[Rd] mean(x) != mean(rev(x)) different with x <- c(NA, NaN) for some builds

[Prof Brian Ripley]

 From ?NA

      Numerical computations using ‘NA’ will normally result in ‘NA’: a
      possible exception is where ‘NaN’ is also involved, in which case
      either might result.

and ?NaN

      Computations involving ‘NaN’ will return ‘NaN’ or perhaps ‘NA’:
      which of those two is not guaranteed and may depend on the R
      platform (since compilers may re-order computations).

fortunes::fortune(14) applies (yet again).

On 01/04/2017 04:50, Henrik Bengtsson wrote:
> In R 3.3.3, I observe the following on Ubuntu 16.04 (when building
> from source as well as for the sudo apt r-base build):
>
>> x <- c(NA, NaN)
>> mean(x)
> [1] NA
>> mean(rev(x))
> [1] NaN
>
>> rowMeans(matrix(x, nrow = 1, ncol = 2))
> [1] NA
>> rowMeans(matrix(rev(x), nrow = 1, ncol = 2))
> [1] NaN
>
>> .rowMeans(x, m = 1, n = 2)
> [1] NA
>> .rowMeans(rev(x), m = 1, n = 2)
> [1] NaN
>
>> .rowSums(x, m = 1, n = 2)
> [1] NA
>> .rowSums(rev(x), m = 1, n = 2)
> [1] NaN
>
>> rowSums(matrix(x, nrow = 1, ncol = 2))
> [1] NA
>> rowSums(matrix(rev(x), nrow = 1, ncol = 2))
> [1] NaN
>
> I'd expect NA to trump NaN in all cases (with na.rm = FALSE).  sum()
> does not have this problem and returns NA in both cases (*).
>
> For the same R version build from source on RHEL 6.6 system
> (completely different architecture), I get the expected result (= NA)
> for all of the above cases, e.g.
>
>> x <- c(NA, NaN)
>> mean(x)
> [1] NA
>> mean(rev(x))
> [1] NA
> [...]
>
> Before going insane trying to troubleshoot this, I have a vague memory
> that this, or something related to this, has been discussed
> previously, but I cannot locate it.
>
> Is the above a bug in R, a FAQ, a build error, overzealous compiler
> optimization, and / or ...?
>
> Thanks,
>
> Henrik


-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Emeritus Professor of Applied Statistics, University of Oxford