Primitive and object comparison with the == operator

Question

I am wondering what is the internal Java behavior for the following snippet:

Long a = 123L;
long b = 123;
System.out.println("a equals b?: " + (a == b));

The result true, although the comparison of the two objects Longwill be false(because it compares their reference). Is this converting Java Longto its primitive value because it detects an operator ==against another primitive object?

Answer 1

Is Java converting a long object to its primitive value because it detects the == operator against another primitive object?

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Answer 2

Java:

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List<Integer> li = new ArrayList<>();
for (int i = 1; i < 50; i += 2)
    li.add(i);

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List<Integer> li = new ArrayList<>();
for (int i = 1; i < 50; i += 2)
    li.add(Integer.valueOf(i));

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  public static int sumEven (List li) {     int sum = 0;     for (Integer i: li)          (i% 2 == 0)             sum + = i;          ; }

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public static int sumEven(List<Integer> li) {
    int sum = 0;
    for (Integer i : li)
        if (i.intValue() % 2 == 0)
            sum += i.intValue();
        return sum;
}

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