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Re: Comments on the suggestion to use infinite precision math for wide int.

On Mon, Apr 8, 2013 at 10:39 PM, Lawrence Crowl <> wrote:
> On 4/8/13, Kenneth Zadeck <> wrote:
>> The other problem, which i invite you to use the full power of
>> your c++ sorcery on, is the one where defining an operator so
>> that wide-int + unsigned hwi is either rejected or properly
>> zero extended.  If you can do this, I will go along with
>> your suggestion that the internal rep should be sign extended.
>> Saying that constants are always sign extended seems ok, but there
>> are a huge number of places where we convert unsigned hwis as
>> the second operand and i do not want that to be a trap.  I went
>> thru a round of this, where i did not post the patch because i
>> could not make this work.  And the number of places where you
>> want to use an hwi as the second operand dwarfs the number of
>> places where you want to use a small integer constant.
> You can use overloading, as in the following, which actually ignores
> handling the sign in the representation.
> class number {
>         unsigned int rep1;
>         int representation;
> public:
>         number(int arg) : representation(arg) {}
>         number(unsigned int arg) : representation(arg) {}
>         friend number operator+(number, int);
>         friend number operator+(number, unsigned int);
>         friend number operator+(int, number);
>         friend number operator+(unsigned int, number);
> };
> number operator+(number n, int si) {
>     return n.representation + si;
> }
> number operator+(number n, unsigned int ui) {
>     return n.representation + ui;
> }
> number operator+(int si, number n) {
>     return n.representation + si;
> }
> number operator+(unsigned int ui, number n) {
>     return n.representation + ui;
> }

That does not work for types larger than int/unsigned int as HOST_WIDE_INT
usually is (it's long / unsigned long).  When you pass an int or unsigned int

number operator+(unsigned long ui, number n);
number operator+(long ui, number n)

you get an ambiguity.  You can "fix" that by relying on template argument
deduction and specialization instead of on overloading and integer conversion

> If the argument type is of a template type parameter, then
> you can test the template type via
>     if (std::is_signed<T>::value)
>       .... // sign extend
>     else
>       .... // zero extend
> See

Yes, if we want to use the standard library.  For what integer types
is std::is_signed required to be implemented in C++98 (long long?)?
Consider non-GCC host compilers.


> If you want to handle non-builtin types that are asigne dor unsigned,
> then you need to add a specialization for is_signed.
> --
> Lawrence Crowl

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