Java中的==和.equals之间的差异。 [英] Difference between == and .equals in Java.

问题描述

因此，如果我有：

  String a =apple2e; 
 String b =apple2e; 
 
 System.out.println（a == b？+ a == b）; 
  

我得到 FALSE 。

根据我的理解，这是因为 a 和 b 是两个不同的引用 apple2e ）。

所以我会有：

  a（reference_id 123）------ 
 ---------apple2e
b（reference_id 456）--- --- 
  

现在，如果我只想比较内容两个字符串，我会使用 a.equals（b）

这意味着JVM简单地返回如果两个引用指向同一个对象？

感谢

EDIT

strong>

握住手机。感谢 delnan 指出 + 优先权！

当我把它改为：

  System.out.println（a == b）; 
  

我真的得到 true 。

这更有意义。

EDIT 2

我不敢相信我没有抓住那个。 lol

我在做：

 a == b？ + a == b 
  

这转换为

 a == b？apple2e==apple2e
  

$ b b

难怪这是错误的。

解决方案

a和b是对同一对象（apple2e）的两个不同引用。

strong>因为字符串实现 a 和 b 是不同的引用相同的 code> object。

---

不幸的是，你的代码不能做你想象的。尝试此操作：

  String a =apple2e; 
 String b =apple2e; 
 
 System.out.println（a == b？+ a == b）; //false
 System.out.println（a == b？+（a == b））; //a == b？true
  

Java自动实现所有字符串字面量。 是第二个sysout打印它的作用的原因。第一个sysout只打印false，因为字符串连接（ + ）的优先级高于 == ，因此它等效于：

  System.out.println b？apple2e==apple2e）; 
  

我不认为这是你想测试的！

另一方面，这将给你两个单独的 String 实例：

  String a = new String（apple2e）; 
 String b = new String（apple2e）; 
 
 System.out.println（a == b？+（a == b））; //a == b？false
  

$ b

  a（reference_id 123）---------------apple2e
 
b reference_id 456）---------------apple2e
  

并可以使用 String＃intern（）：

  String a = new String（apple2e）。intern（）; 
 String b = new String（apple2e）。intern（）; 
 
 System.out.println（a == b？+（a == b））; //a == b？true
  

例如

  a（reference_id 123）------ + 
 + ---------apple2e
b ）------ + 
  

I know this has been covered but I've seen inconsistent arguments here on SO.

So if I have:

String a = "apple2e";
String b = "apple2e";

System.out.println("a==b? " + a == b);

I get FALSE.

As I understand it, it's because a and b are two different references to the same object (apple2e).

So I would have something like:

a (reference_id 123) ------
                           ---------  "apple2e"
b (reference_id 456) ------

Now, if I just want to compare the contents of the two strings, I would use a.equals(b)

Does that mean that the JVM is simply returning if the two references are pointing to the same object? So it's not really doing a character-by-character comparison?

Thanks

EDIT

Hold the phones. Thanks delnan for pointing out the + precedence!!!

When I change it to:

System.out.println(a == b);

I indeed get true.

This makes more sense.

EDIT 2

I can't believe I didn't catch that. lol

I was doing:

"a==b? " + a == b

Which translates to

"a==b? apple2e" == "apple2e"

No wonder it was false!!

解决方案

As I understand it, it's because a and b are two different references to the same object (apple2e).

Because of string interning, and only because of string interning a and b are different references to the same String object.

---

Unfortunately, your code does not do what you think it does. Try this:

String a = "apple2e";
String b = "apple2e";

System.out.println("a==b? " + a == b);    // "false"
System.out.println("a==b? " + (a == b));  // "a==b? true"

Java automatically interns all string literals. That is why the second sysout prints what it does. The first sysout prints only "false" because string concatenation (+) has higher precedence than ==, so it's equivalent to this:

System.out.println("a==b? apple2e" == "apple2e");

I don't think that's what you meant to test!

This, on the other hand, will give you two separate String instances:

String a = new String("apple2e");
String b = new String("apple2e");

System.out.println("a==b? " + (a == b));  // "a==b? false"

Which would schematically look like

a (reference_id 123) ---------------  "apple2e"

b (reference_id 456) ---------------  "apple2e"

and can be reduced to the original situation using String#intern():

String a = new String("apple2e").intern();
String b = new String("apple2e").intern();

System.out.println("a==b? " + (a == b));  // "a==b? true"

e.g.

a (reference_id 123) ------+
                           +---------  "apple2e"
b (reference_id 456) ------+

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