Question

所以我决定制作一个脚本，告诉我有多少时间，直到Unix时间重置（在32位系统上它是2038年的问题，2038-01-19 03:14:07）我会很酷想通了，但这段代码看起来非常糟糕，我知道必须有更好的方法来做到这一点。我想过使用divmod，但是看不出有什么帮助。这是我的代码，我很想知道如何重构这一点，并提高可读性。

time = Time.now

end_of_unix_time = Time.gm(2038, 1, 19, 3, 14, 7).getlocal

diff = end_of_unix_time - time

years = (diff / 60 / 60 / 24 / 365).floor
days  = (diff / 60 / 60 / 24 - (years * 365)).floor
hours = (diff / 60 / 60      - (years * 365 * 24)           - (days * 24)).floor
min   = (diff / 60           - (years * 365 * 24 * 60)      - (days * 24 * 60)      - (hours * 60)).floor
sec   = (diff                - (years * 365 * 24 * 60 * 60) - (days * 24 * 60 * 60) - (hours * 60 * 60) - (min * 60)).floor

puts "#{years} years"
puts "#{days} days"
puts "#{hours} hours"
puts "#{min} minutes"
puts "#{sec} seconds"

Answer 1

end_of_unix_time = Time.at(2 ** 31 - 1)
diff = end_of_unix_time - Time.now
years, days, hours, min, sec =
[60, 60, 24, 365].each_with_object([diff]){|d, a| a[0..0] = a[0].divmod(d)}

puts "#{years} years"
puts "#{days} days"
puts "#{hours} hours"
puts "#{min} minutes"
puts "#{sec} seconds"

如何重构这段代码，找到Unix时间的结束？

1 个答案: