Does remquo work?

[Ulrich Drepper]

On Sun, 2002-06-09 at 10:37, Stephen L Moshier wrote:

> Responding to an inquiry, I tried to use "remquo" in a sentence
> and it doesn't seem to make sense.  The standard says (I think) that
> along with the remainder it returns at least 3 low-order bits of the
> integer-valued quotient.  So if you put, say, 4.1 in the denominator
> and 7 * 4.1 in the numerator, the quotient should read 7.
> Instead, it returns 2:

There were three mistakes in the implementation.  I hope I fixed all of
them.  At least I now get the correct result with your example code.

Thanks,

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