CODEWITHSUNDEEP
javaarraysindexingjlsunboxing
Jeffrey K.A
Jeffrey K.A

Reputation: 45

Why does unboxing occur in this case?

According to the Java Tutorial, the

Converting an object of a wrapper type (Integer) to its corresponding primitive (int) value is called unboxing. The Java compiler applies unboxing when an object of a wrapper class is:

  • Passed as a parameter to a method that expects a value of the corresponding primitive type.
  • Assigned to a variable of the corresponding primitive type.

Why does unboxing occur in this case?

char l = 0;
int arr[] = new int[]{1,2,3};
System.out.println(arr[new Integer(1)]);

Where in this scenario does either of those things happen? Is there an underlying method that governs element access in an array? Or does [] imply some sort of variable?

Upvotes: 2

Views: 112

Answers (3)

Turing85
Turing85

Reputation: 20195

The JLS 15, §15.10.3 is pretty clear on this one:

...

The index expression undergoes unary numeric promotion (§5.6). The promoted type must be int, or a compile-time error occurs.

...

Similar paragraphs can be found in older JLSes, e.g. JLS 8, §15.10.3.

Upvotes: 4

Edwin Buck
Edwin Buck

Reputation: 70939

The unboxing occurs on line three

System.out.println(arr[new Integer(1)]);

arr is an array as declared on line two

int arr[] = int[]{1, 2, 3};

Note that the type of arr is an "array of int". All arrays accept an int for the index being accessed. In line 3, you are passing an Integer, these two types are not the same. One is a primitive type, while the other is an Object type. Since there exists a "unboxing conversion" to change the Integer to an int, unboxing occurs just before the value is passed as an index into the int array.

Upvotes: 1

Patricia
Patricia

Reputation: 2865

In (arr[new Integer(1)] the wrapper Integer gets converted to primitive type, because it is used as an array index.

Upvotes: 0

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