StringBuffer

[mi te]

I have a question about the length() method. Please read the code below.

public class BuffString{
public static void main(String args[]){
StringBuffer s1 = new StringBuffer("12345");
StringBuffer s2 = new StringBuffer("abcde");
StringBuffer s3 = s1.append(s2);
System.out.println(s1);
}
}

//output

12345abcde

I don't understand why s1 should contain the addition of s1 and s2? Is it impossible to keep the original value for s1? Why can't we have three different StringBuffers with differrent values? Why is the rule this way?
Please someone explain about this so I shut up and stop whining.

[Keith Lynn]

According to the Javadocs, the append method of StringBuffer when called with a StringBuffer will append the parameter to the caller of the method.

[Barry Gaunt]

s1.append(s2) appends the contents of StringBuffer s2 onto the contents of s1. It changes s1's state. The method also returns a StringBuffer reference referencing the target of the method. That's what you are assigning to s3. So s3 refers to s1's StringBuffer.

It works this way to enable the chaining of method calls, for example:
String s = new StringBuffer().append("abc").append("cde").append("fgh").toString()
[ March 14, 2006: Message edited by: Barry Gaunt ]

[Kotto Bass]

Apparently, you are confusing Strings and StringBuffers. StringBuffers are mutable - their values can be changed by using the append() function. However, Strings are immutable. If S1, S2 and S3 were Strings, and you used S3 = S1 + S2, S1 will remain the same. It is immutable (the same as if you used S1.toUpperCase(); S1 will still not change).

[ March 14, 2006: Message edited by: Kotto Bass ]
[ March 14, 2006: Message edited by: Kotto Bass ]