Re: [Boost-users] [review][constrained_value] Review of Constrained Value Library begins today

Subject: Re: [Boost-users] [review][constrained_value] Review of Constrained Value Library begins today
From: Robert Kawulak (robert.kawulak_at_[hidden])
Date: 2008-12-04 16:02:58

Hi Jesse,

From: "Jesse Perla" 

>4. If it is true in general, then this statement:
>http://student.agh.edu.pl/~kawulak/constrained_value/constrained_value/
>rationale.html#constrained_value.rationale.no_floats
>makes this library practically useless.

This is the problem of limited usefulness of built-in floating point arithmetic
rather than of this library. If you need constrained floats, use a library for
reliable, repeatable floating-point computations.

>typedef bounded<double> interest_rate;
>interest_rate r1, r2;
>change_bounds(r1, .8, .99);
>
>Now the interest rate bounds on r2 don't exist, right?  This is a big
>problem for me... It would mean that the bounds information is not kept on
>the type, but on the instantiation...

Sorry, but this makes no sense to me in the context of constrained values. If
the current bounds are [0, 1] and you have r1 == .5 and r2 == .9, then what
changing the bounds for the whole type to [.8, .99] would mean for r1? How the
type is supposed to know that there is an instance that is not valid according
to the new bounds? The type should keep handles to all its instances? Or the
bounds should change anyway, leaving the instance in invalid state? So what
would be the point of bounds checking, if you can't guarantee that once you
assign a valid value then the object will remain valid?

>6) Is there any way to get the library to compile: std::max(2, i) instead of
>using i.value() as mentioned in the tutorial for a constrained i?

No. These are just the rules for template functions. Implicit conversions are
not taken into consideration in such case, so both the arguments have to be of
identical type.

>This
>really is essential to make the library equivalent to normal operations, but
>with extra type information.  Again, I am willing to forgo ease of use on
>the bounds checking for this...  What I really want is the instantiation to
>behave EXACTLY the same as an underlying type.  So overloading the ++, +,
>etc. are useful, but I would go further to say that perhaps everything could
>be done by just going back to the underlying type...

What you ask for is not possible in C++. You cannot have one type and use it
exactly the same way as other type. I'll give you some examples (having
constrained<string> cs):

        const char * s = cs.c_str(); // not possible

Type constrained<string> can't have members that string has.

        char c = cs[0]; // not possible

Type constrained<string> can't have operators that string has. How would you
define such operator for constrained class template? How would you know what
should be its return type in general case (without decltype available, of
course)?

Best regards,
Robert