Rex Eastbourne wrote:
Thanks. I adapted it a bit:
def debug(foo):
print foo, 'is:'
exec('pprint.pprint(' + foo + ')')
But I'm getting NameError: name 'foo' is not defined, since foo is
not defined in this scope. (The function works beautifully when I'm
dealing with global
Rex -
If what you are looking for is a monitor of calls to a certain
function, check out this decorator example from the Python Wiki:
http://wiki.python.org/moin/PythonDecoratorLibrary?highlight=%28Decorator%29#head-d4ce77c6d6e75aad25baf982f6fec0ff4b3653f4
This will allow you to very quickly
Hi all,
I've written the following simple macro called debug(aname, avalue)
that prints out the name of an expression and its value:
def debug(aname, avalue):
print aname, 'is':
pprint.pprint(avalue)
An example call is:
debug('compose(f1,f2)', compose(f1,f2))
Writing the exact same
def debug(s):
print s
exec(s)
The line thing i'm not so sure about. Er. Hmmm.
On Thu, 2005-08-11 at 14:04 -0700, Rex Eastbourne wrote:
Hi all,
I've written the following simple macro called debug(aname, avalue)
that prints out the name of an expression and its value:
def
[Format recovered from top posting]
Jeremy Moles [EMAIL PROTECTED] writes:
On Thu, 2005-08-11 at 14:04 -0700, Rex Eastbourne wrote:
Hi all,
I've written the following simple macro called debug(aname, avalue)
that prints out the name of an expression and its value:
def debug(aname,
Rex Eastbourne wrote:
def debug(aname, avalue):
print aname, 'is':
pprint.pprint(avalue)
use eval:
def debug(s):
print s, 'is'
pprint.pprint(eval(s))
(it does mean the arg is a string not code..)
On a
slightly different topic, is it also possible to make
Thanks. I adapted it a bit:
def debug(foo):
print foo, 'is:'
exec('pprint.pprint(' + foo + ')')
But I'm getting NameError: name 'foo' is not defined, since foo is
not defined in this scope. (The function works beautifully when I'm
dealing with global variables, which is very rarely).
