Re: Exception throwing and guarentees about execution order
On Feb 2, 10:29 am, DerTop...@web.de wrote:
On 1 Feb., 10:12, "Dick Brown" <inva...@invalid.invalid> wrote:
As far as I understand 11.6 (6) of the Ada 95 Reference
Manual, Ada implementations may "lose" information, where
exactly an exception has been raised. I'm not sure whether I
understand Ada in that regard, but that's off-topic anyways
(if someone wants to correct me there, I'd be grateful
nonetheless).
Something like:
try
{
int i = 0;
if (...) throw(i);
int i = 1;
if (...) throw(i);
int i = 2;}
catch (...)
{
// i may be 0 or 1 or 2, no matter which throw(i) has brought us here
}
Now the real question: Does C++ have any similar provisions?
I've found nothing in the standard, am I right there?
IMHO, "losing" the information about where exactly an
exception has been raised is what exception handling is all
about:
Yes and no. You generally shouldn't be concerned about the
exact location in itself, but you often have to know how far
you've progressed in order to do a rollback. This is typically
handled by the memento pattern or some variation of transaction
handling, but his code (modified so it makes sense) can be
understood as a simplified stand-in for the memento pattern.
[...]
AFAIK, exception handling will be same under
C++/Java/C#/Ada95. Since the Ada95 compiler uses C++ code
internally, it would be quite surprising if Ada didn't provide
a feature that is present in C++.
There's no fundamental reason to suppose that exception handling
will be the same in different languages (and of course, Ada95
has nothing to do with C++, and there's certainly no requirement
that it use the same compiler). I couldn't find the liberty the
original poster talked about in the Ada standard, but it makes
sense from an optimizer point of view. Given something like:
int i = 0 ;
try {
f1() ;
i = 1 ;
f2() ;
x = f3( ++i ) + f4() ;
} catch ( ... ) {
// ...
}
, on many architectures, it might be advantageous to put i in a
register during certain parts (but not all) of the try block.
If the compiler does this, both C++ and Java require that if an
exception is thrown in f2, i == 1; most of the time, this means
that the compiler will have to write the value back to memory
before any function call (unless that function has an empty
exception specification in C++). Note that Java also guarantees
that if an exception is thrown in f4, i will be equal to 2 in
the catch block; in C++, it's unspecified whether i will be
equal to 1 or to 2. (In other words, exception handling is
different under Java and C++.)
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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