How does my teacher know how many parameters to put in her closures and how does she know what type Swift will infer?

How does my teacher know how many parameters to put in her closures and how does she know what type Swift will infer?

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How does my teacher know how many parameters to put in her closures and how does she know what type Swift will infer?
Tag : swift , By : Derek
Date : January 11 2021, 05:14 PM

To fix this issue As rmaddy stated, check the documentation as a first step. If you don't feel like Googling it, you can press the command key and click on the the thing you're looking for and it'll take you to the definition.
@IBAction func logInPressed(_ sender: AnyObject) {
    //TODO: Log in the user
        let emailText = emailTextField.text,
        let passwordText = passwordTextField.text else { return }

    Auth.auth().signIn(withEmail: emailText, password: passwordText) { (user, error) in
        if error != nil {
            print("Log in unsuccessful:", error)
        else {
            print("Log in successful")

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Is the C# compiler unable to infer method type parameters by expected return type?

Tag : chash , By : user165781
Date : March 29 2020, 07:55 AM
Hope that helps Check C# Language Specification §7.5.2, the declaring type of a variable is not an attestation for type inference, and obviously it shouldn't be. Consider the following code:
Base b = Test<Derived>();
Derived d = Test<Derived>();

Why doesn't Scala fully infer type parameters when type parameters are nested?

Tag : scala , By : Ian Badcoe
Date : March 29 2020, 07:55 AM
To fix the issue you can do Kipton got close with his higher-kinded solution. Unfortunately he tripped over what appears to be a bug in Scala < 2.9.1.RC1. The following works as expected with 2.9.1.RC1 and trunk,
Welcome to Scala version 2.9.1.RC1 (Java HotSpot(TM) Server VM, Java 1.7.0).
Type in expressions to have them evaluated.
Type :help for more information.

scala> abstract class A
defined class A

scala> abstract class B[T <: A]
defined class B

scala> class ConcreteA extends A
defined class ConcreteA

scala> class ConcreteB[T <: A] extends B[T]
defined class ConcreteB

scala> class Example[T <: A, U[X <: A] <: B[X]](resolver: U[T])
defined class Example

scala> new Example(new ConcreteB[ConcreteA])
res0: Example[ConcreteA,ConcreteB] = Example@ec48e7

parameters with optional closures in swift

Tag : swift , By : Ruchi
Date : March 29 2020, 07:55 AM
hope this fix your issue Firstly, to use closures as an argument for a function, you should declare them like so:
func myFunc(closure: (Int) -> Void) {
    // Now I can call closure like so:
    let myInt = 10
func myFunc(closure: ((Int) -> Void)?) {
    // Now when calling the closure you need to make sure it's not nil.
    // For example:
func myFunc(closure: ((Int) -> Void)? = nil) {
    // Still need to make sure it's not nil.
    if let c = closure {
func myFunc(closure: ((value: Int) -> Void)) {
    closure(value: 10)
typealias MyClosureType = () -> Void

func myFunc(closure: MyClosureType) {

Infer teacher-student relationships via courses?

Tag : development , By : iyogee
Date : March 29 2020, 07:55 AM
wish help you to fix your issue
Is determining that Hechinger is a student of Cadbury something that I should be able to do with OWL reasoning or inference in Protege?
@prefix :      <urn:ex:#> .
@prefix ex:    <urn:ex:#> .
@prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix owl:   <http://www.w3.org/2002/07/owl#> .
@prefix xsd:   <http://www.w3.org/2001/XMLSchema#> .
@prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#> .

ex:isStudentOf  a               owl:ObjectProperty ;
        owl:propertyChainAxiom  ( ex:studies _:b0 ) .

ex:studies  a   owl:ObjectProperty .

_:b0    owl:inverseOf  ex:teaches .

ex:EL101  a     owl:Thing , owl:NamedIndividual .

<urn:ex:>  a    owl:Ontology .

ex:Hechinger  a     owl:Thing , owl:NamedIndividual ;
        ex:studies  ex:EL101 .

ex:teaches  a   owl:ObjectProperty .

ex:Cadbury  a       owl:Thing , owl:NamedIndividual ;
        ex:teaches  ex:EL101 .

Swift closures: Is it possible to specify the return type, without giving up implicit parameters?

Tag : swift , By : Thomas Gueze
Date : March 29 2020, 07:55 AM
To fix this issue Looking into Swift Grammar, specifically the syntax for closure-expression, we can see that closure-parameter-clause is the only required part of closure-signature.
Therefore the answer is no. If you want to specify function-result you have to also specify the parameters.
let fib: Int = ...
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