Hi,
?x <- c(rep(1800000003L, 10000000), -rep(1200000002L, 15000000))
This is correct:
?> sum(as.double(x))
?[1] 0
This is not:
?> sum(x)
?[1] 4996000
Returning NA (with a warning) would also be acceptable for the latter.
That would make it consistent with cumsum(x):
?> cumsum(x)[length(x)]
?[1] NA
?Warning message:
?Integer overflow in 'cumsum'; use 'cumsum(as.numeric(.))'
Thanks!
H.
R version 2.14.0 (2011-10-31)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
?[1] LC_CTYPE=en_CA.UTF-8 ? ? ? LC_NUMERIC=C
?[3] LC_TIME=en_CA.UTF-8 ? ? ? ?LC_COLLATE=en_CA.UTF-8
?[5] LC_MONETARY=en_CA.UTF-8 ? ?LC_MESSAGES=en_CA.UTF-8
?[7] LC_PAPER=C ? ? ? ? ? ? ? ? LC_NAME=C
?[9] LC_ADDRESS=C ? ? ? ? ? ? ? LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats ? ? graphics ?grDevices utils ? ? datasets ?methods ? base
--
Herv? Pag?s
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024
E-mail: hpages at fhcrc.org
Phone: ?(206) 667-5791
Fax: ? ?(206) 667-1319