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Kirill Smelkov
cpython
Commits
58abc5bc
Commit
58abc5bc
authored
Jan 05, 2013
by
Ezio Melotti
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#13094: add Programming FAQ entry about the behavior of closures.
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Doc/faq/programming.rst
Doc/faq/programming.rst
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Misc/NEWS
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Doc/faq/programming.rst
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58abc5bc
...
@@ -352,6 +352,58 @@ an imported module. This clutter would defeat the usefulness of the ``global``
...
@@ -352,6 +352,58 @@ an imported module. This clutter would defeat the usefulness of the ``global``
declaration for identifying side-effects.
declaration for identifying side-effects.
Why do lambdas defined in a loop with different values all return the same result?
----------------------------------------------------------------------------------
Assume you use a for loop to define a few different lambdas (or even plain
functions), e.g.::
squares = []
for x in range(5):
squares.append(lambda: x**2)
This gives you a list that contains 5 lambdas that calculate ``x**2``. You
might expect that, when called, they would return, respectively, ``0``, ``1``,
``4``, ``9``, and ``16``. However, when you actually try you will see that
they all return ``16``::
>>> squares[2]()
16
>>> squares[4]()
16
This happens because ``x`` is not local to the lambdas, but is defined in
the outer scope, and it is accessed when the lambda is called --- not when it
is defined. At the end of the loop, the value of ``x`` is ``4``, so all the
functions now return ``4**2``, i.e. ``16``. You can also verify this by
changing the value of ``x`` and see how the results of the lambdas change::
>>> x = 8
>>> squares[2]()
64
In order to avoid this, you need to save the values in variables local to the
lambdas, so that they don't rely on the value of the global ``x``::
squares = []
for x in range(5):
squares.append(lambda n=x: n**2)
Here, ``n=x`` creates a new variable ``n`` local to the lambda and computed
when the lambda is defined so that it has the same value that ``x`` had at
that point in the loop. This means that the value of ``n`` will be ``0``
in the first lambda, ``1`` in the second, ``2`` in the third, and so on.
Therefore each lambda will now return the correct result::
>>> squares[2]()
4
>>> squares[4]()
16
Note that this behaviour is not peculiar to lambdas, but applies to regular
functions too.
How do I share global variables across modules?
How do I share global variables across modules?
------------------------------------------------
------------------------------------------------
...
...
Misc/NEWS
View file @
58abc5bc
...
@@ -693,6 +693,9 @@ Tools/Demos
...
@@ -693,6 +693,9 @@ Tools/Demos
Documentation
Documentation
-------------
-------------
-
Issue
#
13094
:
add
"Why do lambdas defined in a loop with different values
all return the same result?"
programming
FAQ
.
-
Issue
#
14901
:
Update
portions
of
the
Windows
FAQ
.
-
Issue
#
14901
:
Update
portions
of
the
Windows
FAQ
.
Patch
by
Ashish
Nitin
Patil
.
Patch
by
Ashish
Nitin
Patil
.
...
...
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