Java IO.pdf - Nguyen Dang Binh

}
}
catch (IOException e) {System.err.println(e);}
Here's a sample run:
% java RootFinder
Enter a number (-1 to quit):
87
The square root of 87 is 9.327379053088816
Enter a number (-1 to quit):
3.151592
The square root of 3.151592 is 1.7752723734683644
Enter a number (-1 to quit):
2,345,678
The square root of 2,345,678 is 1531.5606419596973
Enter a number (-1 to quit):
2.998E+8
The square root of 2.998E+8 is 1.7314733610425546
Enter a number (-1 to quit):
299800000
The square root of 299800000 is 17314.733610425545
Enter a number (-1 to quit):
0.0
The square root of 0.0 is 0.0
Enter a number (-1 to quit):
four
four is not a number I understand.
Enter a number (-1 to quit):
4
The square root of 4 is 2.0
Enter a number (-1 to quit):
(12)
(12) is not a number I understand.
Enter a number (-1 to quit):
-1
Java I/O
These results tell you a few things about the default number format on the platform where I
ran it (U.S. English Solaris, JDK 1.1.4). First, it doesn't understand exponential notation. The
square root of 2.998E+8 is not 1.7314733610425546; it's 1.7314733610425546E+4. The
number format parsed up to the first character it didn't recognize (E) and stopped, thus
returning the square root of 2.998 instead. You can also see that this number format doesn't
understand negative numbers represented by parentheses or words like "four." On the other
hand, it can parse numbers with thousands separators like 2,345,678. This is more than the I/O
libraries in most other languages can do. With the appropriate, nondefault number format,
Java could parse (12), four, and 2.998E+8 as well.
16.6 Decimal Formats
The java.text package contains a single concrete subclass of NumberFormat,
DecimalFormat. The DecimalFormat class provides even more control over how floating
point numbers are formatted:
public class DecimalFormat extends NumberFormat
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