Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!mailrus!ames!amdcad!nucleus!tim
From: tim@nucleus.amd.com (Tim Olson)
Newsgroups: comp.arch
Subject: Re: IEEE 754 "rem" query
Message-ID: <28412@amdcad.AMD.COM>
Date: 15 Dec 89 14:40:39 GMT
References: <1291@quintus.UUCP>
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Reply-To: tim@amd.com (Tim Olson)
Organization: Advanced Micro Devices, Inc., Austin, Texas
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In article <1291@quintus.UUCP> ok@quintus.UUCP (Richard A. O'Keefe) writes:
| I have read the IEEE 754 and IEEE 854 standards as carefully as I can,
| and it is not clear to me what the sign of x REM y should be when x=0.
| x REM y is "defined ... by the mathematical relation r = x - y.n"
| in section 5.1.  If we thought of this as involving an subtraction in
| the implementation, section 6.3's rule that "when the sum of two operands
| with opposite signs (or the difference of two operands with like signs)
| is exactly zero, the sign of that sum (or difference) shall be "+" in
| all rounding modes except round towards -oo, in which mode that sign 
| shall be "-".  However, x+x = x-(-x) retains the same sign as x even
| when x is zero."  This would mean that the sign of -0.0 REM y might be
| "+" or "-" depending on the sign of "y.n".  However, (a) the sign of y.n
| is not defined because this is a mathematical multiplication, not an IEEE
| one, and (b) if we took this reading the result of (-0.0 REM y) would
| depend on the rounding mode, and x REM y is "defined regardless of the
| rounding mode".
| 
| I think the result of (-0.0 REM y) should be -0.0 for any non-zero y, and
| that's what I get when I try it, but is that the intent of the standard?

Yes.  My copy of the Standard 754 (admittedly old) says:

      "... r = x - y*n, where n is the integer nearest the exact value
      x/y... If r=0, its sign shall be that of x."


	-- Tim Olson
	Advanced Micro Devices
	(tim@amd.com)