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作者 [转贴]StringBuffer versus String [英文]
iforem



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于 2003-12-12 13:12 user profilesend a private message to usersearch all posts byselect and copy to clipboard. 
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StringBuffer versus String
What is the performance impact of the StringBuffer and String classes?

Summary
Reggie illuminates the underlying performance impact of using the StringBuffer and String classes when performing concatenations. (600 words)
By Reggie Hutcherson

Java provides the StringBuffer and String classes, and the String class is used to manipulate character strings that cannot be changed. Simply stated, 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. In String manipulation code, character strings are routinely concatenated. Using the String class, concatenations are typically performed as follows:

String str = new String ("Stanford ");
str += "Lost!!";

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

StringBuffer str = new StringBuffer ("Stanford ");
str.append("Lost!!");

Developers usually assume 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 innocent, but the code generated produces some surprises. 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
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("Stanford ");

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

str += "Lost!!";

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("Stanford ");

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

str.append("Lost!!");

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.

About the author
Reggie Hutcherson is a Sun technology evangelist. He evangelizes Sun's Java 2 Platform technologies around the world concentrating on J2SE and the HotSpot performance engine.

Resources

"JavaWorld debuts new weekly Java performance column," Reggie Hutcherson (JavaWorld, March 2000):
http://www.javaworld.com/jw-03-2000/jw-03-javaperf.html

"The basics of Java performance," Reggie Hutcherson (JavaWorld, March 2000):
http://www.javaworld.com/jw-03-2000/jw-03-javaperf_2.html

"Performance problem or design problem?" Reggie Hutcherson (JavaWorld, March 2000):
http://www.javaworld.com/jw-03-2000/jw-03-javaperf_3.html

"Compiler optimizations," Reggie Hutcherson (JavaWorld, March 2000):
http://www.javaworld.com/jw-03-2000/jw-03-javaperf_4.html




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7030 [转贴]StringBuffer versus String [英文] iforem 5449 2003-12-12 13:12

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