Carlo, "Before aborting, assert() outputs the name of its source file and the number of the line on which it appears. This makes assert() a useful debugging aid." Now as users will be not interested in my source code, I propose to let the run-time error be an exception. Regards, Maarten. "Maarten Kronenburg" <M.Kronenburg@inter.nl.net> wrote in message news:e5uf5p$53j$1@sea.gmane.org...
Carlo, Yes the unsigned_integer decrement does a test in the predecrement operator, but it is only the test of the sign, which will be fast. Users that are VERY performance critical then still have the option of not using unsigned_integer. Yes the unsigned_integer is not a proper group; on the other hand, when a user runs the following program: unsigned int x = -2; cout << x << endl; and the user gets 4294967294 then most users will be surprised, and not uderstand what is going on. The following program: unsigned_integer x = -2; cout << x << endl; will generate an error, which in my opinion is still more intuitive for most users who don't know what a modular integer is. Regards, Maarten.
"Carlo Wood" <carlo@alinoe.com> wrote in message news:20060604020720.GA25970@alinoe.com...
On Sat, Jun 03, 2006 at 02:54:53PM +0200, Maarten Kronenburg wrote:
The assertion for unsigned_integer is an interesting possibility. In "C++ in a nutshell" I read: "If the expression evaluates to 0, assert prints a message to the standard error file and calls abort." Personally I would rather throw an exception, so that the user can determine what should happen if an unsigned_integer becomes negative.
There is nothing to decide, it is a bug when that happens. The best you can do is to dump a core (which is what assert does) so that the developer can examine as close as possible to the bug what has happened.
Once the program works, an integer that should never become negative will never become negative anymore, and no testing for less than zero is needed. You can still use a normal signed integer and just omit the test (assertion).
Thus,
void Foo::release(void) { assert(x > 0); if (--x == 0) do_something(); }
where x can be perfectly a signed integer, and no testing overhead is done in the final application.
rather than
void Foo::release(void) { if (--x == 0) do_something(); }
where x is a special type, solely to throw an exception for the case that the program is bugged, needing not only a total new type (with no mathematical meaning, and no simularity with builtin-types*) ), but also the cost of testing as part of the pre-decrement operator, which can't be removed from the final application.
*) Because, at least unsigned int is modular: both int (assuming no overflow occurs) and unsigned int are rings. While an infinite precision unsigned integer wouldn't even be a proper group.
-- Carlo Wood <carlo@alinoe.com> _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
_______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost