Java里面instanceof怎么实现的
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开始完全一头雾水呀,后面看了Java指令集的介绍,逐渐理解了。
https://www.zhihu.com/question/21574535/answer/18998914
下面这个答案比较直白
你在面月薪10000的Java高级研发职位。面试官对JVM有一些了解,想让你说说JVM会如何实现 instanceof 指令。
但他可能也没看过实际的JVM是怎么做的,只是臆想过一下而已。JVM的规定就是“底层”。这种情况就给他JVM规范对 instanceof 指令的定义就好: Chapter 6. The Java Virtual Machine Instruction Set, JVM规范Java SE 7版 根据规范来臆想一下实现就能八九不离十的混过这题了。 地址: http://docs.oracle.com/javase/specs/jvms/se7/html/jvms-6.html#jvms-6.5.instanceof
上面指令的定义详细解释如下(解释和重要点已经飘红指出):
instanceof
The objectref, which must be of type reference
, is popped from the operand stack. The unsigned indexbyte1 and indexbyte2 are used to construct an index into the run-time constant pool of the current class (§2.6), where the value of the index is (indexbyte1 <<
8) | indexbyte2. The run-time constant pool item at the index must be a symbolic reference to a class, array, or interface type.
If objectref is null
, the instanceof instruction pushes an int
result of 0 as an int
on the operand stack.
意思是左侧是null,就返回false.(另外,右侧不能为null)
Otherwise, the named class, array, or interface type is resolved (§5.4.3.1). If objectref is an instance of the resolved class or array or implements the resolved interface, the instanceof instruction pushes an int
result of 1 as an int
on the operand stack; otherwise, it pushes an int
result of 0.
这个只是概述,下面是详细解释。分为多钟情况。
The following rules are used to determine whether an objectref that is not null
is an instance of the resolved type: If S is the class of the object referred to by objectref and T is the resolved class, array, or interface type, instanceof determines whether objectref is an instance of T as follows:
-
If S is an ordinary (nonarray) class, then:
-
If T is a class type, then S must be the same class as T, or S must be a subclass of T;
-
If T is an interface type, then S must implement interface T.
-
-
If S is an interface type, then:
-
If T is a class type, then T must be
Object
. -
If T is an interface type, then T must be the same interface as S or a superinterface of S.
-
-
If S is a class representing the array type SC
[]
, that is, an array of components of type SC, then:-
If T is a class type, then T must be
Object
. -
If T is an interface type, then T must be one of the interfaces implemented by arrays (JLS §4.10.3).
-
If T is an array type TC
[]
, that is, an array of components of type TC, then one of the following must be true:-
TC and SC are the same primitive type.
-
TC and SC are reference types, and type SC can be cast to TC by these run-time rules.
-
-
During resolution of the symbolic reference to the class, array, or interface type, any of the exceptions documented in §5.4.3.1 can be thrown.
The instanceof instruction is very similar to the checkcast instruction (§checkcast). It differs in its treatment of null(上面提到,左侧null会直接返回false)
, its behavior when its test fails (checkcast throws an exception, instanceof pushes a result code), and its effect on the operand stack.
下面这个答案比较“晦涩”
你在面试月薪10000以上的Java资深研发职位,注重性能调优啥的。这种职位虽然不直接涉及JVM的研发,但由于性能问题经常源自“抽象泄漏”,
对实际使用的JVM的实现的思路需要有所了解。面试官对JVM的了解可能也就在此程度。
对付这个可以用一篇论文:Fast subtype checking in the HotSpot JVM。
之前有个讨论帖里讨论过对这篇论文的解读:请教一个share/vm/oops下的代码做fast subtype check的问题
其中提到了 “抽象泄露”,后面会单开文章分析。
再看下面的代码例子(http://jurisp.iteye.com/blog/815704)
String s = null; System.out.println(s instanceof Object); // false System.out.println(s instanceof String); // false // 下面的不能通过编译 Incompatible conditional operand types String and String[] // boolean as = (s instanceof String[]); // 不能编译 Syntax error on token "null", invalid ReferenceType // System.out.println(s instanceof null); // 不能编译 Incompatible conditional operand types TypeInstanceTest and String//System.out.println(new Solution() instanceof String); System.out.println(new String() instanceof String); //true System.out.println(new String() instanceof Object); //true // 会抛出异常 Solution test = (Solution) new Object();
而运行结果如下:
false false true true Exception
说明不是随便两个类型之间都能用instanceof的,如果两个操作数的类型都是类,其中一个必须是另一个的子类型。
另外,注意下面这段晦涩的表述。
访问静态方法,使用表达式作为限定符。表达式的值所引用的对象的运行期类型在确定哪一个方法被调用的时候不起作用,而且对象有标识符,其标识符也不起任何作用。
代码示例如下:
class Solution { public static boolean canFinish() { System.out.println("here"); return true; } public static void main(String[] args) { ((Solution) null).canFinish(); } } 运行得到正确调用结果: here
说明,用null是可以调用 static 方法的。
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