Why doesn\'t \'example\'[999:9999]
result in error? Since \'example\'[9]
does, what is the motivation behind it?
From this behavior I can a
Slicing is not bounds-checked by the built-in types. And although both of your examples appear to have the same result, they work differently; try them with a list instead.
You're correct! 'example'[3:4]
and 'example'[3]
are fundamentally different, and slicing outside the bounds of a sequence (at least for built-ins) doesn't cause an error.
It might be surprising at first, but it makes sense when you think about it. Indexing returns a single item, but slicing returns a subsequence of items. So when you try to index a nonexistent value, there's nothing to return. But when you slice a sequence outside of bounds, you can still return an empty sequence.
Part of what's confusing here is that strings behave a little differently from lists. Look what happens when you do the same thing to a list:
>>> [0, 1, 2, 3, 4, 5][3]
3
>>> [0, 1, 2, 3, 4, 5][3:4]
[3]
Here the difference is obvious. In the case of strings, the results appear to be identical because in Python, there's no such thing as an individual character outside of a string. A single character is just a 1-character string.
(For the exact semantics of slicing outside the range of a sequence, see mgilson's answer.)
For the sake of adding an answer that points to a robust section in the documentation:
Given a slice expression like s[i:j:k]
,
The slice of s from i to j with step k is defined as the sequence of items with index
x = i + n*k
such that0 <= n < (j-i)/k
. In other words, the indices arei
,i+k
,i+2*k
,i+3*k
and so on, stopping when j is reached (but never including j). When k is positive, i and j are reduced tolen(s)
if they are greater
if you write s[999:9999]
, python is returning s[len(s):len(s)]
since len(s) < 999
and your step is positive (1
-- the default).