== operator with Strings
The code below should not print "Bye", since == operator is used to compare references, but oddly enough, "Bye" is still printed. Why does this happen? I'm using Netbeans 6.9.1 as the IDE.
public class Test {
public static void main(Str开发者_StackOverflow中文版ing [] args) {
String test ="Hi";
if(test=="Hi"){
System.out.println("Bye");
}
}
}
This behavior is because of interning. The behavior is described in the docs for String#intern
(including why it's showing up in your code even though you never call String#intern
):
A pool of strings, initially empty, is maintained privately by the class
String
.When the
intern
method is invoked, if the pool already contains a string equal to thisString
object as determined by theequals(Object)
method, then the string from the pool is returned. Otherwise, thisString
object is added to the pool and a reference to thisString
object is returned.It follows that for any two strings
s
andt
,s.intern() == t.intern()
istrue
if and only ifs.equals(t)
is true.All literal strings and string-valued constant expressions are interned. String literals are defined in §3.10.5 of the Java Language Specification.
So for example:
public class Test {
private String s1 = "Hi";
public static void main(String [] args) {
new Test().test();
System.exit(0);
}
public void test() {
String s2 ="Hi";
String s3;
System.out.println("[statics] s2 == s1? " + (s2 == s1));
s3 = "H" + part2();
System.out.println("[before interning] s3 == s1? " + (s3 == s1));
s3 = s3.intern();
System.out.println("[after interning] s3 == s1? " + (s3 == s1));
System.exit(0);
}
protected String part2() {
return "i";
}
}
Output:
[statics] s2 == s1? true [before interning] s3 == s1? false [after interning] s3 == s1? true
Walking through that:
- The literal assigned to
s1
is automatically interned, sos1
ends up referring to a string in the pool. - The literal assigned to
s2
is also auto-interned, and sos2
ends up pointing to the same instances1
points to. This is fine even though the two bits of code may be completely unknown to each other, because Java'sString
instances are immutable. You can't change them. You can use methods liketoLowerCase
to get back a new string with changes, but the original you calledtoLowerCase
(etc.) on remains unchanged. So they can safely be shared amongst unrelated code. - We create a new
String
instance via a runtime operation. Even though the new instance has the same sequence of characters as the interned one, it's a separate instance. The runtime doesn't intern dynamically-created strings automatically, because there's a cost involved: The work of finding the string in the pool. (Whereas when compiling, the compiler can take that cost onto itself.) So now we have two instances, the ones1
ands2
point to, and the ones3
points to. So the code shows thats3 != s1
. - Then we explicitly intern
s3
. Perhaps it's a large string we're planning to hold onto for a long time, and we think it's likely that it's going to be duplicated in other places. So we accept the work of interning it in return for the potential memory savings. Since interning by definition means we may get back a new reference, we assign the result back tos3
. - And we can see that indeed,
s3
now points to the same instances1
ands2
point to.
Hard-coded Strings are compiled into the JVM's String Table, which holds unique Strings - that is the compiler stores only one copy of "Hi", so you are comparing that same object, so ==
works.
If you actually create a new String using the constructor, like new String("Hi")
, you will get a different object.
There is a String cache in java. Where as in this case the same object is returned from the cache which hold the same reference.
The main reason for that is because the "Hi"
is picked up from String Pool
. The immutable object must have some sort of cache so it can perform better. So String
class is immutable and it uses String Pool
for basic cache.
In this case, "Hi"
is in the String pool and all the String having values of "Hi"
will have the same reference for String Pool.
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