如何避免来自 Java 算术运算符的 NullPointerException？

Question

Given the following:

Integer var1 = null;
Integer var2 = 4;
Integer result = var1 + var2; // throws NullPointerException

The requirement for my use case is that result should be null whenever either operand is null (and the same applies for other operators). I know I can use an if statement to do this but is there a smarter way?

Answer 1

The best way is not to use Boxed types for normal arithmetic operations. Use primitive types instead.

Only if you are using them in the collections somewhere you should resort to Boxed types.

EDIT:

Incorporating the suggestion from @Ingo there is a good utility class Optional in Guava, which explains on how to avoid nulls.

Now use of this class makes it explicit that the value can be null .

Answer 2

Integer var1 = null;
Integer var2 = 4;
Integer result = (var1 == null || var2 == null) ? null : var1 + var2; // returns null if either argument is null

Answer 3

Unfortunately no, there is no other way in Java to prevent NullPointerException on boxed types other than to explicitly check for null .

The same applies to .equals() , you can do:

Integer var1 = null;
Integer var2 = 4;
var1 == var2

But if you want to compare values:

Integer var1 = null;
Integer var2 = 4;
var1.equals(var2)  //throws NullPointerException

This is why a two argument static Object::equals (see here ) was introduced in 2011 to Java.

There are no such methods for boxed numbers (like Integer ). In fact during var1 + var2 you get automatic unboxing which causes exception. So you are not adding two Integer s, but .intValue() is called instead on both of them, and then a new Integer is constructed from the result.

See this Stack Overflow answer for further information on boxed numbers: Unboxing Long in java .

Sorry, this is one of the Java's nuisance. If you work on Integer s, check for null every time.

Answer 4

I guess you could catch it instead of using if :

Integer result = null;
try {
  result = var1 + var2;
} catch(NullPointerException ex) {}

The good news is, you don't have to test the values of var1 or var2, and the value is already null by default (so you don't need setting it in the catch block).

Answer 5

int and Integer has its own properties and they are used for specified functions.

int is a primitive datatype and its not considered as an object. ie; the values in an int can only be from a predefined set of elements.

Integer are wrapped in an object so they have the behaviour of an object.

So if you want to have use just the arithmetic operations , you better use int instead of Integer.

When you are using the Integer for arithmetic operation , it UnBox the Integer to find the int value and then do the operations and get it boxed. so the overhead is more and hence its a bad practice.

Answer 6

Technically, this doesn't have an if per se:

Integer result = var1 == null || var2 == null ? null : var1 + var2;

You could refactor this into a utility method for reuse:

public static Integer add(Integer var1, Integer var2) {
    return var1 == null || var2 == null ? null : var1 + var2;
}