Kotlin中arrayListOf函数参数前星号(*)语法解析求助

Hey there! That asterisk you're scratching your head over is Kotlin's spread operator—a nifty feature for bridging arrays and variable-length function arguments. Let's break down exactly what it does in your code:

What the Spread Operator Does #

First, take a look at the arrayListOf function: its signature uses vararg elements: T, which means it accepts any number of individual arguments of type T.

When you have an existing array (like your list2: Array<Int?>), the * operator "spreads" the array's elements into separate, individual arguments. So:

arrayListOf(*list2)

is exactly equivalent to passing every element in list2 as a standalone parameter to arrayListOf. For example, if list2 held values [0, 1, null, 3, ...], this line would be the same as writing:

arrayListOf(0, 1, null, 3, ...) // all 8 elements from list2

What Happens Without the Spread Operator? #

If you skipped the * and wrote arrayListOf(list2) instead, you'd end up with an ArrayList<Array<Int?>>—a list that contains one single element (the entire list2 array), not a list of the array's individual elements. That's definitely not what you want here!

Putting It All Together in Your Code #

Your full line of code:

arrayListOf(*list2) + arrayListOf(*list1)

does two key things:

1. Uses the spread operator to convert each array (list2 and list1) into an ArrayList<Int?> containing their respective elements.
2. Uses Kotlin's built-in + operator for collections to merge these two array lists into one new ArrayList<Int?>, with all elements from list2 first, followed by all elements from list1.

Quick Additional Notes #

• The spread operator only works with Array types (like Array<Int?>), not with Kotlin's standard List types. If you had a List, you'd first convert it to an array with toTypedArray() (e.g., arrayListOf(*myList.toTypedArray())).
• Since your arrays are created with arrayOfNulls<Int>, the resulting array lists will hold nullable Int? values—this matches the original arrays' ability to contain null.

内容的提问来源于stack exchange，提问作者Teddy Consultant