unexpected behaviour of lambda expression

[leonhard.vogt@gmx.ch]

Please consider that example:
Python 2.4.3 (#69, Mar 29 2006, 17:35:34) [MSC v.1310 32 bit (Intel)]
on win32
Type "help", "copyright" , "credits" or "license" for more information.

>>s = 'foo'
>>f = lambda x: s
>>f(None)

'foo'

>>s = 'bar'
>>f(None)

'bar'

>>del(s)
>>f(None)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 1, in <lambda>
NameError: global name 's' is not defined

It seems to me, that f is referencing the name s instead of the string
object bound to it
i would expect the analogous behaviour to the following example:
Python 2.4.3 (#69, Mar 29 2006, 17:35:34) [MSC v.1310 32 bit (Intel)]
on win32
Type "help", "copyright" , "credits" or "license" for more information.

>>s = 'foo'
>>f = s
>>f

'foo'

>>s = 'bar'
>>f

'foo'

I could work around this but I am interested why there is that
difference.
Leonhard

[Fredrik Lundh]

Re: unexpected behaviour of lambda expression

leonhard.vogt@g mx.ch wrote:

Please consider that example:
Python 2.4.3 (#69, Mar 29 2006, 17:35:34) [MSC v.1310 32 bit (Intel)]
on win32
Type "help", "copyright" , "credits" or "license" for more information.

>>>s = 'foo'
>>>f = lambda x: s
>>>f(None)

'foo'

>>>s = 'bar'
>>>f(None)

'bar'

>>>del(s)
>>>f(None)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 1, in <lambda>
NameError: global name 's' is not defined
>
It seems to me, that f is referencing the name s instead of the string
object bound to it

that's how lexical scoping works, of course.

if you want to bind to the object instead of the name, use explicit binding:

f = lambda x, s=s: s

</F>

[Duncan Booth]

Re: unexpected behaviour of lambda expression

leonhard.vogt@g mx.ch wrote:

>>>f = lambda x: s

....

>>>f(None)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 1, in <lambda>
NameError: global name 's' is not defined
>
It seems to me, that f is referencing the name s instead of the string
object bound to it

Of course it is. Why would you expect a function to lookup its global
variables before it is called? Remember "f = lambda x: s" is just a
confusing way to write:

def f(x):
return s

[leonhard.vogt@gmx.ch]

Re: unexpected behaviour of lambda expression

Fredrik Lundh schrieb:

leonhard.vogt@g mx.ch wrote:
>

Please consider that example:
Python 2.4.3 (#69, Mar 29 2006, 17:35:34) [MSC v.1310 32 bit (Intel)]
on win32
Type "help", "copyright" , "credits" or "license" for more information.

>>s = 'foo'
>>f = lambda x: s
>>f(None)

'foo'

>>s = 'bar'
>>f(None)

'bar'

>>del(s)
>>f(None)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 1, in <lambda>
NameError: global name 's' is not defined

It seems to me, that f is referencing the name s instead of the string
object bound to it

>
that's how lexical scoping works, of course.
>
if you want to bind to the object instead of the name, use explicit binding:
>
f = lambda x, s=s: s
>
</F>

Thank you, together with the response of Duncan it is clear to me now.
I will use something like

>>def makefunc(t):

.... return lambda x: t
....

>>s = 'foo'
>>f = makefunc(s)
>>f(None)

'foo'

>>s = 'bar'
>>f(None)

'foo'

Leonhard