blocks|key|593949|text|您需要使用s1.equals(s2)。对String对象使用==会比较对象引用本身。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|593950|编辑:当我运行你的第二个代码片段时，我没有打印出“两者都相等”。|593951|Edit2:澄清了使用'==‘时会比较引用。|593952|entityMap^0|5|D|K|6|U|2|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|K|$]]

You need to use <code>s1.equals(s2)</code>. Using <code>==</code> with <code>String</code> objects compares the object references themselves.

Edit: When I run your second code snippet, I do not get "both are equal" printed out.

Edit2: Clarified that references are compared when you use '=='.

blocks|key|809558|text|当你比较两个字符串时，不要使用==而使用eqauls()，因为你是在比较对象而不是引用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|809559|string1.equals(string2);|code-block|syntax|javascript|809560|entityMap^0|F|2|K|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Whenever you are comparing between two String, don't use <code>==</code> and use <code>eqauls()</code> becaue you are comparing objects not references:

<pre><code>string1.equals(string2);
</code></pre>

blocks|key|809654|text|在jdk6中：String+s1="Good";在常量池中创建一个字符串对象"Good“。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|809655|s1=s1%2B"morning";在常量池中创建了另一个字符串对象"morning“，但这次实际上是这样做的：s1=new+StringBuffer().append(s1).append("morning").toString();。|809656|现在，由于new操作符在堆中创建了一个对象，因此s1中的引用是堆而不是常量池的，并且String+s2="Goodmorning";在常量池中创建了一个字符串对象"Goodmorning“，其引用存储在s2中。|809657|因此，if(s1==s2)条件为false。|809658|但是在jdk7中会发生什么呢？|809659|entityMap^0|7|H|0|0|G|1I|1Q|0|5|3|O|2|16|O|2T|2|0|3|A|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|19|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|O|$]]

In jdk6:
<code>String s1="Good";</code> creates a String object "Good" in constant pool.

<code>s1=s1+"morning";</code> creates another String object "morning" in constant pool but this time actually JVM do: <code>s1=new StringBuffer().append(s1).append("morning").toString();</code>.

Now as the <code>new</code> operator creates an object in heap therefore the reference in <code>s1</code> is of heap not constant pool 
and the <code>String s2="Goodmorning";</code> creates a String object "Goodmorning" in constant pool whose reference is stored in <code>s2</code>.

Therefore, <code>if(s1==s2)</code> condition is false.

But what happens in jdk7?

blocks|key|594165|text|让我们从示例中省略不必要的细节：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|594166|class+Test+{
++++public+static+void+main(String...+args)+{
++++++++String+s1+=+"Good";
++++++++s1+=+s1+%2B+"morning";
++++++++System.out.println(s1+==+s1.intern());+//+Prints+true+for+jdk7,+false+-+for+jdk6.
++++}
}|code-block|syntax|javascript|594167|让我们把String#intern看作是一个黑盒。根据运行的几个测试用例，我可以得出如下结论：|offset|length|style|CODE|594168|Java+6：|594169|如果池中包含的对象等于this，则返回对该对象的引用，否则创建新的字符串(等于this)，放入池中，然后返回对创建的实例的引用。|594170|Java+7：|594171|如果池中包含的对象等于this，则返回对该对象的引用，否则将this放入池中，并返回this。|594172|Java6和Java7都不会破坏contract+of+the+method。|594173|似乎新的实习生方法行为是修复这个bug的结果：http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6962931。|594174|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/lang/String.html#intern%2528%2529|1|http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6962931^0|0|0|4|D|0|0|B|4|13|4|0|0|B|4|U|4|16|4|0|G|M|0|0|N|1M|1|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@$I|1B|J|1C|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1D|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1E|8|@$I|1F|J|1G|K|L]|$I|1H|J|1I|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1J|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1K|8|@$I|1L|J|1M|K|L]|$I|1N|J|1O|K|L]|$I|1P|J|1Q|K|L]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1R|8|@]|9|@$I|1S|J|1T|1|1U]]|A|$]]|$1|W|3|X|5|6|7|1V|8|@]|9|@$I|1W|J|1X|1|1Y]]|A|$]]|$1|Y|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]|16|$5|11|12|13|A|$14|17]]]]

Let's omit unnecessary details from the example:

<pre><code>class Test {
 public static void main(String... args) {
 String s1 = "Good";
 s1 = s1 + "morning";
 System.out.println(s1 == s1.intern()); // Prints true for jdk7, false - for jdk6.
 }
}
</code></pre>

Let's consider <code>String#intern</code> as a black box. Based on a few test cases run, I would conclude that implementation is as following:

Java 6: 
if the pool contains object equals to <code>this</code>, then return reference to that object,
else create new string (equal to <code>this</code>), put to the pool, and return reference to that created instance.

Java 7: 
if the pool contains object equals to <code>this</code>, then return reference to that object,
else put <code>this</code> to the pool, and return <code>this</code>.

Neither Java 6 nor Java 7 breaks the <a href="http://download.oracle.com/javase/6/docs/api/java/lang/String.html#intern%28%29">contract of the method</a>.

It seems that new intern method behavior was a result of the fix of this bug: <a href="http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6962931">http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6962931</a>.

blocks|key|594200|text|结果代码依赖于运行时：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|594201|class+Test+{
+++++public+static+void+main(String...+args)+{
++++++++String+s1+=+"Good";
++++++++s1+=+s1+%2B+"morning";
++++++++System.out.println(s1+==+s1.intern());+//+Prints+true+for+jdk7,+false+-+for+jdk6.
++++}
}|code-block|syntax|javascript|594202|如果你是这样写的：|594203|class+Test+{
+++++public+static+void+main(String...+args)+{
++++++++String+s+=+"GoodMorning";
++++++++String+s1+=+"Good";
++++++++s1+=+s1+%2B+"morning";
++++++++System.out.println(s1+==+s1.intern());+//+Prints+false+for+both+jdk7+and+jdk6.
++++}
}|594204|原因是‘ldc+#N’(从常量池中加载字符串)和String.intern()都将在hotspot+JVM中使用StringTable。为了获得更多细节，我写了一篇英文文章：http://aprilsoft.cn/blog/post/307.html|offset|length|594205|entityMap|0|LINK|mutability|MUTABLE|url|http://aprilsoft.cn/blog/post/307.html^0|0|0|0|0|2F|12|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

The result code dependents runtime:

<pre><code>class Test {
 public static void main(String... args) {
 String s1 = "Good";
 s1 = s1 + "morning";
 System.out.println(s1 == s1.intern()); // Prints true for jdk7, false - for jdk6.
 }
}
</code></pre>

If you write like this:

<pre><code>class Test {
 public static void main(String... args) {
 String s = "GoodMorning";
 String s1 = "Good";
 s1 = s1 + "morning";
 System.out.println(s1 == s1.intern()); // Prints false for both jdk7 and jdk6.
 }
}
</code></pre>

the reason is ' ldc #N ' (Load string from constant pool) and String.intern() both will use StringTable in hotspot JVM. For detail I wrote a pool english article: <a href="http://aprilsoft.cn/blog/post/307.html" rel="nofollow">http://aprilsoft.cn/blog/post/307.html</a>

<pre><code>class Test {
 public static void main(String...args) {
 String s1 = "Good";
 s1 = s1 + "morning";
 System.out.println(s1.intern());
 String s2 = "Goodmorning";
 if (s1 == s2) {
 System.out.println("both are equal");
 }
 }
}
</code></pre>

This code produces different outputs in Java 6 and Java 7.
In Java 6 the <code>s1==s2</code> condition returns <code>false</code> and in Java 7 the <code>s1==s2</code> returns <code>true</code>. Why?

Why does this program produces different output in Java 6 and Java 7?

intern() behaving differently in Java 6 and Java 7

Java

class Test {    public static void main(String...args) {        String s1 = "Good";        s1 = s1 + "morning";        System.out.println(s1.intern());        S...

问intern()在Java 6和Java 7中的行为不同
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