bpo-29710: Clarify documentation for Bitwise binary operation (GH-1691)
Mathematically, bitwise operations on integers behave as if there were an
infinite number of sign bits. Pragmatically, that gives the same answer as
using one extra sign bit for the bitwise logical operations.
diff --git a/Doc/library/stdtypes.rst b/Doc/library/stdtypes.rst
index 2e551dc..e6f7b2c 100644
--- a/Doc/library/stdtypes.rst
+++ b/Doc/library/stdtypes.rst
@@ -382,7 +382,7 @@
.. _bitstring-ops:
Bitwise Operations on Integer Types
---------------------------------------
+-----------------------------------
.. index::
triple: operations on; integer; types
@@ -396,9 +396,9 @@
operator: >>
operator: ~
-Bitwise operations only make sense for integers. Negative numbers are treated
-as their 2's complement value (this assumes that there are enough bits so that
-no overflow occurs during the operation).
+Bitwise operations only make sense for integers. The result of bitwise
+operations is calculated as though carried out in two's complement with an
+infinite number of sign bits.
The priorities of the binary bitwise operations are all lower than the numeric
operations and higher than the comparisons; the unary operation ``~`` has the
@@ -409,13 +409,13 @@
+------------+--------------------------------+----------+
| Operation | Result | Notes |
+============+================================+==========+
-| ``x | y`` | bitwise :dfn:`or` of *x* and | |
+| ``x | y`` | bitwise :dfn:`or` of *x* and | (4) |
| | *y* | |
+------------+--------------------------------+----------+
-| ``x ^ y`` | bitwise :dfn:`exclusive or` of | |
+| ``x ^ y`` | bitwise :dfn:`exclusive or` of | (4) |
| | *x* and *y* | |
+------------+--------------------------------+----------+
-| ``x & y`` | bitwise :dfn:`and` of *x* and | |
+| ``x & y`` | bitwise :dfn:`and` of *x* and | (4) |
| | *y* | |
+------------+--------------------------------+----------+
| ``x << n`` | *x* shifted left by *n* bits | (1)(2) |
@@ -438,6 +438,12 @@
A right shift by *n* bits is equivalent to division by ``pow(2, n)`` without
overflow check.
+(4)
+ Performing these calculations with at least one extra sign extension bit in
+ a finite two's complement representation (a working bit-width of
+ ``1 + max(x.bit_length(), y.bit_length()`` or more) is sufficient to get the
+ same result as if there were an infinite number of sign bits.
+
Additional Methods on Integer Types
-----------------------------------