. But if you want a formal reference, from the language specs, Instantiations:
A generic function that is is not called requires a type argument list for instantiation
So when you attempt to initialize a variable of function type, the function hello.IsZero is not called, and therefore requires instantiation with specific type parameters:
// not called, instantiated with int64
var IsZero = hello.IsZero[int64]
At this point the variable (let's give it a different name for clarity) zeroFunc has a concrete function type:
var zeroFunc = IsZero[int64]
fmt.Printf("type: %T\n", zeroFunc)
Prints:
type: func(int64) bool
This might or might not be what you want, as this effectively monomorphises the function.
If you just want to have a local symbol, with the same implementation (or a tweaked version thereof), declaring a "wrapper" function works. Just remember that the type parameters of your wrapper can only be as strict or stricter than the wrapped one's
E.g. Given
IsZero[T int64 | float64](v T)
your wrapper can not be
WrapIsZeroPermissive[T int64 | float64 | complex128](v T) bool {
return IsZero(v) // does not compile, T's type set is a superset
}
but can be
WrapIsZeroStricter[T int64](v T) bool {
return IsZero(v) // ok, T's type set is a subset
}
If the function is small, like in the question, it's probably easier to just vendor it:
package vendor
func thisIsJustCopy[T int64|float64](value T) bool {
return value == 0
}
but if the function is big, you can do it like this:
package world
import "hello"
func IsZero[T int64|float64](value T) bool {
return hello.IsZero(value)
}
I try to alias that function in another package
Aliases work for types only. Your code just tries to declare a variable.
Is it possible to do this, using some other syntax?
No.
I think an extension lambda requires that you pass in the correct argument, but i seems not to do so in the following example.
open class Base {
open fun f() = 1
}
class Derived : Base() {
override fun f() = 99
}
fun Base.g(): Int { return f()}
fun Base.h(xl: Base.() -> Int): Int { return xl()}
fun main() {
val b: Base = Derived() // Upcast
println(b.g())
println(b.h { f()}) // [1]
}
I understand that Base.h takes a function that takes a Base object as its parameter. But line [1] shows that it accepts f(), which is a function that takes no parameter. I was thinking hard about this and I prefixed it with this.f() and it still worked. Not convinced, I modified the code as follows:
open class Base {
open fun f() = 1
}
class Derived : Base() {
override fun f() = 99
}
fun Base.g(): Int { return f()}
fun Base.h(xl: (Base) -> Int): Int { return xl(Base())}
fun test(i:Int) = 1
fun main() {
val b: Base = Derived() // Upcast
println(b.g())
println(b.h { test(1) })
}
This code works. I've run it to verify. And as you can see, b.h() accepts test(), which takes an Int. And this is contrary to the fact that Base.h() takes a Base.
Could you explain this? Thank you for reading.
Note the curly brackets around the functions that are passed in! They change everything.
In the second code, b.h { test(1) } is not passing the function test to b.h. The syntax to pass test to b.h would be b.h(::test), and that does produce an error as you would expect.
b.h { test(1) } passes a function (a lambda expression) that takes a Base as parameter, ignores that parameter, calls test(1) and returns the result. You are basically passing a function that looks like this to b.h:
fun foo(p: Base) = test(1)
You might be wondering how Kotlin knows about Base when you did not write the word Base in the call at all. Well, it can just look at the declaration of b.h, and see that { test(1) } must take a parameter of Base.
The first code snippet is a bit different, because b.h accepts a Base.() -> Int in that case. Base.() -> Int represents a function whose receiver type is Base, that is, a function func that can be called like someBaseObject.func(). Compare this to a function func that takes a Base object as parameter, which can be called like func(someBaseObject).
Again, { f() } is not passing the function f. It is a lambda expression that does nothing but calls f. In this case though, f itself can be passed to b.h (b.h(Base::f)), because it is a function with a receiver type of Base! You can do someBaseObject.f(), can't you? Passing the lambda is similar to passing an extension function that is declared like this (you're just "wrapping" f in another function):
fun Base.foo() = f()
And since the receiver of the function is Base, you are able to access other functions that has Base as the receiver (such as f) in the lambda. You can also specify the receiver (which is this) explicitly.
I noticed that I get the same effect if I define this trivial function:
fun double ( i: Int ) = i*2
and if I define a variable and assign a lambda (with an identical body) to it:
var double = { i : Int -> i*2 }
I get the same result if I call double(a) with either declaration.
This leaves me confused. When is it needed, recommended, advantageous to define a variable as a lambda rather than define a function to it?
When is it needed, recommended, advantageous to define a variable as a lambda rather than define a function to it?
Whenever you have the choice of either, you should use a fun declaration. Even with a fun you can still get a first-class callable object from it by using a function reference.
On the JVM, a fun is significantly more lightweight, both in terms of RAM and invocation overhead. It compiles into a Java method, whereas a val compiles into an instance field + getter + a synthetic class that implements a functional interface + a singleton instance of that class that you must fetch, dereference, and invoke a method on it.
You should consider a function-typed val or var only when something is forcing you to do it. One example is that you can dynamically replace a var and effectively change the definition of the function. You may also receive function objects from the outside, or you may need to comply with an API that needs them.
In any case, if you ever use a function-typed property of a class, you'll know why you're doing it.
First, if I understand you right, your question is "Why are functions first-class citizens in Kotlin -- And when to use them as such?", right?
Kotlin functions are first-class, which means that they can be stored in variables and data structures, passed as arguments to and returned from other higher-order functions. You can operate with functions in any way that is possible for other non-function values. (see here)
As stated in the docs, one use case are higher-order functions. As a first step, I will leave the wikipedia link here: https://en.wikipedia.org/wiki/Higher-order_function
Basically, a higher-order function is a function that takes functions as parameters, or returns a function.
This means that a higher-order function has at least one parameter of a function type or returns a value of a function type.
Following a short example of a higher-order function that receives a parameter of function type (Int) -> Boolean:
fun foo(pred: (Int) -> Boolean) : String = if(pred(x)) "SUCCESS" else "FAIL"
This higher-order function can now be called with any (Int) -> Boolean function.
The docs also state ... [can be used] in any way that is possible for other non-function values.
This means that you can, for example, assign different functions to a variable, depending on your current context.
For example:
// This example is verbose on purpose ;)
var checker: (Int) -> Boolean
if (POSITIVE_CHECK) {
checker = { x -> x > 0 } // Either store this function ...
} else {
checker = { x -> x < 0 } // ... or this one ...
}
if (checker(someNumber)) { // ... and use whatever function is now stored in variable "checker" here
print("Check was fine")
}
(Code untested)
You can define variable and assign it lambda when you want change behaviour for some reason. For example, you have different formula for several cases.
val formula: (Int) -> Int = when(value) {
CONDITION1 -> { it*2 }
CONDITION2 -> { it*3 }
else -> { it }
}
val x: Int = TODO()
val result = formula(x)
If you simply need helper function, you should define it as fun.
If you pass a lambda as a parameter of a function it will be stored in a variable. The calling application might need to save that (e.g. event listener for later use). Therefore you need to be able to store it as a variable as well. As said in the answer however, you should do this only when needed!
For me, I would write the Lambda variable as followed:
var double: (Int) -> Int = { i -> //no need to specify parameter name in () but in {}
i*2
}
So that you can easily know that its type is (i: Int) -> Int, read as takes an integer and returns an integer.
Then you can pass it to somewhere say a function like:
fun doSomething(double: (Int) -> Int) {
double(i)
}
I'm just learning TypeScript from here , at the Function Overloads section, got an 'undefined' error, Like I said I'm very new on TypeScript, don't know what is going on and how to fix it, please help!
Here is the snippet:
function foo_overload(s: string): void;
function foo_overload(n: number, s: string): void;
function foo_overload(x: any, y?: any): void {
console.log(x);
console.log(y);
}
foo_overload("Jack")
foo_overload(50, "Zhao");
And when I compile and execute this, the output like this:
Jack
undefined
50
Zhao
What the "undefined" it is?
Your pattern matching works well until you hit function foo_overload(x: any, y?: any):.
In that function, while you declare y to be optional, you invoke y with console.log(y), so it returns undefined, as in the first function invocation, you did not offer a y value (there is nothing after Jack).
Hope it helps!
I always thought functions and methods were the same, until I was learning Swift through the "Swift Programming Language" eBook. I found out that I cannot use greet("John", "Tuesday") to call a function that I declared inside a class, as shown in the eBook in the screen shot below:
I received a error saying that "Missing argument label 'day:' in call" as per this screen shot:
Here is the code:-
import Foundation
import UIKit
class ViewController2: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
//var dailyStatement = greet("John", "Tuesday")
var dailyStatement = greet("John", day: "Tuesday")
println(dailyStatement)
}
func greet(name: String, day: String) -> String {
return "Hello \(name), today is \(day)."
}
}
After some research, I found this post: Difference between a method and a function, and it seems to me that the function that I declared inside a class is actually called a method. So, the syntax that I use to call the method is different compared to the syntax that I use to call a function.
I never realized this difference when I was programming in Objective-C.
What are the differences between functions and methods in Swift?
When do we use functions and when do we use methods in Swift?
After a few hours of reading and experimenting, here are the things that I found out:-
Functions in Swift
Functions are self-contained chunks of code that perform a specific
task. You give a function a name that identifies what it does, and
this name is used to “call” the function to perform its task when
needed.
Resource: Official Apple Documentation on Functions in Swift
Function Parameter Names
However, these parameter names are only used within the body of the
function itself, and cannot be used when calling the function. These
kinds of parameter names are known as local parameter names, because
they are only available for use within the function’s body.
It means that by default, all the parameters for Function are local parameters.
But, sometimes we want to indicate the purpose of each parameter. So, we can actually define an external parameter name for each parameter. Example Code:
func someFunction(externalParameterName localParameterName: Int) {
// function body goes here, and can use localParameterName
// to refer to the argument value for that parameter
}
Another way to make the external parameter name is using hash symbol (#) to shorten the name.
func someFunction(#localParameterName: Int) {
// function body goes here, and can use localParameterName
// to refer to the argument value for that parameter
}
To call the above functions with external parameter, you may use
someFunction(localParameterName:10)
Methods in Swift
Methods are functions that are associated with a particular type.
Classes, structures, and enumerations can all define instance methods,
which encapsulate specific tasks and functionality for working with an
instance of a given type.
Resource: Official Apple Documentation on Methods in Swift
However, the default behavior of local names and external names is
different for functions and methods.
Specifically, Swift gives the first parameter name in a method a local
parameter name by default, and gives the second and subsequent
parameter names both local and external parameter names by default.
Code below shows the differences for default and non-default parameters for method in Swift.
import Foundation
import UIKit
class ViewController2: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
//Default methods calling
var dailyStatement = greet("Rick", day: "Tuesday")
println(dailyStatement)
//First parameter is also an external parameter
var dailyStatement2 = greet2(name:"John", day: "Sunday")
println(dailyStatement2)
}
//Default: First Parameter is the local parameter, the rest are external parameters
func greet (name: String, day: String) -> String {
return "Hello \(name), today is \(day)."
}
//Use Hash symbol to make the First parameter as external parameter
func greet2 (#name: String, day: String) -> String {
return "Hello \(name), today is \(day)."
}
}
I might miss some important details. Hope someone can provide a better answer.
As you said yourself, methods are functions, but in a class. In objective-c you never realized this, because we were only coding in classes. Every function that we wrote was a method of a class (ViewController or some other class we created).
In Swift we have the ability to create functions that are not inside some class. The main reason for doing this is to write functions that are not tied to any class, and can be used wherever we need them. So if you have a function that is related to a class you write it inside the class and you can access is from every instance of the class:
class Square {
var length: Double
func area() -> Double {
return length * length
}
}
But if you need to access the function from everywhere, then you don't write it inside a class. For example:
func squared(number: Int) -> Int {
return number * number
}
About your syntax issues between functions and methods: You guessed it right, methods and functions are called a little bit differently. That is because in Objective-C we had long method names and we liked them because we could read what the methods were doing and what the parameters were for. So the first parameter in a method is in most cases described by the function name itself. And the other parameters shouldn't only be some numbers or strings or instances, they should be described as well, so Swift writes the name of the variable automatically. If you want to describe it by yourself you can do that as well:
class Something {
func desc(firstString string1: String, secondString string2:String) {...}
}
Well, #Ricky's answer says it pretty much. I was confused what exactly they are. So here is my thought:
Functions could be defined outside of classes or inside of classes/structs/enums, while Methods have to be defined inside of and part of classes/structs/enums.
We could define a Function outside of any Type's definition and could use it within Methods of any Type's definition.
Just my understanding and illustration here, hope this helps someone else or you may edit if you feel there is an improvement needed OR let me know if anything is wrong:
//This is a Function which prints a greeting message based on the category defined in an 'enum'
func greet(yourName name: String, category: GreetingsCategory) {
switch category {
case .Person:
print("Hello, " + name + " Today is Tuesday")
case .Vehicle:
print("Hello, " + name + " your Vehicle is a Car")
}
}
//This is an 'enum' for greetings categories
enum GreetingsCategory: String {
case Person
case Vehicle
}
//Type: Person
class Person {
//This is a method which acts only on Person type
func personGreeting() {
greet(yourName: "Santosh", category: .Person)
}
}
//Type: Vehicle
class Vehicle {
//This is a method which acts only on Vehicle type
func vehicleGreeting() {
greet(yourName: "Santosh", category: .Vehicle)
}
}
//Now making use of our Function defined above by calling methods of defferent types.
let aPerson = Person()
aPerson.personGreeting()
//prints : Hello, Santosh Today is Tuesday
let aVehicle = Vehicle()
aVehicle.vehicleGreeting()
//prints: Hello, Santosh your Vehicle is a Car
//We can also call the above function directly
greet(yourName: "Santosh", category: .Person)
Mainly the names are used interchangeably without people having a real intent of distinguishing them. But ultimately they do have a difference.
someFile.swift:
func someFunc{
//some code
}
class someClass{
func someMethod{
//some code
}
}
Note: someClass != someFile
someMethod works only on its associated type which is 'someClass'. However the same can't be said for someFunc. someFunc is only in the someClass.Swift because semantically it is better suited to be written in that file. It could have been written in any other class as long as it's marked with private
And obviously the method can access self. With functions, there is no self.. For more see: What's the difference between a method and a function?
Here is a simple answer on the difference between functions and methods:
Some folks use “function” and “method” interchangeably, but there’s a
small difference: both of them are reusable chunks of code, but
methods belong to classes, structs, and enums, whereas functions do
not.
So:
func thisIsAFunction() {
}
struct Person {
func thisIsAMethod() {
}
}
Because methods always belong to a data type, they have a concept of
self that functions do not.
source: https://www.hackingwithswift.com/example-code/language/whats-the-difference-between-a-function-and-a-method
Lots of great answers, but let me use Xcode to show something visually from the UIKit module:
That is a function because it's written at the global level. It's not a method. Methods are scoped to a class.
Screenshot to show that it's at the global level.
The following function is at the global level:
public func UIApplicationMain(_ argc: Int32, _ argv: UnsafeMutablePointer<UnsafeMutablePointer<Int8>>!,
_ principalClassName: String?, _ delegateClassName: String?) -> Int32
Icons for the different symbols. (Class, Method, Property, Protocol, Function, Extensions are all different symbols)
The function has an icon like 𝓯
The method has an icon of M
functional principle as a part of functional language
function is a first-class type (first-class citizen) in Swift. Higher order functions
assign to a variable
pass as an argument
return
Function
Function is a block of code that is created for executing some task. Function consists of name, optional parameters(name, type), optional return type, body.
func name(parameterName1: Int, parameterName2: String) -> Bool {
//statements
return true
}
Function type - function’s parameter type + return type[Java about]. For example it is used as function's parameter
//Function type for the sample above
(Int, String) -> Bool
Method
Method - is a function which is associated with a type - class, structure, enum [About]:
Instance method - method which belongs to instance
MyClass().foo()
Type method - method which belongs to type itself. class or static is used[About]
MyClass.foo()
Closure
As official doc says that Closure in Swift has three next forms:
global function(with name, without capturing) - is a function that is declared in a global scope(out of class scope). Usually it is defined as a first level of .swift file and does not have a big memory foot print
nested function(with name, with capturing enclosing function variables) - function inside other function
closure expression(without name, with capturing enclosing context)
Closure(closure expression) - anonymous function - is a block of code(functionality). Closure is a type of function without name. Closure is a function in terms of Functional programming. It can support capturing concept(if it doesn't capture it is lambda). It is similar to block in Objective-C.
[Closure vs Lambda]
They can be used for:
non-escaping closure - sync operations - click events, sort...
escaping closure - async operations - e.g.completion handler - it is a callback/notification which is called when task is done
class ClassA {
var variable = "Hello"
func fooA() {
print(variable)
let b = ClassB() //1
b.fooB(completionHandler: { [weak self] (input) -> String? in //2 pass closure //3 capture list or any calls from closure to outher scope
guard let self = self else { return nil }
self.variable = "World" //capturing self.variable
return String(input)
})
}
}
class ClassB {
var myCompletionHandler: ((Int) -> String?)? = nil //5
func fooB(completionHandler: #escaping (Int) -> String?) { //4
self.myCompletionHandler = completionHandler //6
let result = completionHandler(7)
}
func fooB2(completionHandler: #escaping (Int) -> String?) { //if you re-pass function to #escaping function you must mark it by #escaping too
self.fooB(completionHandler: completionHandler)
}
}
func testClosure() {
ClassA().fooA()
}
(Int) -> String? //Function type
//Closure Expression
{ [weak self] (input) -> String? in
//logic
}
[non-escaping vs escaping closure]
[#autoclosure]
[Init/customize stored property by closure]
[JVM Memory model]