问题
Integer integer1 = 127;
Integer integer2 = 127;
System.out.println(integer1 == integer2);//true
integer1 = 128;
integer2 = 128;
System.out.println(integer1 == integer2);//false
I found it returns == (if it is) under the range of -128 - 127 , why is there such specification ?
回答1:
Because of this code in Integer.valueOf(int):
public static Integer valueOf(int i) {
if(i >= -128 && i <= IntegerCache.high)
return IntegerCache.cache[i + 128];
else
return new Integer(i);
}
Explanation:
Integer integer1 = 127 is a shortcut for Integer integer1 = Integer.valueOf(127), and for values between -128 and 127 (inclusive), the Integers are put in a cache and returned multiple times, while higher and lower numbers generate new Integers each time.
回答2:
== will return true if it's the exact same object. Boxing integers in Java 'intern' numbers within that that range, so any boxed version of such a number will result in the exact same object.
To get avoid this effect in comparisons, use .equals()
System.out.println(integer1.equals(integer2));
来源：https://stackoverflow.com/questions/5581913/integer-wrapper-class-and-operator-where-is-behavior-specified