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。

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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!!

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.

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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) ------+