问题
While this code will raise indexError:
In [1]: lst = [1, 2, 3]
In [2]: lst[3]
IndexError: list index out of range
Slicing the list with "out of range index" will not produce any error.
In [3]: lst[3:]
Out[3]: []
What is the rationale of this design?
回答1:
When you are accessing an element in a list whose index is beyond its length, we cannot return anything. (There is no way we can represent an element which is not there). That's why the error is thrown. But when you are slicing, you are making a sliced COPY of the original list and that new list can be empty if the start or end are not valid.
回答2:
It's nice being able to test if an element exists:
if sys.argv[2:]:
# do something with sys.argv[2], knowing it exists
来源：https://stackoverflow.com/questions/20033201/python-index-out-of-range-in-list-slicing