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Type inference seems to fail vavr's Try works on jOOQ's fetchOne() function


Type inference seems to fail vavr's Try works on jOOQ's fetchOne() function

By : user2185633
Date : November 22 2020, 04:01 AM
Hope that helps Given the code you've provided so far, and assuming there's no typo, this is probably caused because of your raw type reference to ResultQuery. Use ResultQuery or ResultQuery> instead.
Never use raw types, unless you really have to. And you probably don't.
code :


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Why does method type inference fail to infer a type parameter?

Why does method type inference fail to infer a type parameter?


By : Carsten Blaser
Date : March 29 2020, 07:55 AM
wish helps you Your problem is caused by the fact that constraints are not considered part of the signature and are never used to make deductions during type inference. You are expecting the inference to go:
TEnumerable is determined by taking the type of the first argument. TElement is determined by taking the IList implementation information from TElement TProperty is determined by the type of the body of the lambda
code :
void AssertPropertyValues<TElement, TProperty>(
  IList<TElement> sequence, 
  Func<TElement, TProperty> projection, 
  params TProperty[] expectedValues)
Why Scala fail to recognize Null type as a subtype of T in generic function type parameter inference?

Why Scala fail to recognize Null type as a subtype of T in generic function type parameter inference?


By : Wafflez
Date : March 29 2020, 07:55 AM
Any of those help Why would Null be a subtype of a generic R? It's not the compiler being off the mark, it's the underlying assumption that Null <: R is always true. Here are a few fun examples, and they have to do with primitives.
Some(5).orNull will yield error: Cannot prove that Null <:< Int
code :
implicitly[Null <:< AnyRef] // will compile
implicitly[Null <:< AnyVal] // blows up, primitives are not included.
Why does type inference fail here?

Why does type inference fail here?


By : PTZ-M
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further So, I made this relatively simple code, and neither me nor IntelliJ IDEA see anything wrong with it, but javac keels over on the marked line, complaining it can't infer the types: , javac doesn't like that DefaultEdge is a raw type.
code :
    return createEdges(nodes, DefaultEdge<N>::new);
Why does type inference fail for a polymorphic function applied to different inputs withing the same function

Why does type inference fail for a polymorphic function applied to different inputs withing the same function


By : Michael Nuccio
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You've correctly identified that you need universal quantification. In fact, you already have universal quantification – your signature, like any polymorphic signature, is basically shorthand for
code :
{-# LANGUAGE ExplicitForall, UnicodeSyntax #-}
calculate :: ∀ a . Num a => (a -> a -> a) -> Val -> Val -> Val
calculate :: (Int -> Int -> Int) -> Val -> Val -> Val
{-# LANGUAGE Rank2Types #-}
calculate :: (∀ a . Num a => a -> a -> a) -> Val -> Val -> Val
data NumDict a = NumDict {
        addition :: a -> a -> a
      , subtraction :: a -> a -> a
      , multiplication :: a -> a -> a
      , abs :: a -> a
      ...
      }

calculate' :: (∀ a . NumDict a -> a -> a -> a) -> Val -> Val -> Val
calculate' f (I i1) (I i2) = I (f ndict i1 i2)
 where ndict = NumDict ((+) :: Integer -> Integer -> Integer)
                       ((-) :: Integer -> Integer -> Integer)
                       ...
Type inference only works for extension function

Type inference only works for extension function


By : Suraj Gundi
Date : September 28 2020, 12:00 PM
I wish this helpful for you The extension function returns the result of the Foo type parameter. So the result type can be inferred from the receiver type.
And the member function result type has nothing in common with Foo type parameter except the name, which means nothing for a compiler. You can see that T in method and T in class are different types by writing and compiling the following code:
code :
Foo<BarImpl>().get<BarImpl2>()
class Foo<T : Bar>(private val clazz: KClass<T>) {
    fun get(): T {
        return SomeMap(this).get(clazz)
    }

    companion object {
        inline operator fun <reified T : Bar> invoke() = Foo(T::class)
    }
}
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