小编Bra*_*ney的帖子

该文档描述.Machine$double.eps为最小的正浮点数x1 + x != 1.所以我希望以下产生1:

options(digits=17)
1 + .Machine$double.eps
# [1] 1.0000000000000002

但事实并非如此.似乎我必须尽可能低到0.5*.Machine$double.eps达到预期的四舍五入之前.我误解了一些事情.Machine$double.eps吗？这个平台依赖吗？(下面的sessionInfo())

> sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_CA.UTF-8/en_CA.UTF-8/en_CA.UTF-8/C/en_CA.UTF-8/en_CA.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
[1] tools_3.0.2

谢谢