时间:2019年8月4日14:17:06
问题描述:
看下边的小例子:
data class Man(val name: String, val age: Int, val type: Int)
fun main(args: Array<String>) {
val list = mutableListOf<Man>()
list.add(Man("wzc", 31,2))
list.add(Man("wzj", 32,1))
list.add(Man("wcx", 3,1))
list.add(Man("wcg", 7,1))
println("before sort")
for (man in list) {
println(man)
}
list.sortedWith(Comparator {lh, rh ->
if (lh.type.compareTo(rh.type) == 0) {
lh.age.compareTo(rh.age)
} else {
lh.type.compareTo(rh.type)
}
})
println("after sort")
}
/*
打印结果:
before sort
Man(name=wzc, age=31, type=2)
Man(name=wzj, age=32, type=1)
Man(name=wcx, age=3, type=1)
Man(name=wcg, age=7, type=1)
after sort
*/
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31
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38
可以看到排序前后,打出的内容没有丝毫变化。
解决方法:
看一下 sortedWith 的代码:
/**
* Returns a list of all elements sorted according to the specified [comparator].
*
* The sort is _stable_. It means that equal elements preserve their order relative to each other after sorting.
public fun <T> Iterable<T>.sortedWith(comparator: Comparator<in T>): List<T> {
if (this is Collection) {
if (size <= 1) return this.toList()
@Suppress("UNCHECKED_CAST")
return (toTypedArray<Any?>() as Array<T>).apply { sortWith(comparator) }.asList()
return toMutableList().apply { sortWith(comparator) }
可以排序后的结果是在返回值里面。
修改代码: