?最簡單的java單例源代碼：

?

class Singleton
{
?private Singleton()
?{

?}

?public static Singleton instance = null;

?public static Singleton getInstance()
?{
??if(instance == null)
???instance = new Singleton();
??return instance;
?}?

?public static void main(String args[])
?{
??Singleton s1, s2;

??s1 = Singleton.getInstance();
??s2 = Singleton.getInstance();

??System.out.println(s1 == s2);
??System.out.println(s1.equals(s2));

??System.out.println(s1);
??System.out.println(s2);
?}
}

/*
class Singleton
{
?private static int count = 0;

?private Singleton()
?{

?}

?public static Singleton instance = null;

?public static Singleton getInstance()
?{
??if(instance == null)
???instance = new Singleton();
??count++;
??return instance;
?}

?public static void display()
?{
??System.out.println(count);
?}

?public static void main(String args[])
?{
??Singleton s1, s2;

??s1 = Singleton.getInstance();
??s1.display();

??s2 = Singleton.getInstance();
??s2.display();

??System.out.println(s1 == s2);
??System.out.println(s1.equals(s2));

??System.out.println(s1);
??System.out.println(s2);
?}
}

*/

?

上述代碼對于單線程的程序是不會有問題的，但是如果用于多線程，很可能產生問題。改進方法：

將public static Singleton getInstance()改成public static synchronized Singleton getInstance()

總結

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