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Re: [PATCH][4.3] Expand lround inline for x86_64/i?86 SSE math
On 10/19/06, Brooks Moses <firstname.lastname@example.org> wrote:
Toon Moene wrote:
> Richard Guenther wrote:
>>I wonder if fortran specifies round differently, as the frontend
>>explicitly converts NINT(x) = INT(x + ((x > 0) ? 0.5 : -0.5)).
> Yep - sorry, don't have the reference handy.
For what it's worth, since I do have the reference handy, that's
essentially a direct translation of how the Fortran 95 standard defines
the NINT intrinsic.
So, does it define how the x + 0.5 is carried out with respect to
intermediate rounding before converting to INT? Literaly writing
the above yields 1.0 for NINT ( 0.5 - epsilon ) assuming the hardware
rounds to nearest even for the addition.