CODEWITHSUNDEEP
pythonstack
Milano
Milano

Reputation: 18725

Why inspect.stack() does not work properly?

I'm trying to write name of the current executed function. But it does not work properly. When I put 'print inspect.stack()[0][3]' after the def func(), it works, but when I try to put this command after the if, it prints nothing.

import inspect

debug = True

def debug():
    print inspect.stack()[0][3]
    if debug==True:
        print "test"
        print inspect.stack()[0][3]

debug()

returns 'debug' but it should return 

'debug'\n'test'\n'debug'

Where is the problem?

Upvotes: 1

Views: 1546

Answers (2)

BartoszKP
BartoszKP

Reputation: 35891

When you define the function:

def debug():

you are loosing the last reference to the previously created variable debug from the global scope. You redefine the debug variable from being a boolean holding True to a function reference. So the condition debug==True is not met (because debug is a function now, not a boolean). And thus only the first print statement works (see an illustrative demo).

This will work as intended, for example:

import inspect

debug = True

def f():
    print inspect.stack()[0][3]
    if debug==True:
        print "test"
        print inspect.stack()[0][3]

f()

This is a common mistake in Python, especially from people used to other languages, in which function names are treated in a special way. In Python they are not - after you define a function, you can use its name as any other variable. Which has many advantages, but as seen here sometimes can be confusing.

Upvotes: 1

Cos
Cos

Reputation: 129

In python 3 and hight you need to use print () if. It becomes a syntax error and won't print any thing if you don't have the ()

Upvotes: 0

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