Bounded integer types (was: C bug of the day)

[Ed Avis <ed@membled.com>]

James Rogers <jimmaureenrogers@att.net> writes:

>Compare this with defining a numeric type with all integers from
>1 through 101:
>
>type My_Ints is range 1..101;
>
>In this case Ada automatically provides all the checking and
>numeric operators for you.

FWIW, I have been working on an equivalent for C++, as

    bounded<int, 1, 10> i;

to say that i has values in the range 1..10 (I use .. to mean an
inclusive range, I hope that it doesn't mean something different in
Ada-land).  Then operations like addition are implemented so that all
range checking can be done statically - no checks at run time, except
for an explicit range_cast<int, x, y>() function which turns an int
into a bounded<int, x, y>.

So for example the type of (i + i) would be bounded<int, 2, 20>.  If
you wanted to assign this to a variable that holds only values 1..10
you would need a range_cast and a possible exception at run time.  But
assigning it to a variable that can take a larger range (or to a plain
old int) is fine.

Apart from operator=, there are no mutating operations like ++ or +=,
because these cannot be statically checked.  And operator= takes only
another bounded<int, ...> or a compile-time constant.  The aim of the
bounded<int, ...> type is to have no overhead beyond an ordinary int
(if the compiler is reasonably good) and to have no exceptions thrown
except when the programmer asks for them with range_cast.

To emphasize one of those points: you cannot assign or initialize a
bounded<int, ...> with a value that's not known at compile time.  The
default value of an uninitialized bounded type is the minimum value of
the range.  If I wanted to declare i with range 1..10 and initialize
it to 5, I'd have to write

    bounded<int, 1, 10> i = bounded<int, 6, 6>();

This is a horrible syntax but I couldn't find any much cleaner way of
specifying that the constructor must take a compile-time constant.

I realized this was a bit limiting so I have also started work on a
throwing_bounded class which does have operator=, operator++ and
friends, and throws exceptions at run time if you go outside the
range.

Followups set to the C++ newsgroup only.

-- 
Ed Avis <ed@membled.com>

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