phone numbers that start with one 0 get treated as int

[pcasaes]

Hello,

Here in brazil we have phone numbers that start with 0800 and they get treated as ints thereby losing the first 0.

Tracked down the error to this bit of code in XML.java

    try {
        char initial = string.charAt(0);
        boolean negative = false;
        if (initial == '-') {
            initial = string.charAt(1);
            negative = true;
        }
        if (initial == '0' && string.charAt(negative ? 2 : 1) == '0') {
            return string;
        }
        if ((initial >= '0' && initial <= '9')) {
            if (string.indexOf('.') >= 0) {
                return Double.valueOf(string);
            } else if (string.indexOf('e') < 0 && string.indexOf('E') < 0) {
                Long myLong = new Long(string);
                if (myLong.longValue() == myLong.intValue()) {
                    return new Integer(myLong.intValue());
                } else {
                    return myLong;
                }
            }
        }
    }  catch (Exception ignore) {
    }

It seems to want to convert any value 0XNNNN... (where X isn't 0 and N is any digit) as an int.
Any particular reason for this?

[douglascrockford]

Phone numbers should be represented as strings. Wrap them in quotes.

[pcasaes]

Hello,

The method doesn't convert 00222 into an int. If it the second character is a 0 is keeps it as a string.
Is this behaviour wrong? If not why not?

[douglascrockford]

Integers in JSON may not start with a leading zero. See http://json.org/

[pcasaes]

Ah! now I understand.
I was focusing to much on ints that I forgot doubles.

I do believe the code is not following that specification correctly though.

changed this bit of code on XML.java, line 330:
if (initial == '0' && string.charAt(negative ? 2 : 1) == '0') {
return string;
}
to
if (initial == '0' && string.charAt(negative ? 2 : 1) != '.') {
return string;
}

tested it on:
-123
123
-0.123
0.123

worked perfectly. thanks.