为什么 Scala 定义了一个“+="?Short 和 Byte 类型的运算符? [英] Why does Scala define a "+=" operator for Short and Byte types?

问题描述

给定以下 Scala 代码:

Given the following scala code:

var short: Short = 0
short += 1        // error: type mismatch
short += short    // error: type mismatch
short += 1.toByte // error: type mismatch

我不质疑底层类型 - 很明显Short + value == Int".

I don't questioning the underlying typing - it's clear that "Short + value == Int".

我的问题是:
1.有没有什么方法可以使用运算符?
2. 如果没有，那么为什么运营商可以在 Short & 上使用?字节?

My questions are:
1. Is there any way at all that the operator can be used?
2. If not, then why is the operator available for use on Short & Byte?

[并且通过扩展 *=、|= &= 等]

[And by extension *=, |= &=, etc.]

推荐答案

问题好像是Short class上的+(Short)"定义为:

The problem seems to be that "+(Short)" on Short class is defined as:

def +(x: Short): Int

所以它总是返回一个 Int.

So it always returns an Int.

鉴于此，您最终无法使用 += "operator"，因为 + 操作计算为一个 Int，该 Int(显然)不能分配给脱糖版本中的short"var:

Given this you end up not being able to use the += "operator" because the + operation evaluates to an Int which (obviously) can not be assigned to the "short" var in the desugared version:

short = short + short

至于你的第二个问题，它是可用的"，因为当 Scala 编译器找到如下表达式时:

As for your second question, it is "available" because when the scala compiler finds expressions like:

x K= y

如果 x 是一个 var 并且 K 是任何符号运算符，并且在 x 中有 K 方法，那么编译器将其翻译或脱糖"为:

And if x is a var and K is any symbolic operator and there is K method in x then the compiler translates or "desugar" it to:

x = x K y

然后尝试继续编译.

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