About Question enthuware.ocpjp.v11.2.3488 :

[OliviaJohnson]

Hello, option B i1 == i3 does not compile. The error is

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Syntax error on token "==", invalid AssignmentOperator

I just copy and pasted your code into IDE

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public class main {
    public static void main(String[] args) throws Exception  {
        Integer i1 = 1;
        Integer i2 = new Integer(1);
        int i3 = 1;
        Byte b1 = 1;
        Long g1 = 1L;
        Double d = 1.0;

        i1 == i3;

        i1.equals(i2);
    }
}

[admin]

i1 == i3; is indeed an invalid statement. But it is a valid expression that returns true, which is why this option is correct. (All of the given options are just expressions.)

You may try this:
boolean b = (i1 == i3);
System.out.println(b);

[Badem48]

Hi,

You know

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new Integer(int num)

is depreciated.

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Integer.valueOf(int num)

factory method is being use.
Additionally,

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        Integer i1 = 1;
        Integer i2 = Integer.valueOf(1);
        System.out.println(i1 == i2);

Returns true.

[ShabnamLK]

Hi,
for Integer i2 = new Integer(1);

I can see that the answer explains in the end - Note that constructors of wrapper objects such as Integer have been deprecated (marked for removal since Java 9), but moreover it also gives compilation error in IDE.
So boolean expressions involving i2 may want to be disregarded in the correct options I would think?

[admin]

>but moreover it also gives compilation error in IDE.
Don't use IDE for testing code while preparing for certification exam because they give different errors than command line tools. https://enthuware.com/oca-ocp-java-cert ... cation-ide

[ShabnamLK]

Many thanks, you are right. From command line tool this usage is only shown as a compiler warning. Much appreciated.