[英]Java Wrapper equality test
public class WrapperTest {
public static void main(String[] args) {
Integer i = 100;
Integer j = 100;
if(i == j)
System.out.println("same");
else
System.out.println("not same");
}
}
The above code gives the output of same
when run, however if we change the value of i
and j
to 1000 the output changes to not same
. 上面的代码在运行时给出same
的输出,但是如果我们将i
和j
的值更改为1000,则输出变为not same
。 As I'm preparing for SCJP, need to get the concept behind this clear. 在我为SCJP做准备时,需要明确这个概念。 Can someone explain this behavior.Thanks. 有人可以解释这种行为。谢谢。
In Java, Integers between -128 and 127 (inclusive) are generally represented by the same Integer object instance. 在Java中,-128和127(包括)之间的整数通常由相同的Integer对象实例表示。 This is handled by the use of a inner class called IntegerCache (contained inside the Integer class, and used eg when Integer.valueOf() is called, or during autoboxing): 这是通过使用一个名为IntegerCache的内部类(包含在Integer类中,并在调用Integer.valueOf()时或在自动装箱期间使用)来处理的:
private static class IntegerCache {
private IntegerCache(){}
static final Integer cache[] = new Integer[-(-128) + 127 + 1];
static {
for(int i = 0; i < cache.length; i++)
cache[i] = new Integer(i - 128);
}
}
See also: http://www.owasp.org/index.php/Java_gotchas 另见: http : //www.owasp.org/index.php/Java_gotchas
Basically Integers between -127 and 127 are 'cached' in such a way that when you use those numbers you always refer to the same number in memory, which is why your ==
works. 基本上-127和127之间的整数是“缓存”的,这样当你使用这些数字时,你总是在内存中引用相同的数字,这就是你的==
有效的原因。
Any Integer outside of that range are not cached, thus the references are not the same. 超出该范围的任何整数都不会被缓存,因此引用不相同。
@tunaranch is correct. @tunaranch是对的。 It is also the same issue as in this Python question . 它也与此Python问题中的问题相同 。 The gist is that Java keeps an object around for the integers from -128 to 127 (Python does -5 to 256) and returns the same object every time you ask for one. 要点是Java为-128到127之间的整数保留一个对象(Python确实为-5到256),并且每次请求时都返回相同的对象。 If you ask for an Integer outside of this fixed range, it'll give you a new object every time. 如果你要求在这个固定范围之外的整数,它每次都会给你一个新的对象。
(Recall that ==
returns whether two objects are actually the same, while equals
compares their contents.) (回想一下, ==
返回两个对象是否实际相同,而equals
比较它们的内容。)
Edit : Here's the relevant paragraph from Section 5.1.7 of the Java Language Specification : 编辑 :这是Java语言规范的 5.1.7节中的相关段落:
If the value p being boxed is
true
,false
, abyte
, achar
in the range\
to\
, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. 如果被装箱的值p为true
,false
,一个byte
,范围为\
到\
的char
,或者介于-128和127之间的int或短号,则让r1和r2为任意两个装箱转换的结果p。 It is always the case that r1 == r2 . 始终是r1 == r2的情况。
Note that this also describes what happens with other types. 请注意,这也描述了其他类型的情况。
这与平等和自动装箱有关: http ://web.archive.org/web/20090220142800/http://davidflanagan.com/2004/02/equality-and-autoboxing.html
Your code doesn't compile. 您的代码无法编译。 This is what I get: 这就是我得到的:
Exception in thread "main" java.lang.Error: Unresolved compilation problems: Type mismatch: cannot convert from int to Integer Type mismatch: cannot convert from int to Integer 线程“main”中的异常java.lang.Error:未解决的编译问题:类型不匹配:无法从int转换为Integer类型不匹配:无法从int转换为Integer
at WrapperTest.main(WrapperTest.java:5)
Variables i and j are instances of Integer object. 变量i和j是Integer对象的实例。 Don't compare instances of object using "==" operator, use "equals" method instead. 不要使用“==”运算符比较对象的实例,而是使用“equals”方法。
Greetings 问候
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