理解HashSet
一、新建一个类Userfromboss ，其中有id和name属性，都是String类型的二、写一个函数： @Test public void test(){ String a ="123"; String b ="123"; Set<String> set = new HashSet<String>(); set.add(a); set.add(b); System.out.println(set.size()); Set<Userfromboss> set1 = new HashSet<Userfromboss>(); Userfromboss u1 = new Userfromboss(); Userfromboss u2 = new Userfromboss(); u1.setId("1"); u1.setName("zhangsan"); u2.setId("1"); u2.setName("zhangsan"); set1.add(u1); set1.add(u2); System.out.println(set1.size()); } 打印结果为1,2。结果并不相同三、修改Userfromboss ，重写这个类的equals和hashcode两个函数 @Override public boolean equals(Object obj) { // TODO Auto-generated method stub return this.getId().equals(((Userfromboss)obj).getId())&&this.getName().equals(((Userfromboss)obj).getName()); } @Override public int hashCode() { // TODO Auto-generated method stub return 1; }四。再回去打印一下，发现已经是1，1了五。任意删掉一个重写的函数，打印结果是1,2六。结论：hashset是同时比较equals和hashcode两个函数的。。同时打开String的源码。可以看到equals和hashcode两个函数如下： public boolean equals(Object anObject) { if (this == anObject) { return true; } if (anObject instanceof String) { String anotherString = (String)anObject; int n = count; if (n == anotherString.count) { char v1[] = value; char v2[] = anotherString.value; int i = offset; int j = anotherString.offset; while (n-- != 0) { if (v1[i++] != v2[j++]) return false; } return true; } } return false; } public int hashCode() { int h = hash; if (h == 0) { int off = offset; char val[] = value; int len = count; for (int i = 0; i < len; i++) { h = 31*h + val[off++]; } hash = h; } return h; }下次重新这两个函数的时候，不要自己重写了，要仿照string类的两个方法来重写哦。O(∩_∩)O~?