Re: Interrupted exception chaining
On 9/25/2012 3:25 AM, raphfrk@gmail.com wrote:
Is there a recommended way of "chaining" interrupted exceptions?
I mostly agree with Eric. You shouldn't suppress interrupts unless you
must. If for example you are implementing a Runnable, you can't throw
InterruptedException but must deal with it somehow. Then catching an
interrupted exception is reasonable.
In this case however it's 100% your code and you could throw
InterruptedException. That would defeat the purpose of the code you
show, of course, but then that is also our point: what you are doing is
rather questionable.
There's no standard pattern for this because it's not done. I'd say the
closest we have to to use a finally block to make certain the interrupt
flag is restored.
/** A questionable method: wait without throwing InterruptedException. */
public void uninterruptableWait(Object c) {
boolean interrupted = false;
try {
synchronized(c) {
for(;;) {
try {
c.wait();
return;
} catch(InterruptedExcpetion ex ) {
interrupted = true;
}
}
}
} finally {
if(interrupted == true ) Thread.currentThread().interrupt();
}
}
If that interrupt was unexpected, and causes a stack trace, then it
would be nice if it could include the details from the thrown
exception.
If you really need the details of the exception but can't throw from
your local use site, I'd say wrapping the exception in a
RuntimeExcpetion and throwing that is best. (Exception don't ever
throwing RuntimeException; subclass it and throw your specific exception.)
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From: Roedy Green <see_website@mindprod.com.invalid>
Newsgroups: comp.lang.java.softwaretools
Subject: Re: replacement IDE for IntelliJ
Date: Fri, 23 Dec 2011 13:04:41 -0800
Organization: Canadian Mind Products
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On Fri, 23 Dec 2011 03:14:01 GMT, eric@invalid.com (EricF) wrote,
quoted or indirectly quoted someone who said :
But if it
doesn't work for you, then it is time to move on.
I have got it going again with a virgin install and not restoring
anything from before.
--
Roedy Green Canadian Mind Products
http://mindprod.com
If you can't remember the name of some method,
consider changing it to something you can remember.
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From: Eric Sosman <esosman@ieee-dot-org.invalid>
Newsgroups: comp.lang.java.help
Subject: Re: Beginner Java question - need help writing a Fraction class
Date: Sun, 05 Dec 2010 09:10:44 -0500
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On 12/5/2010 5:21 AM, Erik wrote:
I am trying to write a method class in java that handles fractions. I
have a driver that allows the user to enter the numerator and
denominator each separately. Then my program asks for another
fraction and attempts to add the two fractions together. In the driver
program I commented out the code after adding the fractions because I
figured once I got the fractions to add up I could figure out the
rest. Currently when I run the program the only value I get for the
sum of the two fractions is 0/1.
The basic requirements are to write a driver and fraction class that
performs addition, multiplication, prints the fraction, and prints as
a double.
*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*
*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*==*
Here is the code for my driver program:
public static void main(String[] args)
{
Scanner stdIn = new Scanner(System.in);
Fraction c, d, x; // Fraction objects
System.out.println("Enter numerator; then denominator.");
c = new Fraction(stdIn.nextInt(), stdIn.nextInt());
c.print();
System.out.println("Enter numerator; then denominator.");
d = new Fraction(stdIn.nextInt(), stdIn.nextInt());
d.print();
x = new Fraction(); // create a fraction for number 0
System.out.println("Sum:");
x.add(c).add(d);
x.print();
Your add() method does not change the value of "this", but
instead creates and returns a brand-new Fraction and leaves "this"
as it was before. When you calculate x.add(c).add(d) the first
thing that happens is x.add(c), which generates a new Fraction
representing the sum of x and c -- let's just call it "u" for the
moment, OK? The next step is u.add(d), which generates yet another
Fraction representing the sum of u and d; let's call this one "v".
And then you print "x", but the sum you wanted was in "v". You
can't actually get at "v" (or at "u") because you never kept a
reference to either of those Fractions, but they're the Fractions
that held the answer (intermediate answer). "x" is still just as it
was to begin with.
You don't really want "x" to be a brand-new Fraction with the
default value; you want it to be the sum of "c" and "d". So try
something more like
Fraction x = c.add(d);
// This method finds the sum of 2 fractions.
public Fraction add(Fraction otherFraction)
{
Fraction sum = new Fraction(1,1);
int newdenom = this.denominator * otherFraction.denominator;
sum.denominator = newdenom;
sum.numerator = this.numerator * otherFraction.denominator +
otherFraction.numerator * this.denominator;
return new Fraction(sum.numerator, sum.denominator); // originally
return sum;
Returning "sum" would have been just fine. Incidentally, you
will probably want to make add() a little bit smarter. As things
stand, if you add 1/2 and 1/6 you will get 8/12 -- which is correct,
but less pleasing than 2/3. Nor is this just a matter of aesthetics:
when you start doing lengthy calculations with fractions you will
find that the sizes of numerator and denominator can become large,
and if you don't do something about common factors they will become
*very* large. For example, if you try to compute
1/1 + 1/2 + 1/2 + ... + 1/13
you will arrive at 19802759040/6227020800, whose numerator and
denominator are both too large for "int", and your calculation will
go astray. The reduced value 1145993/360360 is comfortably in
range, and will cause no unpleasant surprises.
--
Eric Sosman
esosman@ieee-dot-org.invalid