比较整数对象与原始int

Question

I am comparing integer object with primitive int but getting null pointer exception if Integer object is null

public static void main(String[] args) {

        Integer a = null;
        int b = 1;

        if (a != b) {
            System.out.println("True");
        }

    }


Output : Exception in thread "main" java.lang.NullPointerException
    at com.nfdil.loyalty.common.Test.main(Test.java:10)

I am getting this because its trying to convert null integer object (a.intValue()) into primitive int .

Is there any other way to avoid this?

Answer 1

您可以使用Objects.equals ：

if (!Objects.equals(a, b)) { ... }

Answer 2

For the sake of being able to compare the two, you can avoid this exception by casting b to an Integer .

Integer a = null;
int b = 0;
System.out.println(a != (Integer) b);

This will always result in true however, since a and (Integer) b evaluate each evaluate to different objects.

The reason that your original code threw an exception is because the JVM tried to convert a to an int . a had a value of null however, and the JVM can't convert that to an int . Casting the int to an Integer beforehand ensures that both of the compared terms are Integer objects so they can be compared normally, without the JVM having to do any unsafe implicit casting.

If you want to compare the numerical value of the two items, you could use an approach such as this:

Integer a = null;
int b = 0;
System.out.println(!((Integer) b).equals(a));

Answer 3

您可以先将b转换为Integer ，然后与.equals进行比较：

Integer.valueOf(b).equals(a)

Answer 4

Use if (a != null && a == b) to test of objects should be considered equal, or if (a == null || a != b) to test of objects should be considered unequal. Since b is an int rather than Integer it can't be null. Thus, if a is null it can't match b .

While one could convert b to an Integer and then use Object.equals() upon it, a literal interpretation of such code would ask the compiler and runtime to do a lot of needless work. While some versions of the JVM (runtime) might be able to recognize that one of the arguments to equals will always be a non-null Integer , and that it can thus skip a lot of the needless work and instead generate code equivalent to the if statement above, there's no guarantee that all versions would do so (and in fact, I would be very surprised if there weren't some versions that can't).

Incidentally, this approach will test the numerical value of a against b even if b happens to be numeric primitive type other than int [eg short , long or double ]. Behavior when b is float might be a little unexpected [it compares b not with the number in a , but with the float value nearest to a ], but when using long or double it will test for an exact numerical match. By contrast, approaches using equals would report that a was not equal to any number of any type other than int or Integer .