考虑以下代码片段:
public static Object o = new Object();
public static Callable x1() {
Object x = o;
return () -> x;
}
public static Callable x2() {
return () -> o;
}
方法 x2()
将始终返回相同的 Lamba 对象,而 x1()
将始终创建新对象:
Method x2()
will always return the same lamba object, while x1()
will always create new one:
System.out.println(x1());
System.out.println(x1());
System.out.println(x2());
System.out.println(x2());
会打印出如下内容:
TestLambda$$Lambda$1/821270929@4a574795
TestLambda$$Lambda$1/821270929@f6f4d33
TestLambda$$Lambda$2/603742814@7adf9f5f
TestLambda$$Lambda$2/603742814@7adf9f5f
在哪里(我猜是在 JVM 规范中?)描述了这个 lambda 重用规则?JVM如何决定在哪里重用?
Where (in JVM specification I guess?) is this rule of lambda reuse described? How does JVM decide where do reuse or not?
您无法确定为 lambda 表达式返回的对象的身份.它可以是新实例,也可以是预先存在的实例.
You can't be sure about the identity of the object returned for a lambda expression. It can be a new instance, or a pre-existing instance.
这在 中指定JLS §15.27.4:
在运行时,对 lambda 表达式的求值类似于对类实例创建表达式的求值,只要正常完成产生对对象的引用.lambda 表达式的求值不同于 lambda 主体的执行.
At run time, evaluation of a lambda expression is similar to evaluation of a class instance creation expression, insofar as normal completion produces a reference to an object. Evaluation of a lambda expression is distinct from execution of the lambda body.
要么分配并初始化具有以下属性的类的新实例,要么引用具有以下属性的类的现有实例.如果要创建一个新实例,但没有足够的空间来分配对象,则 lambda 表达式的计算会通过抛出 OutOfMemoryError 突然完成.
Either a new instance of a class with the properties below is allocated and initialized, or an existing instance of a class with the properties below is referenced. If a new instance is to be created, but there is insufficient space to allocate the object, evaluation of the lambda expression completes abruptly by throwing an OutOfMemoryError.
这篇关于JVM 什么时候决定重用旧的 lambda?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!