On Fri, Sep 19, 2014 at 5:44 AM, Mark Dickinson

On Fri, Sep 19, 2014 at 3:16 AM, Alexander Belopolsky

wrote: The standards are influenced by the limitation inherent in many languages where ints have finite range and cannot represent floor() of many finite floating point values. Python does not have this limitation. (Granted - PEP 3141 could do a better job explaining why floor, ceil, round, //, etc. should return Integer rather than Real.)

Indeed. FWIW, I think it was a mistake to change the return type of math.floor and math.ceil in Python 3. There's no longer any way to spell the simple, fast, float->float floor operation. It would be nice to have those basic floating-point math operations accessible again.

It looks like the Scheme standards committee agree with you. PEP-3141 references: http://groups.csail.mit.edu/mac/ftpdir/scheme-reports/r5rs-html/r5rs_8.html#... There is a later version that specifically discusses the behavior for +-Inf and Nan. http://www.r6rs.org/final/html/r6rs/r6rs-Z-H-14.html#node_sec_11.7.2 Here is the pertinent section: Although infinities and NaNs are not integer objects, these procedures return an infinity when given an infinity as an argument, and a NaN when given a NaN. (floor -4.3) ⇒ -5.0 (ceiling -4.3) ⇒ -4.0 (truncate -4.3) ⇒ -4.0 (round -4.3) ⇒ -4.0 (floor 3.5) ⇒ 3.0 (ceiling 3.5) ⇒ 4.0 (truncate 3.5) ⇒ 3.0 (round 3.5) ⇒ 4.0 (round 7/2) ⇒ 4 (round 7) ⇒ 7 (floor +inf.0) ⇒ +inf.0 (ceiling -inf.0) ⇒ -inf.0 (round +nan.0) ⇒ +nan.0 Also note that a floating point representation is used for the results when the argument is floating point number. Scheme does not use the internal representation (IEEE-754 or long int or....) to determine if a number is an integer or real but rather its value. Here is an example from a Scheme session: (integer? 2) ;Value: #t (integer? 2.0) ;Value: #t (integer? 2.1) ;Value: #f (#t and #f are equivalent to True and False.) I think PEP-3141 incorrectly interpreted the meaning of "integer" to imply the use of an "integer representation" while I think the Scheme standard implies an "integer value"; i.e. x is an "integer" iff x-round(x) == 0.

-- Mark