Why is it that I can declare a numerical variable as a double but still use .nextInt()?

Question

Okay, I'm new to coding so I am unfamiliar with everything. Here's my question:

Why is it that .nextInt(); will process a double even though the name is .nextInt();?

double max = scan.nextInt();

This works, but why?

`scan.nextInt()` is an `int`; `int` can be widened to `double`. — Andy Turner, Jun 14 '18 at 22:58
[JLS-5.1.2. Widening Primitive Conversion](https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.2) - *`int` to `long`, `float`, or `double`*. — Elliott Frisch, Jun 14 '18 at 22:59

score 0 · Accepted Answer · answered Jun 14 '18 at 23:00

0

If you peek into the java-doc you'll actually see that nextInt() returns int as the method name suggests.

In your specific example, the value returned from nextInt() is widened to a double as every int can fit into a double type.

answered Jun 14 '18 at 23:00

Ousmane D.

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score 0 · Answer 2 · answered Jun 14 '18 at 23:01

The things after the period (.nextInt()) are called methods or functions.

Now to your actual question:

The method .nextInt() will return an integer, which will then be converted internally to a double, since every int is also a double if you ask for it explicitly (by saying double max=)

Also see: Why does Java implicitly (without cast) convert a `long` to a `float`?

score -1 · Answer 3 · answered Jun 14 '18 at 23:19

As others have indicated, this is an example of an implicit cast - you can assign an int to a double and it'll automatically be typecast (have the type changed) for you.

The basic idea here is that there's not really any serious harm in assigning an int to a double (other than consuming marginally more memory).

Note that the converse is not true - there could be serous harm in casting a double to int, so there is no implicit cast from double to int.

Why is it that I can declare a numerical variable as a double but still use .nextInt()?

3 Answers3