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From: "redi at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug libstdc++/94823] modulo arithmetic bug in random.tcc
Date: Fri, 09 Oct 2020 14:59:33 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D94823

--- Comment #7 from Jonathan Wakely <redi at gcc dot gnu.org> ---
Despite the code being correct, I think it would be better to hoist this br=
anch
out of the loop:

          if (__k =3D=3D 0)
            __r2 +=3D __s;
          else if (__k <=3D __s)
            __r2 +=3D __kn + _M_v[__k - 1];
          else
            __r2 +=3D __kn;

We can do the k=3D0 case first, where we know that (k+p)%n=3D=3Dp and (k+q)=
%n=3D=3Dq,
and we also know that for that first iteration begin[0]^begin[q]^begin[n-1]=
 is
simply 0x8b8b8b8bu because every element has that value:

    // k =3D=3D 0, every element in [begin,end) equals 0x8b8b8b8bu
      {
        uint_least32_t __r1 =3D 1371501266u;
        uint_least32_t __r2 =3D __r1 + __s;
        __begin[__p] +=3D __r1;
        __begin[__q] +=3D __r2;
        __begin[0] =3D __r2;
      }

The we can loop up to __s+1

    for (size_t __k =3D 1; __k <=3D __s; ++__k)
      ...

And then up to m

    for (size_t __k =3D __s + 1; __k < __m; ++__k)
      ...

This unasks the question of whether begin[-1ul % n] is the right element or=
 has
the right value.=