The strong exception guarantee guaranties that if an exception is
thrown, the function will have no side effect. Of course, not all
function can support this (a file I/O error in the middle of writing
will have side effects), but often it can and it's generally good
practice to offer the guaranty whenever possible.
<http://en.wikipedia.org/wiki/Exception_guarantees>
But if one of your function has an 'out' parameter, it's impossible to
implement the strong guarantee, as illustrated by this trivial example:
void testOut(out int a) {
throw new Exception("hello!");
}
void main() {
int a = 2;
try
testOut(a);
finally
writeln(a);
}
Prints:
0
object.Exception: hello!
This happens because the out parameter gets reset to its default value
as soon as you enter the function, so you can't throw an exception
before it has been changed.
So should 'out' be reformed to behave more like a return value? I'm not
sure. But I think this is something to keep in mind when using out
parameters.
--
Michel Fortin
[email protected]
http://michelf.com/
