Why double is converted to int in JSON string

Question

I just coded to put an array of double values in the JsonObject. But, all my double values are converted to int values, when i print it. can someone help me understand what is happening behind? Please let me know the best way to put primitive arrays in JsonObject

public class JsonPrimitiveArrays {        
    public static void main(String[] args) {
        JSONObject jsonObject = new JSONObject();
        double[] d = new double[]{1.0,2.0,3.0};
        jsonObject.put("doubles",d);
        System.out.println(jsonObject);            
    }        
}

Output:

{"doubles":[1,2,3]}

Is it localization issues? – ChiefTwoPencils Feb 23 '14 at 08:20 — ChiefTwoPencils, Feb 23 '14 at 08:20
I'm sorry.What do you mean by localization issues? – Keerthivasan Feb 23 '14 at 08:21 — Keerthivasan, Feb 23 '14 at 08:21

score 7 · Accepted Answer · answered Feb 23 '14 at 08:31

Its in real not getting converted into int. Only thing happening is as JS Object its not showing .0 which is not relevant.

In your sample program, change some of the value from double[] d = new double[]{1.0,2.0,3.0} to

double[] d = new double[]{1.0,2.1,3.1} and run the program.

You will observer its in real not converting into int. The output you will get is {"doubles":[1,2.1,3.1]}

score 7 · Answer 2 · answered Feb 23 '14 at 08:40

Looking at toString of net.sf.json.JSONObject it eventually calls the following method to translate the numbers to String (source code here):

public static String numberToString(Number n) {
        if (n == null) {
            throw new JSONException("Null pointer");
        }
        //testValidity(n);

        // Shave off trailing zeros and decimal point, if possible.

        String s = n.toString();
        if (s.indexOf('.') > 0 && s.indexOf('e') < 0 && s.indexOf('E') < 0) {
            while (s.endsWith("0")) {
                s = s.substring(0, s.length() - 1);
            }
            if (s.endsWith(".")) {
                s = s.substring(0, s.length() - 1);
            }
        }
        return s;
    }

It clearly tries to get rid of the trailing zeroes when it can, (s = s.substring(0, s.length() - 1) if a String is ending in zero).

System.out.println(numberToString(1.1) + " vs " + numberToString(1.0));

Gives,

1.1 vs 1

score 6 · Answer 3 · answered Feb 23 '14 at 08:28

6

All numbers are floats in Javascript. So, 1.0 and 1 are the same in JS. There is no differenciation of int, float and double.

Since JSON is going to end up as a JS object, there is no use in adding an extra '.0' since '1' represents a float as well. I guess this is done to save a few bytes in the string that is passed around.

So, you will get a float in JS, and if you parse it back to Java, you should get a double. Try it.

In the mean time, if you are interested in the way it displays on screen, you can try some string formatting to make it look like '1.0'.

answered Feb 23 '14 at 08:28

Teddy

4,009
2
33
55

2

+1, I learnt that all numbers in Javascript are floats and the zeros are removed to save a few bytes, thanks :) – Keerthivasan Feb 23 '14 at 09:08
1

Not sure if the answer is accurate since the question is about Java, not JavaScript. – Jason Capriotti Aug 25 '17 at 02:24
3

@JasonCapriotti In the question, when a Java double array is converted into JSON, it looks like an integer array in the JSON string. I explained that there is no difference between float/double/int in Javascript. So, the ".0" is unnecessary. So the JSON library in java doesn't bother to add the ".0". I hope you understand that **JS**ON is **Javascript** Object Notation. – Teddy Aug 25 '17 at 03:30

Why double is converted to int in JSON string

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