Why doesn't Python throw an error for this? [duplicate]

Answer 1

Because [2:3] is from 4 to the next ele - 1, which returns 4.

Slicing never raises an error. The least it can do is return an empty list/tuple/string (depending on the type of course):

>>> a[12312312:]
[]

[start:end:step]

So index 2 is 4, then end - 1 is index 2 which is 4.

Answer 2

As others answered, Python generally doesn't raise an exception for out-of-range slices. However, and this is important, your slice is not out-of-range. Slicing is specified as a closed-open interval, where the beginning of the interval is inclusive, and the end point is exclusive.

In other words, [2:3] is a perfectly valid slice of a three-element list, that specifies a one-element interval, beginning with index 2 and ending just before index 3. If one-after-the-last endpoint such as 3 in your example were illegal, it would be impossible to include the last element of the list in the slice.

Answer 3

Slicing never raise error in python for out of bound indexes..

>>> s =[1,2,3]
>>> s[-1000:1000]
[1, 2, 3]

From the docs on string(applies to lists, tuples as well):

Degenerate slice indices are handled gracefully: an index that is too large is replaced by the string size, an upper bound smaller than the lower bound returns an empty string.

Docs(lists):

The slice of s from i to j is defined as the sequence of items with index k such that i <= k < j. If i or j is greater than len(s), use len(s). If i is omitted or None, use 0. If j is omitted or None, use len(s). If i is greater than or equal to j, the slice is empty.

Out-of-range negative slice indices are truncated, but don’t try this for single-element (non-slice) indices:

>>> word = 'HelpA'
>>> word[-100:]
'HelpA'

Answer 4

You have a range there. As soon as one index from the range goes outside the bounds the process of extracting elements stops.

There are no errors in slicing in Python.