How would you write a @debuggable decorator in python?

Question

When debugging, I like to print out all the inputs and outputs of a function (I know I need a better IDE, but humour me, this could be used for error reporting). So, I\'d id

Answer 1

I think what you're after isn't really a debugging decorator, but more of a logging decorator.

It might make sense to use Python's logging module so you can have more fine grained control over the logging itself. For example you would be able to output to a file for later analysing the output.

The decorator might then look something more like:

import logging

logger = logging.getLogger('TraceLog')
# TODO configure logger to write to file/stdout etc, it's level etc


def logthis(level):
    def _decorator(fn):
        def _decorated(*arg,**kwargs):
            logger.log(level, "calling '%s'(%r,%r)", fn.func_name, arg, kwargs)
            ret=fn(*arg,**kwargs)
            logger.log(level, "called '%s'(%r,%r) got return value: %r", fn.func_name, arg, kwargs, ret)
            return ret
        return _decorated
    return _decorator

@logthis(logging.INFO)
def myfunc(this,that):
    return this+that

Then if you configure the logger to output to stderr you'd see:

>>> logger.setLevel(logging.INFO)
>>> handler=logging.StreamHandler()
>>> logger.addHandler(handler)
>>> myfunc(1,2)
calling 'myfunc'((1, 2),{})
called 'myfunc'((1, 2),{}) got return value: 3