How do I force a number with unknown digits behind the decimal mark?

Question

Is there a way to force a number to be placed behind the decimal mark of a number?

Say I have a number = 2, and another number = 23. Is there a way for me to force 23 into 0.23, so that when I add the numbers I end up with 2.23? And is there a way to do this when the number of digits in the second number, in this case 23, are unknown?

EDIT: I realize this was badly written. I am currently working on a program that converts from imperial units to metric units. Part of the code looks like this:

double feet = nextDouble();
double inches = nextDouble();
double heightInMeters = (feet + (inches/10)) / 3.2808;

The problem with this code is that I anticipate that the user only enters a value <0, 9> for feet. Is there a way to force the input for inches to something like 0.x where x = inches so that it doesn't matter if the number is greater than 9?

Would be lovely if it was possible without using toString() and parseInt().

Not wanting to be pedantic, but there are 12 inches in a foot, not 10, so `( feet + (inches / 10))/3.2808` gives the wrong answer. — Dawood ibn Kareem, Sep 06 '16 at 20:59
I feel like the easier way to do this is probably something like: http://ideone.com/heEEQi — A_Elric, Sep 06 '16 at 21:00

Andy Turner · Accepted Answer · 2016-09-06T21:07:14.903

2

You can get the number of digits in an integer, i, using:

1 + Math.floor(Math.log10(i))

(not ceil(log10(i)), since that calculates that 1 has zero digits)

You then need to divide i by 10 to the power of that number:

i / Math.pow(10, 1 + Math.floor(Math.log10(i)))

e.g.

23 / Math.pow(10, 1 + Math.floor(Math.log10(23))) == 0.23

Ideone demo

Alternatively, if you think those floating point operations log and pow are too expensive, you can determine the number of digits with a loop:

int d = 1;
while (d < i) d *= 10;

then

System.out.println(i / (double) d);

(noting that you need to cast at least one of the numerator or denominator to a floating point type, otherwise it will use integer division).

edited Sep 06 '16 at 21:07

answered Sep 06 '16 at 20:54

Andy Turner

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Couldn't you also use `ceil` instead of `floor` and leave off the `1 +`? – HeatfanJohn Sep 06 '16 at 21:02
"(not ceil(log10(i)), since that calculates that 1 has zero digits)". – Andy Turner Sep 06 '16 at 21:03
Thank you! The first option was exactly what I was looking for. – Anders Tuft Sep 06 '16 at 21:05
1

Just remember that if you're storing this in a `double`, there'll be no difference between 1 inch and 10 inches. – Dawood ibn Kareem Sep 06 '16 at 22:15

score 0 · Answer 2 · answered Sep 06 '16 at 20:55

Try parsing to double like this, from a string:

Option 1

try
{
    int n = 2;
    int decimal = 23;
    String full = Integer.toString(n) + '.' + Integer.toString(decimal);
    double val = Double.parseDouble(full);
} catch (Exception e) //Or whatever exception 
{
    //Code
}

Option 2

Of course, there are simpler methods, like this:

try
{
    int n = 2; 
    int decimal = 23;
    double val = Double.parseDouble(n + "." + decimal);
} catch (Exception e) //Or whatever exception 
{
    //Code
}

I personally would recommend the solution directly above, since it is the simplest, and requires the least code.

Live Example for Second Option

score 0 · Answer 3 · answered Sep 06 '16 at 20:56

Simple implementation using Strings:

public class Mainclass {
    public static void main(String[] args) {
        Integer num = 1546;
        Integer num2 = 2;

        String s = num.toString();
        s = "0." + s;
        Double d = Double.parseDouble(s);
        System.out.println(num2+d ); // 2.1546
    }
}

How do I force a number with unknown digits behind the decimal mark?

3 Answers3

Option 1

Option 2