是什么?:在科特林做什么? (猫王操作员) [英] What does ?: do in Kotlin? (Elvis Operator)

问题描述

在这种情况下，我不知道?:的作用

I can't figure out what ?: does in for example this case

val list = mutableList ?: mutableListOf() 

以及为什么可以对此进行修改

and why can it be modified to this

val list = if (mutableList != null) mutableList else mutableListOf()

推荐答案

TL; DR:如果生成的对象引用[第一个操作数]不是null，则将其返回.否则，将返回第二个操作数的值(可能为null)

TL;DR: If the resulting object reference [first operand] is not null, it is returned. Otherwise the value of the second operand (which may be null) is returned

Elvis运算符是许多编程语言的一部分，例如Kotlin以及Groovy或C#. 我发现 Wikipedia 定义非常准确:

The Elvis operator is part of many programming languages, e.g. Kotlin but also Groovy or C#. I find the Wikipedia definition pretty accurate:

在某些计算机编程语言中， Elvis运算符 ?:是 binary运算符，如果该操作数为true，则返回其第一个操作数，否则求值并返回其第二个操作数.它是这些语言(以及许多其他语言)中的三元条件运算符的变体 :Elvis运算符是三元运算符，省略了第二个操作数

In certain computer programming languages, the Elvis operator ?: is a binary operator that returns its first operand if that operand is true, and otherwise evaluates and returns its second operand. It is a variant of the ternary conditional operator, ? :, found in those languages (and many others): the Elvis operator is the ternary operator with its second operand omitted.

对于Kotlin，以下情况尤其如此:

The following is especially true for Kotlin:

某些计算机编程语言对此运算符具有不同的语义. 代替第一个操作数必须产生布尔值，它必须导致对象引用.如果生成的对象引用不是null，则将其返回.否则，将返回第二个操作数的值(可能为null).

Some computer programming languages have different semantics for this operator. Instead of the first operand having to result in a boolean, it must result in an object reference. If the resulting object reference is not null, it is returned. Otherwise the value of the second operand (which may be null) is returned.

一个例子:

x ?: y // yields `x` if `x` is not null, `y` otherwise.

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