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String comparison and String interning in Java

Answer 1

+12 A:

You should almost always use equals. You can be certain that string1 == string2 will work if:

You've already made sure you've got distinct values in some other way (e.g. you're using string values fetched from a set, but comparing them for some other reason)
You know you're dealing with compile-time string constants
You've manually interned the strings yourself

It really doesn't happen very often, in my experience.

Jon Skeet 2010-10-07 20:50:22

Thank you Jon, you always put things simply and clearly.

Albus Dumbledore 2010-10-07 21:28:47

Absolutely the right answer. Using '==' because equals() is inefficient is a great example of premature optimization.

DJClayworth 2010-10-07 21:30:06

And String.intern() is slow enough to be useless for performance in most cases.

Darron 2010-10-07 21:30:09

Answer 2

+1 A:

From what I know of Java, string1==string2 will only be true if the references to those objects are the same. Take a look at the following case

String string1 = new String("Bob");
String string2 = new String("Bob");

string1 == string2; // false, they are seperate objects
string1 = string2;  // asigning string1 to string2 object
string1 == string2; // true, they both refer to the same object

Shynthriir 2010-10-07 20:51:21

The first comparison will *not* be false because they are *not* separate objects. String literals are interned in Java.

sepp2k 2010-10-07 20:53:19

The first comparison should return false in this case-String string1 = new("Bob");String string2 = new("Bob");

stratwine 2010-10-07 21:01:40

I'll edit my responce to make it more clearly defined. I thought I might run into this issue with Java.

Shynthriir 2010-10-07 21:05:16

@stratwine: No, this is not true. See http://java.sun.com/docs/books/jls/third_edition/html/lexical.html#3.10.5: Each string literal is a reference (§4.3) to an instance (§4.3.1, §12.5) of class String (§4.3.3). String objects have a constant value. String literals-or, more generally, strings that are the values of constant expressions (§15.28)-are "interned" so as to share unique instances, using the method String.intern.

Dirk 2010-10-07 21:06:20

The answer is correct, and the first comparison is 'false'. "new String(...)" creates (surprise surprise) a new String, which is not the same object as the interned "Bob".

DJClayworth 2010-10-07 21:29:01

@Dirk - Strings created at compile time are interned. Not the Strings that are created on the fly. new operator creates the String at runtime and there's no interning involved here.

stratwine 2010-10-08 00:37:40

@stratwine: Seems I misread your comment. I thought that you'd tried to "expand" the poster's original formulation of the initial assignments instead of providing an alternative case to elaborate on. Apologies.

Dirk 2010-10-08 00:50:58

Answer 3

A:

You can only use the == for comparison if you are sure the objects are the same.

For example, this could occur if you had a final static String variable. You could be certain that a comparison would be between the same object.

Stick with the equals for string comparison.

Starkey 2010-10-07 20:51:44

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