如何使用新的computeIfAbsent函数？ [英] How do I use the new computeIfAbsent function?

问题描述

我非常想使用 Map.computeIfAbsent ，但自从本科以来的兰布斯已经太久了。

I very much want to use Map.computeIfAbsent but it has been too long since lambdas in undergrad.

直接从文档中直接显示：它给出了一个旧的方式来做一个例子：

Almost directly from the docs: it gives an example of the old way to do things:

Map<String, Boolean> whoLetDogsOut = new ConcurrentHashMap<>();
String key = "snoop";
if (whoLetDogsOut.get(key) == null) {
  Boolean isLetOut = tryToLetOut(key);
  if (isLetOut != null)
    map.putIfAbsent(key, isLetOut);
}

新的方式：

map.computeIfAbsent(key, k -> new Value(f(k)));

但是在他们的例子中，我觉得我不是得到它。如何转换代码以使用新的lambda表达方式？

But in their example, I think I'm not quite "getting it." How would I transform the code to use the new lambda way of expressing this?

推荐答案

假设您有以下代码： p>

Suppose you have the following code:

import java.util.Map;
import java.util.concurrent.ConcurrentHashMap;

public class Test {
    public static void main(String[] s) {
        Map<String, Boolean> whoLetDogsOut = new ConcurrentHashMap<>();
        whoLetDogsOut.computeIfAbsent("snoop", k -> f(k));
        whoLetDogsOut.computeIfAbsent("snoop", k -> f(k));
    }
    static boolean f(String s) {
        System.out.println("creating a value for \""+s+'"');
        return s.isEmpty();
    }
}

然后你会看到消息在第二次调用 computeIfAbsent 时，正好创建一个值为snoop的值，该键已经有一个值。 lambda表达式 k中的 k f（k）只是映射将传递给lambda的密钥的一个placeolder（参数），用于计算该值。所以在这个例子中，密钥被传递给函数调用。

Then you will see the message creating a value for "snoop" exactly once as on the second invocation of computeIfAbsent there is already a value for that key. The k in the lambda expression k -> f(k) is just a placeolder (parameter) for the key which the map will pass to your lambda for computing the value. So in the example the key is passed to the function invocation.

或者你可以写： whoLetDogsOut.computeIfAbsent（snoop，k - > ; k.isEmpty（））; 在没有辅助方法的情况下实现相同的结果（但是您将看不到调试输出）。甚至更简单，因为它是对现有方法的简单委派，您可以写入： whoLetDogsOut.computeIfAbsent（snoop，String :: isEmpty）; 此代理不需要要写入的任何参数。

Alternatively you could write: whoLetDogsOut.computeIfAbsent("snoop", k -> k.isEmpty()); to achieve the same result without a helper method (but you won’t see the debugging output then). And even simpler, as it is a simple delegation to an existing method you could write: whoLetDogsOut.computeIfAbsent("snoop", String::isEmpty); This delegation does not need any parameters to be written.

为了更接近你的问题的例子，你可以把它写成 whoLetDogsOut.computeIfAbsent（snoop键 - > tryToLetOut（key））; （无论您是否命名参数 k 或键）。或者写为 whoLetDogsOut.computeIfAbsent（snoop，MyClass :: tryToLetOut）; 如果 tryToLetOut 是 static 或 whoLetDogsOut.computeIfAbsent（snoop，this :: tryToLetOut）; if tryToLetOut 是一个实例方法。

To be closer to the example in your question, you could write it as whoLetDogsOut.computeIfAbsent("snoop", key -> tryToLetOut(key)); (it doesn’t matter whether you name the parameter k or key). Or write it as whoLetDogsOut.computeIfAbsent("snoop", MyClass::tryToLetOut); if tryToLetOut is static or whoLetDogsOut.computeIfAbsent("snoop", this::tryToLetOut); if tryToLetOut is an instance method.

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