关于scala：协变型FParam出现在价值猜测的Seq [FParam]类型的逆变位置

Covariant type FParam occurs in contravariant position in type Seq[FParam] of value guesses

Consider this simple example:

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trait Optimizer[+FParam, FRes] {
def optimize(
fn: (FParam) => FRes,
guesses: Seq[FParam] // <--- error
)
}

它不会编译，因为

Covariant type FParam occurs in contravariant position in type Seq[FParam] of value guesses.

但SEQ是以trait Seq[+A]定义的，所以这一差异的根源是什么？(问题1)

Conversely，consider this simple example with -FParam

ZZU1BLCK1/

同样的矛盾：在Function1[-T1, R]中，第一类参数的变化很明显，所以为什么是FParam？(问题2)

我可以通过惊吓变异来确定这个问题，如描述在下一类边界，但为什么需要这样做是不可理解的。

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trait Optimizer[+FParam, FRes] {
type U <: FParam

def optimize(
fn: (FParam) => FRes,
guesses: Seq[U]
)
}

问题1:+FParam是指协变型FParam，因超型而异。对于guesses，预计cotravariant type，对于Seq。因此，您可以通过明确说明EDOCX1的supertype或FPParam来实现，例如：

1	def optimize[B >: FParam](fn: (B) => FRes, guesses: Seq[B]) // B is the super type of FParam

或者像SeqViewLike。

那么，为什么guesses预期cotravariant type为Seq呢？例如:

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trait Animal

case class Dog() extends Animal

case class Cat() extends Animal
val o = new Optimizer2[Dog, Any]
val f: Dog => Any = (d: Dog) => ...
o.optimize(f, List(Dog(), Cat())) // if we don't bind B in `guesses` for supertype of `FParam`, it will fail in compile time, since the `guesses` it's expecting a `Dog` type, not the `Animal` type. when we bind it to the supertype of `Dog`, the compiler will infer it to `Animal` type, so `cotravariant type` for `Animal`, the `Cat` type is also matched.

问题2:-FParam是指同变异的FParam型，它从超型FParam到亚型不等，对于fn: Function1[-T1, +R]//(FParam) => FRes型，它期望协变型。所以你可以通过反问题1类型的状态来实现，比如：

1	def optimize[B <: FParam](fn: (B) => FRes, guesses: Seq[B]) // B is the sub type of FParam

问题是fparam没有直接使用。它在EDOCX1的论证中(0)，因此它的方差是翻转的。举例来说，让我们看看optimize的类型(参见val optim)：

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trait Optimizer[+FParam, FRes] {
type U <: FParam

def optimize(
fn: (FParam) => FRes,
guesses: Seq[U]
)

val optim: Function2[
Function1[
FParam,
FRes
],
Seq[U],
Unit
] = optimize
}