recursion parameter issue

Question

public void mystery1(int n) {
if (n <= 1) {
    System.out.print(n);
} else {
    mystery1(n / 2);
    System.out.print(", " + n);
}
}

What gives this code for odd numbers. Becuase when we divide it it will not be an integer.

Answer 1

There is not mystery, because result of the integer division in Java is integer.

Answer 2

In Java or most other programming languages, when you divide an integer by an integer, the result will be an integer. If a decimal number occurs, say for example:

5/2=2.5

then, the number before the decimal point will be treated as the integer and 2 will be chosen.

In case you want to explicitly convert the integer into float or double, you can use any of the following conversions:

(float) 3/2;

(double) n/2;

The above explicitly converts it to a decimal.

Answer 3

n / 2, this is an integer division, where the fraction part will be ignored.

System.out.println(3/2); // prints 1
System.out.println(3.0/2); // prints 1.5
System.out.println(3/2.0); // prints 1.5
System.out.println(3.0/2.0); // prints 1.5

Answer 4

Param will rounded to int, for example if param will be 5, the next call the function will be with param 2