Java 8 lambdas，Function.identity（）或t-> t

问题描述

我对 Function.identity（）方法的使用有疑问。

I have a question regarding the usage of the Function.identity() method.

想象一下以下内容代码：

Imagine the following code:

Arrays.asList("a", "b", "c")
          .stream()
          .map(Function.identity()) // <- This,
          .map(str -> str)          // <- is the same as this.
          .collect(Collectors.toMap(
                       Function.identity(), // <-- And this,
                       str -> str));        // <-- is the same as this.

你有什么理由要使用 Function.identity（）而不是 str-> str （反之亦然）。我认为第二种选择更具可读性（当然是品味的问题）。但是，有没有真正的理由为什么应该首选？

Is there any reason why you should use Function.identity() instead of str->str (or vice versa). I think that the second option is more readable (a matter of taste of course). But, is there any "real" reason why one should be preferred?

推荐答案

截至当前的JRE实施，<$ c每次出现标识符时，$ c> Function.identity（）将始终返回相同的实例 - >标识符不仅会创建自己的实例，甚至还会有一个独特的实现类。有关详细信息，请参阅。

As of the current JRE implementation, Function.identity() will always return the same instance while each occurrence of identifier -> identifier will not only create its own instance but even have a distinct implementation class. For more details, see here.

原因是编译器生成一个合成方法，保存该lambda表达式的普通主体（在 x-> x 的情况下，相当于返回标识符; ）并告诉运行时创建调用此方法的函数接口的实现。所以运行时只看到不同的目标方法，当前的实现不会分析方法来确定某些方法是否相同。

The reason is that the compiler generates a synthetic method holding the trivial body of that lambda expression (in the case of x->x, equivalent to return identifier;) and tell the runtime to create an implementation of the functional interface calling this method. So the runtime sees only different target methods and the current implementation does not analyze the methods to find out whether certain methods are equivalent.

所以使用函数.identity（）而不是 x - > x 可能会节省一些内存，但如果你真的认为 x - >这不应该决定你的决定。 x 比 Function.identity（）更具可读性。

So using Function.identity() instead of x -> x might save some memory but that shouldn’t drive your decision if you really think that x -> x is more readable than Function.identity().

您也可以考虑在启用调试信息进行编译时，合成方法将有一个行调试属性指向包含lambda表达式的源代码行，因此您有可能找到特定函数的源代码。 调试时的实例。相反，在调试操作期间遇到 Function.identity（）返回的实例时，您将不知道谁调用了该方法并将实例传递给操作。

You may also consider that when compiling with debug information enabled, the synthetic method will have a line debug attribute pointing to the source code line(s) holding the lambda expression, therefore you have a chance of finding the source of a particular Function instance while debugging. In contrast, when encountering the instance returned by Function.identity() during debugging an operation, you won’t know who has called that method and passed the instance to the operation.

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