索引超出范围的子字符串切片为什么有效？

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 0 <= n < (j-i)/k0 is defined as the sequence of items with index x = i + n*k such that 0 <= n < (j-i)/k. In other words, the indices are i, i+k, i+2*k, i+3*k and so on, stopping when j is reached (but never including j). When 0 <= n < (j-i)/k0 is positive, i and j are reduced to len(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).