Comparison of Java reference types and base types

Reference types and basic types of classification, not written, online there are a lot.

We know:
When two basic types use "= =" comparisons, they compare values .
When two reference types use "= =" comparisons, they compare the addresses .
When two reference types use method Equals () comparisons, they compare values .

But when the base type and his wrapper class (reference type) Use the "= =" comparison,
And what about their results?

Here we use integers and int for instructions.
Integer is the wrapper class for Int. int basic type, Integer is reference type.
Look at an example:

int i = 1234;
Integer i1 = new Integer (1234);
Integer i2 = new Integer (1234);
System.out.print ("I1 = = i2:" + (I1 = = I2));
System.out.println ("\ti1.equals (I2):" + (I1.equals (I2)));
System.out.print ("i = = I1:" + (i = = I1));
System.out.println ("\t\ti1.equals (i):" + (I1.equals (i)));
System.out.print ("i = = I2:" + (i = = I2));
System.out.println ("\t\ti2.equals (i):" + (I2.equals (i)));

Print:
I1 = = I2:false I1.equals (i2): True
i = = I1:true I1.equals (i): true
i = = I2:true I2.equals (i): true

We can see i = = i1, i = = I2, I1!= I2,
But using equals () they are all equal.

Then let's look at the method equals () in the integer class in jdk1.8.

    public boolean equals (Object obj) {
        if (obj instanceof Integer) {return
            value = = ((Integer) obj). Intvalue ()
        ; return
        false;
    }

You can observe the use of equals (), and they compare the values .

Summary: When comparing the base type with his wrapper class (reference type) using "= =" and method equals (), they compare values .

Integer and other basic types are the same, we are interested to be able to prove that I am not in the written.