Weird lambda behavior

[Rüdiger Ranft]

Hi all,

I want to generate some methods in a class using setattr and lambda.
Within each generated function a name parameter to the function is
replaced by a string constant, to keep trail which function was called.
The problem I have is, that the substituted name parameter is not
replaced by the correct name of the function, but by the last name the
for loop has seen.

import unittest

class WidgetDummy:
'''This class records all calls to methods to an outer list'''

def __init__( self, name, calls ):
'''name is the name of the object, which gets included into a
call record. calls is the list where the calls are appended.'''
self.name = name
self.calls = calls
for fn in ( 'Clear', 'Append', 'foobar' ):
func = lambda *y,**z: self.__callFn__( fn, y, z )
setattr( self, fn, func )

def __callFn__( self, fnName, *args ):
'''Add the function call to the call list'''
self.calls.append( ( self.name, fnName, args ) )

class Testcase( unittest.TestCase ):
def testFoo( self ):
calls = []
wd = WidgetDummy( 'foo', calls )
wd.Append( 23 )
self.assertEqual( [ ( 'foo', 'Append', ( 23, {} ) ) ], calls )

unittest.main()

 

[Diez B. Roggisch]

Hi all,

I want to generate some methods in a class using setattr and lambda.
Within each generated function a name parameter to the function is
replaced by a string constant, to keep trail which function was called.
The problem I have is, that the substituted name parameter is not
replaced by the correct name of the function, but by the last name the
for loop has seen.

Python's closures don't capture the values, but only the names they contain
when they are created. So whatever value is bound to that name as last is
used.

to overcome this, you need to create another closure or bind the value to a
name local to the created function via named arguments:

for f in [lambda i=i: i**2 for i in xrange(10)]:

.... f()
....
0
1
4
9
16
25
36
49
64
81


Diez