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从泛型生成无形状记录

shapeless generics scala

1

jamborta · 技术社区 · 8 年前

我有这个标准 Shapeless

case class KeyValue(key: String, value: Int)
val kv = KeyValue("key", 1)

def processCaseClass(p: KeyValue) = {
     val gen = LabelledGeneric[KeyValue] 
     gen.to(p)
}

但是,我不想使用case类的名称,而是想使用泛型,然而,以这种方式重写它是行不通的:

def processCaseClass[KV <: Product, L <: HList](p: KV) = {
     val gen = LabelledGeneric.Aux[KV, L] 
     gen.to(p)
}

如果我改变 gen 对于隐式参数,它工作正常。

1 回复 | 直到 8 年前

1

2

HTNW 8 年前

在中使方法泛型 T 让人摸不着头脑是如果 processCaseClass 不知道是什么 KV ,它没有机会对其结构进行足够的分析,以做出 LabelledGeneric[KV] ,产生隐式分辨率错误。通过使其成为隐式参数,您可以转移生成 LabelledGeneric PK 是

你的第一个定义是

def processCaseClass[KV <: Product, L <: HList](p: KV): Any // Signature of pCC
// Function body doesn't affect sig., aside from return type, so what you write is what you get
// A sig. is the only interface between a method and the world. This method doesn't
// say there's a relationship between KV and L in its signature, so there doesn't
// need to be one.

// Instantiate KV = (Int, Int), L = String :: String :: HNil, p = (5, 5)
val x = processCaseClass[(Int, Int), String :: String :: HNil]((5, 5))
// Perfectly valid to the compiler... but what is this supposed to do?