Question

这里发生了什么？

预期：

>>> datetime.min - timedelta(days=1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: date value out of range

意外的：

>>> datetime.min - timedelta(days=2)
datetime.datetime(1, 0, 255, 0, 0)

>>> datetime.min > (datetime.min - timedelta(days=2))
True

在python中，当你从datetime.min中减去这些值意味着什么？他们代表什么日期？为什么有些情况不会触发OverflowError？

Answer 1

因为您需要升级到修复此错误的Python 2.6或更高版本。

$ python2.5 -c 'import datetime; print(datetime.datetime.min - datetime.timedelta(days=2))'
0001-00-255 00:00:00
$ python2.6 -c 'import datetime; print(datetime.datetime.min - datetime.timedelta(days=2))'
Traceback (most recent call last):
  File "<string>", line 1, in <module>
OverflowError: date value out of range
$ python2.7 -c 'import datetime; print(datetime.datetime.min - datetime.timedelta(days=2))'
Traceback (most recent call last):
  File "<string>", line 1, in <module>
OverflowError: date value out of range
$ python3.3 -c 'import datetime; print(datetime.datetime.min - datetime.timedelta(days=2))'
Traceback (most recent call last):
  File "<string>", line 1, in <module>
OverflowError: date value out of range

您是否需要有人追踪错误编号，补丁和python-dev讨论，或者是否有足够的信息供您使用？

datetime值小于datetime.min的含义是什么？

1 个答案: