Bugs item #1390991, was opened at 2005-12-26 22:00
Message generated for change (Comment added) made by jimjjewett
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Category: Python Interpreter Core
Group: Python 2.4
Status: Open
Resolution: None
Priority: 5
Submitted By: Samuel Hsiung (shsiung)
Assigned to: Nobody/Anonymous (nobody)
Summary: lambda functions confused when mapped in dictionary object
Initial Comment:
***Background***
Lambda functions wrapped around different module
functions overwrite each other when mapped one at a
time in a dictionary object.
Python Version: 2.4.2
***Duplication Instructions***
1. Setup
===module foo.py===
def callme(x):
print "foo called!"
print x
===module bar.py===
def callme(x):
print "bar called!"
print x
===module call.py===
import foo, bar
api = {}
modules = (foo, bar)
for module in modules:
api[module] = lambda x: module.callme(x)
print '%s mapped to %s' % (module, api[module])
api[foo]('above line should be foo')
api[bar]('above line should be bar')
print "foo lambda: %s" % api[foo]
print "bar lambda: %s" % api[bar]
2. Execution
=> python call.py
<module 'foo' from '/home/shsiung/playground/foo.pyc'>
mapped to <function <lambda> at 0xb7f7abc4>
<module 'bar' from '/home/shsiung/playground/bar.pyc'>
mapped to <function <lambda> at 0xb7f7abfc>
bar called!
above line should be foo
bar called!
above line should be bar
foo lambda: <function <lambda> at 0xb7f7abc4>
bar lambda: <function <lambda> at 0xb7f7abfc>
3. Expected behaviour
foo should have been called, followed by bar; instead
bar was called twice.
----------------------------------------------------------------------
Comment By: Jim Jewett (jimjjewett)
Date: 2005-12-28 13:52
Message:
Logged In: YES
user_id=764593
Not a bug, just part of the ugliness of lambda and scopes,
masked by a quirk of for loops.
'module' is not a parameter of the lambda, so it is looked
up in the enclosing scope each time the lambda is called.
for loops leave their index (in this case, 'module') set at
the final value, so that value is found in the globals, and
used by each of the lambdas.
To get what you were expecting, try
api[module]=lambda x,module=module:module.callme(x)
This binds the name 'module' (for that lambda) to whatever
its current value is, at least as a default.
----------------------------------------------------------------------
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