递增Integer变量不会影响另一个引用同一对象的变量 [英] Incrementing an Integer variable doesn't affect another referencing the same object

问题描述

我总是理解静态变量在被引用时共享一个实例.我想对此进行测试，但结果与我预期的不同.

I always understood static variables to share one instance whenever they were referenced. I wanted to put this to the test but the results were different than I had expected.

static Integer counter = 0;
static Integer test = counter;

public static void main(String args[]) {
     counter++;
     System.out.println("counter: " + counter);
     System.out.println("test: " + test);
}

输出:

计数器:1

测试:0

因为 test 引用了 counter ，所以我认为当我增加 counter 时， test 也会自动增加.但是，似乎 test 从某个地方引用了 0 ，问题是在哪里?

Since test references counter I thought that when I increment counter then test will automatically be incremented as well. However, it seems test is referencing 0 from somewhere, question is, where?

推荐答案

由于 test 引用了 counter

这个假设是错误的.在Java中，您不能引用变量.存储在变量中的是一个值.该值可以是原始类型值或引用类型值.对于基元，值是基元的值.对于引用类型，

This assumption is false. In Java, you cannot reference variables. What is stored in a variable is a value. That value can either be a primitive type value or a reference type value. In the case of primitives, the value is the value of the primitive. In the case of reference types,

参考值(通常只是引用)是指向这些的指针对象，以及一个特殊的null引用，该引用不引用任何对象.

在

int a = 0;
int b = a;

对变量 a 求值以生成值 0 ，并将该值存储在 b 中.

the variable a is evaluated to produce a value, 0, and that value is stored in b.

在

Integer a = 0;
Integer b = a;

0 通过 Integer.valueOf(int)和值(对 Integer 对象存储在 a 中.然后对 a 进行求值，生成对该 Integer 对象的引用的值，并将该值存储在 b 中.

0 is converted to an Integer through Integer.valueOf(int) and the value, a reference to an Integer object is stored in a. Then a is evaluated, producing the value of that reference to an Integer object, and that value is stored in b.

这与

• 是Java的通过引用传递"吗?或按值传递"?

此外，变量是 static 的事实是无关紧要的.

Also, the fact that the variables are static is irrelevant.

解决此问题的唯一方法是手动更新 test ，即

The only way around this is to update test manually i.e.

counter++; 
test = counter;

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