The difference between = = and equal of reference types and basic types in Java

1, Java in the "Basic type" and "basic type" comparison = = = Comparison is the value
2, the comparison of "basic type" and "reference type" in Java is also a comparison between the base type and the base type. Long j= 10000l; Long i= New Long (10000);
System. Out. println (I==J); True
So the reference type and the base type are compared, and when compiled into the underlying instruction, the reference type is converted to the base type and the value of the comparison.

System. Out. println (K.equals (j)); True all classes in Java referencing a type inherit from object, and if there is no equal method to override object, then "= =" and "equal" between reference types are the object addresses of the comparisons. For string, the wrapper type of the base type Boolean, Character, Byte, Shot, Integer, Long, Float, Double, applies 1 = = = Table comparison address, equal represents the comparison content
StringBuffer and StringBuilder special, = = and equal are comparison addresses because there is no equal method to rewrite the object.
3. Comparison of "reference type" and "reference type" in Java because the Java privilege mode caches the value of-128 ~ 127 Returns an integer object that is not a new one. Instead, it's an object that's already cached in the heap, so the following two ways to declare it are cached in the objects in the heap , which points to the same address: the object is not re-created. Long m= long.valueof (127); Long n= 127;
Long mk= 128L; Long nk= 128L; So m and n are equal (restricted range -128~127) System for both equal and = = comparisons. Out. println (m==n); True System. Out. println (M.equal (kn);//true
System. Out. println (MK==NK); False System. Out. println (Mk.equal (NK)); True but the new object will still be reassigned in the Java heap, so it is not equal to the above; long M2=new long (127); Long N2=new Long (127); System. Out. println (M2==N2); False System. Out. println (M2.equals (N2)); True