On Tue, Aug 18, 2009 at 1:32 PM, Robert Dailey<rcdai...@gmail.com> wrote:
> On Aug 18, 3:31 pm, Duncan Booth <duncan.bo...@invalid.invalid> wrote:
>> Robert Dailey <rcdai...@gmail.com> wrote:
>> > Hello,
>>
>> > I want to simply wrap a function up into an object so it can be called
>> > with no parameters. The parameters that it would otherwise have taken
>> > are already filled in. Like so:
>>
>> > print1 = lambda: print( "Foobar" )
>> > print1()
>>
>> > However, the above code fails with:
>>
>> > File "C:\IT\work\distro_test\distribute_radix.py", line 286
>> > print1 = lambda: print( "Foobar" )
>> > ^
>> > SyntaxError: invalid syntax
>>
>> > How can I get this working?
>>
>> def print1():
>> print "Foobar"
>>
>> It looks like in your version of Python "print" isn't a function. It always
>> helps if you say the exact version you are using in your question as the
>> exact answer you need may vary.
>
> I'm using Python 2.6. And using the legacy syntax in the lambda does
> not work either. I want to avoid using a def if possible. Thanks.
ch...@morpheus ~ $ python
Python 2.6.2 (r262:71600, May 14 2009, 16:34:51)
[GCC 4.0.1 (Apple Inc. build 5484)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> print1 = lambda: print( "Foobar" )
File "<stdin>", line 1
print1 = lambda: print( "Foobar" )
^
SyntaxError: invalid syntax
>>> from __future__ import print_function
>>> print1 = lambda: print( "Foobar" )
>>> print1()
Foobar
Cheers,
Chris
--
http://blog.rebertia.com
--
http://mail.python.org/mailman/listinfo/python-list
