为什么reduceLeft 的类型参数包含下界?

本文介绍了为什么reduceLeft 的类型参数包含下界?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

reduceLeft 在某些 Seq[A] 上的签名是

The signature of reduceLeft on some Seq[A] is

def reduceLeft [B >: A] (f: (B, A) => B): B

A 的类型是已知的，但是下界 >: 告诉我们 B 可以是 的任何超类型A.

The type of A is known, but the lower bound >: tells us that B can be any supertype of A.

为什么会这样?为什么不

Why is it like this? Why not

def reduceLeft (f: (A, A) => A): A

我们已经知道序列的头部是 A 类型，所以我想不出 B 怎么可能不等于 A.你能提供一个例子，其中 B 是某种超类型吗?

We already know that the head of the sequence is type A and so I can't think of how B could be anything other than equal to A. Can you provide an example where B is some super-type?

推荐答案

假设你的类 B 有一个方法 combine(other:B): B.现在您在 A 列表上调用 reduceLeft((b,a) => bbine(a)).由于combine的返回类型是B，reduceLeft的类型参数需要是B.

Let's say your class B has a method combine(other:B): B. Now you call reduceLeft((b,a) => bbine(a)) on a list of As. Since the return type of combine is B the type parameter to reduceLeft needs to be B.