Integer a = 127; //Integer.valueOf(127)
Integer b = 127; //Integer.valueOf(127)
log.info("\nInteger a = 127;\n" +
"Integer b = 127;\n" +
"a == b ? {}",a == b); // true
Integer c = 128; //Integer.valueOf(128)
Integer d = 128; //Integer.valueOf(128)
log.info("\nInteger c = 128;\n" +
"Integer d = 128;\n" +
"c == d ? {}", c == d); //false
Integer e = 127; //Integer.valueOf(127)
Integer f = new Integer(127); //new instance
log.info("\nInteger e = 127;\n" +
"Integer f = new Integer(127);\n" +
"e == f ? {}", e == f); //false
Integer g = new Integer(127); //new instance
Integer h = new Integer(127); //new instance
log.info("\nInteger g = new Integer(127);\n" +
"Integer h = new Integer(127);\n" +
"g == h ? {}", g == h); //false
Integer i = 128; //unbox
int j = 128;
log.info("\nInteger i = 128;\n" +
"int j = 128;\n" +
"i == j ? {}", i == j); //true
在第一个案例中，编译器会把 Integer a = 127 转换为 Integer.valueOf(127)。查看源码可以发现，这个转换其实在内部做了缓存，使得两个 Integer 指向了同一个对象，所以结果是 true。
public static Integer valueOf(int i) {
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}
String a = "1";
String b = "1";
log.info("\nString a = \"1\";\n" +
"String b = \"1\";\n" +
"a == b ? {}", a == b); //true
String c = new String("2");
String d = new String("2");
log.info("\nString c = new String(\"2\");\n" +
"String d = new String(\"2\");" +
"c == d ? {}", c == d); //false
String e = new String("3").intern();
String f = new String("3").intern();
log.info("\nString e = new String(\"3\").intern();\n" +
"String f = new String(\"3\").intern();\n" +
"e == f ? {}", e == f); //true
String g = new String("4");
String h = new String("4");
log.info("\nString g = new String(\"4\");\n" +
"String h = new String(\"4\");\n" +
"g == h ? {}", g.equals(h)); //true
class Point {
private int x;
private int y;
private final String desc;
public Point(int x, int y, String desc) {
this.x = x;
this.y = y;
this.desc = desc;
}
}
现在我们希望只要 x 和 y 这 2 个属性一致就代表这是同一个点，所以我们需要重写 equals 方法，而不是用 Object 的原始的 equals。
@Override
public boolean equals(Object o) {
PointWrong that = (PointWrong) o;
return x == that.x && y == that.y;
}
@Data
@AllArgsConstructor
class Student implements Comparable<Student>{
private int id;
private String name;
@Override
public int compareTo(Student other) {
int result = Integer.compare(other.id, id);
if (result==0)
log.info("this {} == other {}", this, other);
return result;
}
}
然后，写一段测试代码分别通过 indexOf 方法和 Collections.binarySearch 方法进行搜索。列表中我们存放了两个学生，第一个学生 id 是 1 叫 zhang，第二个学生 id 是 2 叫 wang，搜索这个列表是否存在一个 id 是 2 叫 li 的学生：
@GetMapping("wrong")
public void wrong(){
List<Student> list = new ArrayList<>();
list.add(new Student(1, "zhang"));
list.add(new Student(2, "wang"));
Student student = new Student(2, "li");
log.info("ArrayList.indexOf");
int index1 = list.indexOf(student);
Collections.sort(list);
log.info("Collections.binarySearch");
int index2 = Collections.binarySearch(list, student);
log.info("index1 = " + index1);
log.info("index2 = " + index2);
}
Integer a = 127; //Integer.valueOf(127)
Integer b = 127; //Integer.valueOf(127)
log.info("\nInteger a = 127;\n" +
"Integer b = 127;\n" +
"a == b ? {}",a == b); // true
Integer c = 128; //Integer.valueOf(128)
Integer d = 128; //Integer.valueOf(128)
log.info("\nInteger c = 128;\n" +
"Integer d = 128;\n" +
"c == d ? {}", c == d); //false
Integer e = 127; //Integer.valueOf(127)
Integer f = new Integer(127); //new instance
log.info("\nInteger e = 127;\n" +
"Integer f = new Integer(127);\n" +
"e == f ? {}", e == f); //false
Integer g = new Integer(127); //new instance
Integer h = new Integer(127); //new instance
log.info("\nInteger g = new Integer(127);\n" +
"Integer h = new Integer(127);\n" +
"g == h ? {}", g == h); //false
Integer i = 128; //unbox
int j = 128;
log.info("\nInteger i = 128;\n" +
"int j = 128;\n" +
"i == j ? {}", i == j); //true
在第一个案例中，编译器会把 Integer a = 127 转换为 Integer.valueOf(127)。查看源码可以发现，这个转换其实在内部做了缓存，使得两个 Integer 指向了同一个对象，所以结果是 true。
public static Integer valueOf(int i) {
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}
String a = "1";
String b = "1";
log.info("\nString a = \"1\";\n" +
"String b = \"1\";\n" +
"a == b ? {}", a == b); //true
String c = new String("2");
String d = new String("2");
log.info("\nString c = new String(\"2\");\n" +
"String d = new String(\"2\");" +
"c == d ? {}", c == d); //false
String e = new String("3").intern();
String f = new String("3").intern();
log.info("\nString e = new String(\"3\").intern();\n" +
"String f = new String(\"3\").intern();\n" +
"e == f ? {}", e == f); //true
String g = new String("4");
String h = new String("4");
log.info("\nString g = new String(\"4\");\n" +
"String h = new String(\"4\");\n" +
"g == h ? {}", g.equals(h)); //true
class Point {
private int x;
private int y;
private final String desc;
public Point(int x, int y, String desc) {
this.x = x;
this.y = y;
this.desc = desc;
}
}
现在我们希望只要 x 和 y 这 2 个属性一致就代表这是同一个点，所以我们需要重写 equals 方法，而不是用 Object 的原始的 equals。
@Override
public boolean equals(Object o) {
PointWrong that = (PointWrong) o;
return x == that.x && y == that.y;
}
@Data
@AllArgsConstructor
class Student implements Comparable<Student>{
private int id;
private String name;
@Override
public int compareTo(Student other) {
int result = Integer.compare(other.id, id);
if (result==0)
log.info("this {} == other {}", this, other);
return result;
}
}
然后，写一段测试代码分别通过 indexOf 方法和 Collections.binarySearch 方法进行搜索。列表中我们存放了两个学生，第一个学生 id 是 1 叫 zhang，第二个学生 id 是 2 叫 wang，搜索这个列表是否存在一个 id 是 2 叫 li 的学生：
@GetMapping("wrong")
public void wrong(){
List<Student> list = new ArrayList<>();
list.add(new Student(1, "zhang"));
list.add(new Student(2, "wang"));
Student student = new Student(2, "li");
log.info("ArrayList.indexOf");
int index1 = list.indexOf(student);
Collections.sort(list);
log.info("Collections.binarySearch");
int index2 = Collections.binarySearch(list, student);
log.info("index1 = " + index1);
log.info("index2 = " + index2);
}
