如何仅通过一次地图查找来增加给定键的值? [英] How to increment value of a given key with only one map lookup?

问题描述

说我有一张地图:

 var inventory = mutableMapOf("apples" to 1, "oranges" to 2)

我想将苹果的数量增加一个.

and I want to increment the number of apples by one.

但是这不起作用:

inventory["apples"]!!++ // Error:(9, 4) Variable expected

这都不是

var apples = inventory["apples"]!!
++apples
println(inventory["apples"]) // prints "1" => the value was not incremented.

这令人惊讶，因为在Kotlin中，将数字装箱在实例中时将其装箱通用对象.我不希望正在制作副本.

Which is surprising, since in Kotlin, numbers are boxed when they are stored in instances of generic objects. I would not expect that a copy is being made.

似乎唯一的方法是执行以下操作:

It seems that the only way is to do something like:

var apples = inventory["apples"]!!
++apples
inventory["apples"] = apples
println(inventory["apples"]) 

两者都在地图中使用两次查找，而且非常丑陋且冗长.

Which both uses two lookups in the map and is extremely ugly and lengthy.

有没有一种方法可以仅使用一次查找来增加给定键的值?

Is there a way to increment a value of a given key using only one lookup?

还可以有人解释为什么前两种方法不起作用吗?

Also, could someone explain why the first two methods do not work?

推荐答案

仅使用一个get而不使用put，就无法增加该值.之后，您总是需要执行put，因为Int是不可变的. inc()方法(++是该方法的快捷方式)返回需要存储在映射中的新值.它不会更改存储的值.

There is no way to increment the value with only one get and without a put. You always need to do a put afterwards, because Int is immutable. The inc() method (++ is a shortcut for that) returns a new value that need to be stored in the map. It does not change the stored value.

您可以使用

inventory.computeIfPresent("apples") { _, v -> v + 1 }

要将其写在一行中，但是如果您查看实现，它还将执行get()，然后计算新值，然后执行put().因此，您可以节省必须编写的代码行，但不会节省执行时间上的CPU周期.

To write it in one line, but if you look into the implementation it will also do a get() then compute the new value and do a put() afterwards. So you can save a line of code you have to write, but you will not save CPU cycles on execution time.

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