Does Python do slice-by-reference on strings?
Question:
I want to know if when I do something like
a = "This could be a very large string..."
b = a[:10]
a new string is created or a view/iterator is returned
Answers:
When you slice strings, they return a new instance of String. Strings are immutable objects.
Python does slice-by-copy, meaning every time you slice (except for very trivial slices, such as a[:]), it copies all of the data into a new string object.
According to one of the developers, this choice was made because
The [slice-by-reference] approach is more complicated, harder to implement
and may lead to unexpected behavior.
For example:
a = "a long string with 500,000 chars ..."
b = a[0]
del a
With the slice-as-copy design the string a is immediately freed. The
slice-as-reference design would keep the 500kB string in memory although
you are only interested in the first character.
Apparently, if you absolutely need a view into a string, you can use a memoryview object.
I want to know if when I do something like
a = "This could be a very large string..."
b = a[:10]
a new string is created or a view/iterator is returned
When you slice strings, they return a new instance of String. Strings are immutable objects.
Python does slice-by-copy, meaning every time you slice (except for very trivial slices, such as a[:]), it copies all of the data into a new string object.
According to one of the developers, this choice was made because
The [slice-by-reference] approach is more complicated, harder to implement
and may lead to unexpected behavior.For example:
a = "a long string with 500,000 chars ..." b = a[0] del aWith the slice-as-copy design the string
ais immediately freed. The
slice-as-reference design would keep the 500kB string in memory although
you are only interested in the first character.
Apparently, if you absolutely need a view into a string, you can use a memoryview object.