In Java, why does a Scanner's input.nextInt(3) behave this way?

Question

See code:

Scanner input = new Scanner(System.in);
System.out.println(input.nextInt(3));
//if user inputs 21, then the output is 7

My only understanding is that doing.nextInt(3) turns it into base 3, so only inputs like 10,11,12 or 20,21,22 will be accepted.

But can someone help me understand why nextInt(#) behaves this way, and why inputting 21 outputs 7?

Thanks

Answer 1

The nextInt method you are using accepts an argument for the integer radix that you want to read in. An argument of 3 would read in integers as base 3.

Refer to the Scanner class doc: https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextInt(int)

Answer 2

When you supply an int parameter to nextInt() , that parameter is the radix of the input that's being parsed . In this case, that means the input is parsed in Base-3. And 21 in Base-3 is 7 in Base-10:

00 - 0
01 - 1
02 - 2
10 - 3
11 - 4
12 - 5
20 - 6
21 - 7 // here
22 - 8
// etc.

Answer 3

Basically input.nextInt(x) means that you want to represent the value entered by the user in base x

The output is 7 because 21 is represented as 7 in base 3. You can always use an online calculator to experiment how the same number can be represented in different base.