1. RE: Euphoria versus Java
rforno at tutopia.com wrote:
> I'm using a bit of Java now, and I noticed
> the following: while Java routinely performs
> a subscript checking (unlike C/C++), it does
> not verify if byte, short, integer or long variables
> overflow. So, if you have:
> byte x = 127;
> x = x + 1;
> you end up with x = -128 instead of getting an
> overflow error. Please correct me if there is a
> way of checking for this kind of error
> (someone in the list may know more Java than I do ;)).
> I think this is another point in favor of Euphoria
> that can be stressed.
I don't know Java that well, but I don't think there
is any way to "turn on" integer overflow detection,
or catch it as an exception. If integer values
become greater than what the programmer anticipated,
the program will quietly fail.
In Euphoria, integer values overflow automatically
into floating-point values.
The Translator works hard to eliminate checks
for integer overflow. When it proves to itself
that overflow can't occur, it generates
smaller/faster C code. It can usually eliminate most
of the checks this way.
Regards,
Rob Craig
Rapid Deployment Software
http://www.RapidEuphoria.com
2. RE: Euphoria versus Java
- Posted by rforno at tutopia.com
Apr 08, 2003
Pete:
I didn't know that C checked for integer overflow in some way. Rather, I
thought that it didn't check it at all, for performance reasons.
Testing this hypothesis, I programmed a simple loop in C. When an integer
would exceed the maximum corresponding integer size (say 32767), it simply
(this time actually "simply") reverts to a perfectly fixed value, that is,
the maximum negative number (assuming you were adding 1 to it), and not to
an indeterminate or rubbish value. This means it doesn't disturb what the
hardware does.
I made the same test in Java, with identical results. Moreover, a loop using
pre-defined variables (int, short, long, etc.), not objects, in Java under
Micro$oft Visual J++ 6.0 takes *less* time, believe it or not, than the same
one under Visual C++. This suggests Java performs at least fewer or the same
tests than C++, or maybe it has better optimizations. Strangely enough, the
timings were the same, no matter the resulting executable was .class or
.exe. I have no explanation for this.
Regards.
PS: Do you know if you can do in Java the same that you do in C#, as you
explained?
----- Original Message -----
From: Pete Lomax <petelomax at blueyonder.co.uk>
To: EUforum <EUforum at topica.com>
Sent: Monday, April 07, 2003 3:31 AM
Subject: Re: Euphoria versus Java
On Sun, 6 Apr 2003 15:32:06 -0300, rforno at tutopia.com wrote:
>I'm using a bit of Java now, and I noticed the following: while Java
>routinely performs a subscript checking (unlike C/C++), it does not verify
>if byte, short, integer or long variables overflow. So, if you have:
>byte x = 127;
>x = x + 1;
>you end up with x = -128 instead of getting an overflow error.
>Please correct me if there is a way of checking for this kind of error
>(someone in the list may know more Java than I do ;)).
Java uses the ansi standard C overflow checking model. Basically, when
overflow occurs, some useless rubbish is stored in the result field...
and then program execution continues at the next statement 
In C# instead of z=x+y you can just simply(!!) write:
try {
z = checked(x + y);
}
catch (System.OverflowException e) {
System.Console.WriteLine(e.ToString());
}
==^^===============================================================
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