**Applying the == operator to boxed primitives is almost always wrong. The two boxed primitives can have same value and different identities, whereas the primitives have only values. Check the code below:**

```
int a = new Integer(10);
int b = new Integer(10);
// The values of the boxed primitives are the same, so true.
System.out.println(a == b); // true
// The identities the boxed primitives are different, so false
System.out.println(new Integer(10) == new Integer(10)); // false
```

**When you mix primitives and boxed primitives in an operation, the boxed primitive is auto-unboxed. If a null object reference is auto-unboxed, you get a NullPointerException**

```
static Integer i;
public static void main(String[] args) {
if (i == 42) {
System.out.println("Do not mix primitive and boxed");
}
}
```

**Auto-boxing and auto-unboxing cause significant performance degradation. Compare the performance of the code snippets below:**

```
Integer sumBoxed = 0L;
for (int i = 0; i < Integer.MAX_VALUE; i++) {
sumBoxed += i;
}
...
int sumBoxed = 0L;
for (int i = 0; i < Integer.MAX_VALUE; i++) {
sumBoxed += i;
}
```