Why Comparing Integer with Int Can Throw Nullpointerexception in Java

Why comparing Integer with int can throw NullPointerException in Java?

The Short Answer

The key point is this:

== between two reference types is always reference comparison
- More often than not, e.g. with Integer and String, you'd want to use equals instead
== between a reference type and a numeric primitive type is always numeric comparison
- The reference type will be subjected to unboxing conversion
- Unboxing null always throws NullPointerException
While Java has many special treatments for String, it is in fact NOT a primitive type

The above statements hold for any given valid Java code. With this understanding, there is no inconsistency whatsoever in the snippet you presented.

The Long Answer

Here are the relevant JLS sections:

JLS 15.21.3 Reference Equality Operators == and !=
If the operands of an equality operator are both of either reference type or the null type, then the operation is object equality.

This explains the following:

Integer i = null;
String str = null;

if (i == null) {   // Nothing happens
}
if (str == null) { // Nothing happens
}
if (str == "0") {  // Nothing happens
}

Both operands are reference types, and that's why the == is reference equality comparison.

This also explains the following:

System.out.println(new Integer(0) == new Integer(0)); // "false"
System.out.println("X" == "x".toUpperCase()); // "false"

For == to be numerical equality, at least one of the operand must be a numeric type:

JLS 15.21.1 Numerical Equality Operators == and !=
If the operands of an equality operator are both of numeric type, or one is of numeric type and the other is convertible to numeric type, binary numeric promotion is performed on the operands. If the promoted type of the operands is int or long, then an integer equality test is performed; if the promoted type is float or double`, then a floating-point equality test is performed.
Note that binary numeric promotion performs value set conversion and unboxing conversion.

This explains:

Integer i = null;

if (i == 0) {  //NullPointerException
}

Here's an excerpt from Effective Java 2nd Edition, Item 49: Prefer primitives to boxed primitives:

In summary, use primitives in preference to boxed primitive whenever you have the choice. Primitive types are simpler and faster. If you must use boxed primitives, be careful! Autoboxing reduces the verbosity, but not the danger, of using boxed primitives. When your program compares two boxed primitives with the == operator, it does an identity comparison, which is almost certainly not what you want. When your program does mixed-type computations involving boxed and unboxed primitives, it does unboxing, and when your program does unboxing, it can throw NullPointerException. Finally, when your program boxes primitive values, it can result in costly and unnecessary object creations.

There are places where you have no choice but to use boxed primitives, e.g. generics, but otherwise you should seriously consider if a decision to use boxed primitives is justified.

References

JLS 4.2. Primitive Types and Values
- "The numeric types are the integral types and the floating-point types."
JLS 5.1.8 Unboxing Conversion
- "A type is said to be convertible to a numeric type if it is a numeric type, or it is a reference type that may be converted to a numeric type by unboxing conversion."
- "Unboxing conversion converts [...] from type Integer to type int"
- "If r is null, unboxing conversion throws a NullPointerException"
Java Language Guide/Autoboxing
JLS 15.21.1 Numerical Equality Operators == and !=
JLS 15.21.3 Reference Equality Operators == and !=
JLS 5.6.2 Binary Numeric Promotion

What is the cleanest way to compare an int with a potentially null Integer in Java?

I try to avoid casts whenever possible, so I'd rather use the following, which also looks nicer in my opinion:

Integer.valueOf(8).equals(m.get("null"))

Why do I get NullPointerException when my Integer[] has already been initialized?

You have created the array with

Integer[] numbers = new Integer[input.toString().length()];

Integer is an object type, so all the values in the array start off as null. If it were int, then it would be an array of primitives, which would all be initialized to 0.

As you want to update the array value try

 for (int x = 0; x < numbers.length; x++) {

      numbers [x] = value; // whatever
 }

How to compare Integer with int?

Simplest solution I see is to swap your ternary to the assignment of val (and you might as well use int). Also, I would prefer Map.getOrDefault(Object, V). Like,

int val = 1 + tmap.getOrDefault(key, 0);
tmap.put(key, val);
if (val > maxVal) {
    maxVal = val;
    maxKey = key;
}

Otherwise, you could repeat the ternary in your comparison (but I personally think that's ugly so I'll leave it as a completely optional exercise for the reader).

Why does int num = Integer.getInteger(123) throw NullPointerException?

The Big Picture

There are two issues at play here:

Integer getInteger(String) doesn't do what you think it does
- It returns null in this case
the assignment from Integer to int causes auto-unboxing
- Since the Integer is null, NullPointerException is thrown

To parse (String) "123" to (int) 123, you can use e.g. int Integer.parseInt(String).

References

Java Language Guide/Autoboxing

`Integer` API references

static int parseInt(String)
static Integer getInteger(String)

On `Integer.getInteger`

Here's what the documentation have to say about what this method does:

public static Integer getInteger(String nm): Determines the integer value of the system property with the specified name. If there is no property with the specified name, if the specified name is empty or null, or if the property does not have the correct numeric format, then null is returned.

In other words, this method has nothing to do with parsing a String to an int/Integer value, but rather, it has to do with System.getProperty method.

Admittedly this can be quite a surprise. It's unfortunate that the library has surprises like this, but it does teach you a valuable lesson: always look up the documentation to confirm what a method does.

Coincindentally, a variation of this problem was featured in Return of the Puzzlers: Schlock and Awe (TS-5186), Josh Bloch and Neal Gafter's 2009 JavaOne Technical Session presentation. Here's the concluding slide:

The Moral

Strange and terrible methods lurk in libraries

Some have innocuous sounding names

If your code misbehaves

Make sure you're calling the right methods

Read the library documentation

For API designers

Don't violate the principle of least astonishment

Don't violate the abstraction hierarchy

Don't use similar names for wildly different behaviors

For completeness, there are also these methods that are analogous to Integer.getInteger:

Boolean.getBoolean(String)
Long.getLong(String)

On autounboxing

The other issue, of course, is how the NullPointerException gets thrown. To focus on this issue, we can simplify the snippet as follows:

Integer someInteger = null;
int num = someInteger; // throws NullPointerException!!!

Here's a quote from Effective Java 2nd Edition, Item 49: Prefer primitive types to boxed primitives:

In summary, use primitives in preference to boxed primitive whenever you have the choice. Primitive types are simpler and faster. If you must use boxed primitives, be careful! Autoboxing reduces the verbosity, but not the danger, of using boxed primitives. When your program compares two boxed primitives with the == operator, it does an identity comparison, which is almost certainly not what you want. When your program does mixed-type computations involving boxed and unboxed primitives, it does unboxing, and when your program does unboxing, it can throw NullPointerException. Finally, when your program boxes primitive values, it can result in costly and unnecessary object creations.

There are places where you have no choice but to use boxed primitives, e.g. generics, but otherwise you should seriously consider if a decision to use boxed primitives is justified.

Comparing an int and Integer in Java

For all the relational operators (including therefore < and ==), if one type is the boxed analogue of the other, then the boxed type is converted to the unboxed form.

So your code is equivalent to Foo.intValue() < foo;. This is deeper than you might think: your Foo < foo will throw a NullPointerException if Foo is null.

Why can't I compare null field to a primitive

Because obj.n == 1 is interpreted as primitive integer comparison, so obj.n will be auto-unboxed calling out.n.intValue(), which leads to the exception.

Formally, JLS 15.21 makes it clear that compiler knows your code is a numeric comparison, and then JLS 15.21.1 applies which states:

If the operands of an equality operator are both of numeric type, or one is of numeric type and the other is convertible (§5.1.8) to numeric type, binary numeric promotion is performed on the operands (§5.6.2).
Note that binary numeric promotion performs value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).

NullPointerException using Long after equality check

This is the problem:

value == -1 || value == null

Expressions are evaluated from left to right and since Long must be unboxed first, JVM translates this to:

value.longValue() == -1 || value == null

And value.longValue() throws NullPointerException when value is null argument. It never reaches the second part of the expression.

It works when the order is different though:

value == null || value == -1

because if the value is null, the second part (that can cause NullPointerException when value is null) is never executed due to boolean expression short-circuit evaluation.

Is this on spec or a bug in the JDK?

Of course this is not a bug. The way primitive value wrappers are unboxed is on spec (5.1.8. Unboxing Conversion):

If r is a reference of type Long, then unboxing conversion converts r into r.longValue()

After unboxing is applied, the rest is standard Java.

Compare nullable Integer to 0

x is a field in your class, so when you create it without making it to reference to any Integer object (Integer x = new Integer(7) for example), the compiler gives it a null for you (the default values for Object references). It seems like you have hence: Integer x = null;

So to compare it just use the equals() method that is implemented by Integer wrapper class.

new Integer(0).equals(x)

Why Comparing Integer with Int Can Throw Nullpointerexception in Java