关于String与StringBuffer的.

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都说StringBuffer比String效率高.所以详细看看!

其结论为: 至于推荐用StringBuffer来代替String+拼装字符串，说的是循环方式下

public   class   Bigint   {   
    public   static   void   main(String[]   args)   {   
            String   a   =   "a";   
            String   b   =   "b";   
            String   c   =   "c";   
            String   s   =   a   +   b   +   c;   
    }   
}

生成如下byteCode:

// Compiled from Bigint.java (version 1.5 : 49.0, super bit)
public class com.kinghi.xlink.test.Bigint {

// Method descriptor #6 ()V
// Stack: 1, Locals: 1
public Bigint();
    0 aload_0 [this]
    1 invokespecial java.lang.Object() [8]
    4 return
      Line numbers:
        [pc: 0, line: 5]
      Local variable table:
        [pc: 0, pc: 5] local: this index: 0 type: com.kinghi.xlink.test.Bigint

// Method descriptor #15 ([Ljava/lang/String;)V
// Stack: 3, Locals: 5
public static void main(java.lang.String[] args);
     0 ldc <String "a"> [16]
     2 astore_1 [a]
     3 ldc <String "b"> [18]
     5 astore_2 [b]
     6 ldc <String "c"> [20]
     8 astore_3 [c]
     9 new java.lang.StringBuilder [22]
    12 dup
    13 aload_1 [a]
    14 invokestatic java.lang.String.valueOf(java.lang.Object) : java.lang.String [24]
    17 invokespecial java.lang.StringBuilder(java.lang.String) [30]
    20 aload_2 [b]
    21 invokevirtual java.lang.StringBuilder.append(java.lang.String) : java.lang.StringBuilder [33]
    24 aload_3 [c]
    25 invokevirtual java.lang.StringBuilder.append(java.lang.String) : java.lang.StringBuilder [33]
    28 invokevirtual java.lang.StringBuilder.toString() : java.lang.String [37]
    31 astore 4 [s]
    33 return
      Line numbers:
        [pc: 0, line: 7]
        [pc: 3, line: 8]
        [pc: 6, line: 9]
        [pc: 9, line: 10]
        [pc: 33, line: 11]
      Local variable table:
        [pc: 0, pc: 34] local: args index: 0 type: java.lang.String[]
        [pc: 3, pc: 34] local: a index: 1 type: java.lang.String
        [pc: 6, pc: 34] local: b index: 2 type: java.lang.String
        [pc: 9, pc: 34] local: c index: 3 type: java.lang.String
        [pc: 33, pc: 34] local: s index: 4 type: java.lang.String
}

可知其也是通过StringBuilder来相加的.

至于推荐用StringBuffer来代替String+拼装字符串，说的是循环方式下，如：

  public   class   Test4   {   
          public   static   void   main(String[]   args)   {   
                  String   s   =   "s";   
                  for   (int   i   =   0;   i   <   20;   i++)   {   
                          s   +=   i;   
                  }   
          }   
  }   

对应的bytecode为：

public class Test4 extends java.lang.Object{
public Test4();
Code:
0: aload_0
1: invokespecial #1; //Method java/lang/Object."<init>":()V
4: return

public static void main(java.lang.String[]);
Code:
0: ldc #2; //String s
2: astore_1
3: iconst_0
4: istore_2
5: iload_2
6: bipush 20
8: if_icmpge 36
11: new #3; //class StringBuffer
14: dup
15: invokespecial #4; //Method java/lang/StringBuffer."<init>":()V
18: aload_1
19: invokevirtual #5; //Method java/lang/StringBuffer.append:(Ljava/lang/S
tring;)Ljava/lang/StringBuffer;
22: iload_2
23: invokevirtual #6; //Method java/lang/StringBuffer.append:(I)Ljava/lang
/StringBuffer;
26: invokevirtual #7; //Method java/lang/StringBuffer.toString:()Ljava/lan
g/String;
29: astore_1
30: iinc 2, 1
33: goto 5
36: return

}

如果用String+的方式，每循环一次，就会重新new一个StringBuffer对象，这样的内存消耗完全是不必要的（在数据量大的情况下，还会导致内存不足的错误），所以要这样做：

public class Test5 {
public static void main(String[] args) {
StringBuffer sb = new StringBuffer("s");
for (int i = 0; i < 20; i++) {
sb.append(i);
}
}
}
这样无论循环多少次，都只会生成一个StringBuffer对象