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<>String详解，创建多少对象问题
<>1、String str = new String(“abc”)创建了几个对象？
答案是：两个或一个
情况一：abc 字符串之前没有用过。
这毫无疑问创建了两个对象，一个是new String 创建的一个新的对象，一个是常量“abc”对象的内容创建出的一个新的String对象，
情况二：abc 字符串之前用过。如下情况：
String str1 = "abc"; String str2 = new String("abc");
此时就创建一个对象，而abc 则是从字符串常量缓冲区中取出来的。
<>2、 String s = “a” + “b” + “c” + “d” + “e”，共创建了几个对象？
答案是：一个
“a”、“b”、“c”、“d”、“e"都是常量，对于常量，编译时就直接存储它们的字面值而不是它们的引用，在编译时就直接讲它们连接的结果提取出来变成了"abcde”，该语句在class文件中就相当于String
s = “abcde”，然后当JVM执行到这一句的时候， 就在String pool里找，如果没有这个字符串，就会产生一个。
<>3、String s = a+b+c+d+e，共创建了几个对象？
答案是：三个
java执行，String s = a+b+c+d+e的时候，可以认为步骤如下：
String a = "a"; String b = "b"; String c = "c"; String d = "d"; String e = "e";
StringBuilder builder = new StringBuilder(); builder.append(a).append(b).append(
c).append(d).append(e); String s = builder.toString();
需要结合看以上代码的源码，
StringBuilder builder = new StringBuilder();会产生2个对象，new
StringBuilder();和StringBuilder父类AbstractStringBuilder构造方法里面的new char[capacity];
String s = builder.toString();会产生new String(value,0,count);
其他的动作并不会产生额外的对象。
上诉代码进行字符串拼接的时候，含有变量，则拼接的原理是首先在堆中创建一个StringBuilder对象，这里命名为sb，然后通过StringBuilder对象的append方法将str添加进sb，而后在使用append方法将def字符串也添加进sb中，最后在调用toString方法，将sb转为String类型返回，这样就会额外的创建了两个对象。
