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

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;
        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;

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
      .... // zero extend


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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