API Contents - SCJP Sun Certified Programmer for Java Platform SE6 - page 311

12. System.out.println(fr.parse(s));
13. }catch(ParseException e) {
14. e.printStackTrace();
15. }
The string being parsed is ”123,45” . In the U.S. locale, the comma is treated as a visual
format and is ignored, so the resulting number is the integer 12345 . In the France locale, the
comma is a decimal separator, so the resulting number is the double 123.45 . The output of
the code is
12345
123.45
The parse method only parses the beginning of a string. After it reaches a character that
cannot be parsed, the parsing stops and the value is returned. See if you can determine the
output of the following statements:
NumberFormat nf = NumberFormat.getInstance();
try {
String one = “456abc”;
String two = “-2.5165e10”;
String three = “x85.3”;
System.out.println(nf.parse(one));
System.out.println(nf.parse(two));
System.out.println(nf.parse(three));
}catch(ParseException e) {
e.printStackTrace();
}
The NumberFormat object uses the default locale to parse ”456abc” . When the 'a' char-
acter is reached, the parsing stops and 456 is returned. Similarly, the String two is parsed
into -2.5165 . Parsing ”x85.3” throws a ParseException because the beginning of the
string cannot be parsed. The output of the code is
456
-2.5165
java.text.ParseException: Unparseable number: “x85.3”
I do not think the exam will test your knowledge of such details about the
parse method and the point at which parsing fails, but it is a good trait to
understand. Instead, expect a question that successfully parses a string
to a number.