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From: Maarten Kronenburg (M.Kronenburg_at_[hidden])
Date: 20060530 17:21:25
There is however one catch to the
unsigned_integer: the expression
abc
Then look in [expr.add]:
"The additive operators group from
lefttoright".
The integer binary operator
first clones the rhs, negates it,
and then adds to it the lhs.
This guarantees that when b or c
are nonzero unsigned_integer,
then ALWAYS an exception
is thrown.
BUT what about
a(b+c)
Now (b+c) returns a temporary
integer (not unsigned_integer),
which is negated (NO exception)
and a is added to it.
So although the behaviour is
compilerindependent because of the
[expr.add] remark, the use of braces
in expressions may change the
behaviour of an unsigned_integer
expression, that is while
abc
may throw an exception,
a(b+c)
may not.
Regards, Maarten.
"Maarten Kronenburg" <M.Kronenburg_at_[hidden]> wrote in message
news:e5gvs8$jbm$1_at_sea.gmane.org...
> If you don't mind I start a new thread here.
> The unsigned infinite precision integer is
> different from the base type unsigned int,
> which is actually a modular integer with
> modulus 2^n.
> Therefore two integer derived classes are
> needed: unsigned_integer and modular_integer.
> The unsigned_integer is an infinite precision
> integer which can only be positive or zero.
> The negate() of a nonzero unsigned_integer
> will always throw an exception.
> A subtraction which results in a negative value
> will do the same; therefore in my opinion
> there is no fundamental problem with this,
> as negation is subtraction from zero.
> The modular_integer has a static method
> static void set_modulus( const integer & ).
> When the modulus is not set, it is zero,
> in that case the modular_integer is identical to integer.
> Users that like an unsigned integer with
> a negate() that always works, will have to
> use a modular_integer and set its modulus
> to a positive value.
> In the document I will specify unsigned_integer
> and modular_integer, and thus implementations
> can provide them.
> Regards, Maarten.
>
>
>
>
>
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