String addresses

Question

While going through the mutable and immutable topic in depth. I found that the variable address is different when you simply call "s" v/s when you call it by index. Why is this so?

s = "hello!"
print(id(s))
print(id(s[0:5]))

Answer 1

s[0:5] creates a new, temporary string. That, of course, has a different ID. One "normal" case would be to assign this expression to a variable; the separate object is ready for that assignment.

Also note that a common way to copy a sequence is with a whole-sequence slice, such as

s_copy = s[:]

Answer 2

It's creating a new string, as Python strings are immutable (cannot be changed once created)

However, many implementations of Python will tag the same object if they find they're creating the same string, which makes for a good example. Here's what happens in a CPython 3.9.1 shell (which I suspect is one of the most common at time of writing)

>>> s = "hello!"
>>> print(id(s))
4512419376
>>> print(id(s[0:5]))  # new string from slice of original
4511329712
>>> print(id(s[0:5]))  # refers to the newly-created string
4511329712
>>> print(id(s[0:6]))  # references original string
4512419376
>>> print(id(s[:]))    # trivial slice also refers to original
4512419376

In general you should not rely on this behavior and only consider it an example.
Check for string equality with ==!

See also Why does comparing strings using either '==' or 'is' sometimes produce a different result?

Answer 3

Strings are immutable and thus by creating a new substring by slicing, the new String has a new address in memory. Because the old String cannot be altered due to it being immutable.