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-   -   How Fast Can your Java String Operation's Be? (http://www.go4expert.com/articles/fast-java-string-operations-t23540/)

techexplorer 10Oct2010 19:17

How Fast Can your Java String Operation's Be?
The String class is used to manipulate character strings that cannot be changed. In simple terms, objects of type String are read only and immutable. The StringBuffer class is used to represent characters that can be modified. The significant performance difference between these two classes is that StringBuffer is faster than String when performing simple concatenations.

Operations using Strings in Java

In String manipulation code, character strings are routinely concatenated. Using the String class, concatenations are typically performed as follows:


String str = new String ("Hello ");
str += "World!!";

Operations using StringBuffer in Java

If you were to use StringBuffer to perform the same concatenation, you would need code that looks like this:

StringBuffer str = new StringBuffer ("Hello ");

Comparing String Vs String Buffer Operations

It is usually assumed that the first example above is more efficient because they think that the second example, which uses the append method for concatenation, is more costly than the first example, which uses the + operator to concatenate two String objects. The + operator appears to be easier, but the code generated produces some round about solutions. Using a StringBuffer for concatenation can in fact produce code that is significantly faster than using a String. To discover why this is the case, we must examine the generated bytecode from our two examples. The bytecode for the example using String looks like this:

0 new #7 <Class java.lang.String>
3 dup
4 ldc #2 <String "Stanford ">
6 invokespecial #12 <Method java.lang.String(java.lang.String)>
9 astore_1
10 new #8 <Class java.lang.StringBuffer>
13 dup
14 aload_1
15 invokestatic #23 <Method java.lang.String valueOf(java.lang.Object)>
18 invokespecial #13 <Method java.lang.StringBuffer(java.lang.String)>
21 ldc #1 <String "Lost!!">
23 invokevirtual #15 <Method java.lang.StringBuffer append(java.lang.String)>

26 invokevirtual #22 <Method java.lang.String toString()>
29 astore_1

The bytecode at locations 0 through 9 is executed for the first line of code, namely:

String str = new String("Hello ");

Then, the bytecode at location 10 through 29 is executed for the concatenation:

str += " World!!";

Things get interesting here. The bytecode generated for the concatenation creates a StringBuffer object, then invokes its append method: the temporary StringBuffer object is created at location 10, and its append method is called at location 23. Because the String class is immutable, a StringBuffer must be used for concatenation. After the concatenation is performed on the StringBuffer object, it must be converted back into a String. This is done with the call to the toString method at location 26. This method creates a new String object from the temporary StringBuffer object. The creation of this temporary StringBuffer object and its subsequent conversion back into a String object are very expensive. In summary, the two lines of code above result in the creation of three objects:

A String object at location 0
A StringBuffer object at location 10
A String object at location 26

Now, let's look at the bytecode generated for the example using StringBuffer:

0 new #8 <Class java.lang.StringBuffer>
3 dup
4 ldc #2 <String "Stanford ">
6 invokespecial #13 <Method java.lang.StringBuffer(java.lang.String)>
9 astore_1
10 aload_1
11 ldc #1 <String "Lost!!">
13 invokevirtual #15 <Method java.lang.StringBuffer append(java.lang.String)>
16 pop

The bytecode at locations 0 to 9 is executed for the first line of code:

StringBuffer str = new StringBuffer("Hello ");

The bytecode at location 10 to 16 is then executed for the concatenation:


Notice that, as is the case in the first example, this code invokes the append method of a StringBuffer object. Unlike the first example, however, there is no need to create a temporary StringBuffer and then convert it into a String object. This code creates only one object, the StringBuffer, at location 0.


In conclusion, StringBuffer concatenation is significantly faster than String concatenation. Obviously, StringBuffers should be used in this type of operation when possible. If the functionality of the String class is desired, consider using a StringBuffer for concatenation and then performing one conversion to String.

shabbir 1Nov2010 16:27

Re: How Fast Can your Java String Operation's Be?
Nominate this article for Article of the month - Oct 2010 Before November 15th 2010

baigarman007 8Nov2010 10:30

Re: How Fast Can your Java String Operation's Be?
Will someone, please, explain to a novice what it means when a function returns or doesn't return something? Return where? And what is the difference?

shabbir 17Nov2010 17:12

Re: How Fast Can your Java String Operation's Be?
Vote for this Article for Article of the month - Oct 2010. Voting closes on November 27th, 2010

rameshb 24Dec2010 13:20

Re: How Fast Can your Java String Operation's Be?
this will surly help me alot as am new to JAVA programming

Derby11 28Mar2011 12:53

Re: How Fast Can your Java String Operation's Be?
Thanks for sharing you views on it . Its really helpful to many people and also knowledgeable

andrewgibs 8Apr2011 11:45

Re: How Fast Can your Java String Operation's Be?
The native-method overhead may be high,but they're still fast compared to interpreting byte code.A mutable sequence of characters.This class provides an API compatible with StringBuffer,but with no guarantee of synchronization.

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