786

Is there a way to determine a function's name from within the function?

def foo():
    print("my name is", __myname__)  # <== how do I calculate this at runtime?

In the example above, the body of foo will somehow access the function name "foo" without hard-coding it. The output would be:

>>> foo()
my name is foo
1

27 Answers 27

606
import inspect

def foo():
   print(inspect.stack()[0][3])
   print(inspect.stack()[1][3])  # will give the caller of foos name, if something called foo

foo()

output:

foo
<module_caller_of_foo>
7
  • 117
    You could also use: print(inspect.currentframe().f_code.co_name) or to get the caller's name: print(inspect.currentframe().f_back.f_code.co_name). I think it should be faster since you don't retrieve a list of all the stack frames as inspect.stack() does.
    – Michael
    Commented Feb 23, 2014 at 10:25
  • 19
    inspect.currentframe().f_back.f_code.co_name doesn't work with a decorated method whereas inspect.stack()[0][3] does... Commented Feb 15, 2016 at 18:08
  • 9
    Please note: inspect.stack() can incur heavy performance overhead so use sparingly! On my arm box it took 240ms to complete (for some reason)
    – gardarh
    Commented Dec 8, 2016 at 12:59
  • 3
    @Michael please post your comment as an answer . Commented May 17, 2020 at 19:15
  • 5
    What is the magic number 3 in the code? Commented Oct 18, 2022 at 11:13
303

If you don't want to play with the stack yourself, you should either use "bar" or bar.__name__ depending on context.

Python doesn't have a feature to access the function or its name within the function itself. A magic __function__ had been proposed for Python 3.0 but rejected. See PEP 3130 – Access to Current Module/Class/Function.

The given rejection notice is:

This PEP is rejected. It is not clear how it should be implemented or what the precise semantics should be in edge cases, and there aren't enough important use cases given. response has been lukewarm at best.

9
  • 31
    inspect.currentframe() is one such way.
    – Yuval
    Commented Sep 20, 2014 at 10:47
  • 71
    Combining @CamHart's approach with @Yuval's avoids "hidden" and potentially deprecated methods in @RoshOxymoron's answer as well as numerical indexing into the stack for @neuro/@AndreasJung's answer: print(inspect.currentframe().f_code.co_name)
    – hobs
    Commented Mar 17, 2015 at 21:43
  • 3
    is it possible to summarize why its been rejected? Commented Jun 15, 2017 at 18:27
  • 12
    why is this the chosen answer? Question isn't about accessing the current function or the module itself, just the name. And the stacktrace/debugging features already have this information.
    – nurettin
    Commented Apr 3, 2019 at 11:23
  • 9
    This answer is too old. It is now wrong. Please delete it.
    – Florian
    Commented Jun 23, 2021 at 12:21
302

There are few ways to get the same result:

import sys
import inspect

def what_is_my_name():
    print(inspect.stack()[0][0].f_code.co_name)
    print(inspect.stack()[0][3])
    print(inspect.currentframe().f_code.co_name)
    print(sys._getframe().f_code.co_name)

Note that the inspect.stack calls are thousands of times slower than the alternatives:

$ python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][0].f_code.co_name'
1000 loops, best of 3: 499 usec per loop
$ python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][3]'
1000 loops, best of 3: 497 usec per loop
$ python -m timeit -s 'import inspect, sys' 'inspect.currentframe().f_code.co_name'
10000000 loops, best of 3: 0.1 usec per loop
$ python -m timeit -s 'import inspect, sys' 'sys._getframe().f_code.co_name'
10000000 loops, best of 3: 0.135 usec per loop

Update 08/2021 (original post was written for Python2.7)

Python 3.9.1 (default, Dec 11 2020, 14:32:07)
[GCC 7.3.0] :: Anaconda, Inc. on linux

python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][0].f_code.co_name'
500 loops, best of 5: 390 usec per loop
python -m timeit -s 'import inspect, sys' 'inspect.stack()[0][3]'
500 loops, best of 5: 398 usec per loop
python -m timeit -s 'import inspect, sys' 'inspect.currentframe().f_code.co_name'
2000000 loops, best of 5: 176 nsec per loop
python -m timeit -s 'import inspect, sys' 'sys._getframe().f_code.co_name'
5000000 loops, best of 5: 62.8 nsec per loop
12
  • 16
    inspect.currentframe() seems a good tradeoff between execution time and use of private members Commented Aug 27, 2016 at 9:44
  • 1
    @mbdevpl My numbers are 1.25ms, 1.24ms, 0.5us, 0.16us normal (nonpythonic :) ) seconds accordingly (win7x64, python3.5.1) Commented Jan 5, 2017 at 20:55
  • 8
    I use sys._getframe().f_code.co_name over inspect.currentframe().f_code.co_name simply because I have already imported the sys module. Is that a reasonable decision? (considering the speeds appear quite similar)
    – PatrickT
    Commented Jun 6, 2020 at 19:05
  • 7
    This is fullblow answer and should be the accepted one in my view
    – Nam G VU
    Commented Aug 27, 2020 at 4:29
  • 3
    We can use sys._getframe().f_back.f_code.co_name to get the callers name too Commented Apr 28, 2021 at 18:57
79
functionNameAsString = sys._getframe().f_code.co_name

I wanted a very similar thing because I wanted to put the function name in a log string that went in a number of places in my code. Probably not the best way to do that, but here's a way to get the name of the current function.

5
  • 4
    Totally working, just using sys, don't need to load more modules, but not soo easy to remember it :V
    – m3nda
    Commented Jun 13, 2016 at 20:29
  • Logging supports function name in the message format string. Commented Feb 3, 2023 at 18:09
  • _ indicates a private method. Please don't invoke them publicly. You maybe want to use inspect.currentframe() instead which is a public method.
    – Roland
    Commented Jun 6, 2023 at 9:20
  • Well Roland, you may have a point, but look at the number of upvotes this answer got. Even I'd prefer to use this answer instead of inspect, which a person named gardah (a comment in an answer above) says consumed a lot of time. If the problem is with it being private, the developers need to be coaxed to provide a public function named getCurrentFunctionName() and getParentFunctionName().
    – Nav
    Commented Aug 28, 2023 at 6:54
  • so in my testing I could run sys._getframe().f_code.co_name 1000 times in the length of time it took inspect.stack()[0][3] to run. 1.59 milliseconds verses 1.19 microseconds to put those both in microseconds, inspect took 1590 microseconds, and sys._getframe took 1.19 microseconds. Commented Dec 20, 2024 at 0:42
55

You can get the name that it was defined with using the approach that @Andreas Jung shows, but that may not be the name that the function was called with:

import inspect

def Foo():
   print inspect.stack()[0][3]

Foo2 = Foo

>>> Foo()
Foo

>>> Foo2()
Foo

Whether that distinction is important to you or not I can't say.

4
  • 4
    Same situation as with .func_name. Worth remembering that class names and function names in Python is one thing and variables referring to them is another.
    – Kos
    Commented Dec 13, 2012 at 11:48
  • Sometimes you may want Foo2() to print Foo. For example: Foo2 = function_dict['Foo']; Foo2(). In this case, Foo2 is a function pointer for perhaps a command line parser.
    – Harvey
    Commented Jun 1, 2013 at 20:02
  • What kind of speed implication does this have? Commented Feb 20, 2014 at 20:29
  • 3
    Speed implication with regard to what? Is there a situation where you'd need to have this information in a hard realtime situation or something?
    – bgporter
    Commented Feb 23, 2014 at 15:27
38

This is actually derived from the other answers to the question.

Here's my take:

import sys

# for current func name, specify 0 or no argument.
# for name of caller of current func, specify 1.
# for name of caller of caller of current func, specify 2. etc.
currentFuncName = lambda n=0: sys._getframe(n + 1).f_code.co_name


def testFunction():
    print "You are in function:", currentFuncName()
    print "This function's caller was:", currentFuncName(1)    


def invokeTest():
    testFunction()


invokeTest()

# end of file

The likely advantage of this version over using inspect.stack() is that it should be thousands of times faster [see Alex Melihoff's post and timings regarding using sys._getframe() versus using inspect.stack() ].

2
  • Works in python 3.7
    – Sinux1
    Commented Nov 19, 2021 at 1:51
  • Really like this one. Easy, fast and elegant. I will balantly copy this code ^^ Commented Jan 2, 2024 at 13:10
36

I keep this handy utility nearby:

import inspect
myself = lambda: inspect.stack()[1][3]

Usage:

myself()
4
  • 1
    How would this be done with the alternative proposed here? "myself = lambda: sys._getframe().f_code.co_name" doesn't work (the output is "<lambda>"; I think because the result is determined at definition time, not later at call time. Hmm.
    – nmvega
    Commented May 22, 2017 at 22:42
  • 2
    @prismalytics.io: If you call myself (myself()) and don't just use its value (myself), you'll get what you're looking for.
    – ijw
    Commented Dec 27, 2017 at 19:56
  • 1
    NYCeyes was right, the name is resolved inside the lambda and thus the result is <lambda>. The sys._getframe() and inspect.currentframe() methods MUST be executed directly inside the function you want to get the name of. The inspect.stack() method works because you can specify the index 1, doing inspect.stack()[0][3] also yields <lambda>.
    – kikones34
    Commented Nov 22, 2019 at 13:16
  • 1
    Please do not use anonymous function literal (lambda ...) to create a named function. Define the function the normal way using def and avoid multiple possible problems. Commented Jun 3, 2022 at 20:07
31

I guess inspect is the best way to do this. For example:

import inspect
def bar():
    print("My name is", inspect.stack()[0][3])
1
  • 10
    Instead if using inspect.stack()[0][3], use inspect.stack()[0].function which should be more robust even when semantics in stack traces change.
    – Tom Pohl
    Commented Oct 7, 2020 at 6:16
30

I found a wrapper that will write the function name

from functools import wraps

def tmp_wrap(func):
    @wraps(func)
    def tmp(*args, **kwargs):
        print func.__name__
        return func(*args, **kwargs)
    return tmp

@tmp_wrap
def my_funky_name():
    print "STUB"

my_funky_name()

This will print

my_funky_name

STUB

3
  • 1
    As a decorator noob, I wonder if there is a way to access func.__name__ inside the context of my_funky_name (so I can retrieve its value and use it inside my_funky_name)
    – cowbert
    Commented Jun 29, 2017 at 21:57
  • The way to do that inside the my_funky_name function is my_funky_name.__name__. You could pass the func.__name__ into the function as a new parameter. func(*args, **kwargs, my_name=func.__name__). To get your decorators name from inside your function, I think that would require using inspect. But getting the name of the function controlling my function within my running function ... well that just sounds like the start of a beautiful meme :)
    – cad106uk
    Commented Jul 2, 2017 at 13:00
  • This only answers the question in an overly literal way. One rarely wants to just print the function name out of order.
    – matanox
    Commented Jun 15, 2024 at 14:57
29

Use __name__ attribute:

# foo.py
def bar():
    print(f"my name is {bar.__name__}")

You can easily access function's name from within the function using __name__ attribute.

>>> def bar():
...     print(f"my name is {bar.__name__}")
...
>>> bar()
my name is bar

I've come across this question myself several times, looking for the ways to do it. Correct answer is contained in the Python's documentation (see Callable types section).

Every function has a __name__ parameter that returns its name and even __qualname__ parameter that returns its full name, including which class it belongs to (see Qualified name).

6
  • 27
    What would be the point of this if the function name "bar" would have to be known already to perform this?
    – Stephen B
    Commented Jul 22, 2021 at 13:33
  • 10
    @PyNoob: After renaming bar to foo, print('bar') happily prints (incorrectly) "bar", whereas print(bar.__name__) fails.
    – Gerhard
    Commented Oct 22, 2021 at 21:47
  • 2
    To elaborate on @Gerhard's comment - this is useful to protect yourself later if you rename functions, because the IDE will find this reference more easily. Commented Dec 27, 2022 at 20:07
  • 3
    I still do not understand what is the point here since the name of the function - which was the whole point to avoid - is required to obtain __name__ . Does somene "get " this? btw @StephenB good point - from another StephenB[oesch] Commented Oct 21, 2023 at 17:59
  • 1
    @Gerhard, this is not always true. If I have another function called bar, print(bar.__name__) from within foo it will not break but will report the wrong function. After all, it seems quite natural that if I have bar and make a copy, that I will want to rename the copy to foo. Then we've run into this situation.
    – jpcgt
    Commented Jul 31, 2024 at 14:51
20

print(inspect.stack()[0].function) seems to work too (Python 3.5).

0
20

I am not sure why people make it complicated:

import sys 
print("%s/%s" %(sys._getframe().f_code.co_filename, sys._getframe().f_code.co_name))
2
18

Here's a future-proof approach.

Combining @CamHart's and @Yuval's suggestions with @RoshOxymoron's accepted answer has the benefit of avoiding:

  • _hidden and potentially deprecated methods
  • indexing into the stack (which could be reordered in future pythons)

So I think this plays nice with future python versions (tested on 2.7.3 and 3.3.2):

from __future__ import print_function
import inspect

def bar():
    print("my name is '{}'".format(inspect.currentframe().f_code.co_name))

Update: tested on 3.7.10, 3.8.10, and 3.9.5

1
  • Indexing into the stack will not be reordered in future pythons. Commented Dec 27, 2022 at 20:05
15
import inspect

def whoami():
    return inspect.stack()[1][3]

def whosdaddy():
    return inspect.stack()[2][3]

def foo():
    print "hello, I'm %s, daddy is %s" % (whoami(), whosdaddy())
    bar()

def bar():
    print "hello, I'm %s, daddy is %s" % (whoami(), whosdaddy())

foo()
bar()

In IDE the code outputs

hello, I'm foo, daddy is

hello, I'm bar, daddy is foo

hello, I'm bar, daddy is

13
import sys

def func_name():
    """
    :return: name of caller
    """
    return sys._getframe(1).f_code.co_name

class A(object):
    def __init__(self):
        pass
    def test_class_func_name(self):
        print(func_name())

def test_func_name():
    print(func_name())

Test:

a = A()
a.test_class_func_name()
test_func_name()

Output:

test_class_func_name
test_func_name
11

This is pretty easy to accomplish with a decorator.

>>> from functools import wraps

>>> def named(func):
...     @wraps(func)
...     def _(*args, **kwargs):
...         return func(func.__name__, *args, **kwargs)
...     return _
... 

>>> @named
... def my_func(name, something_else):
...     return name, something_else
... 

>>> my_func('hello, world')
('my_func', 'hello, world')
0
8

You can use a decorator:

def my_function(name=None):
    return name

def get_function_name(function):
    return function(name=function.__name__)

>>> get_function_name(my_function)
'my_function'
7
  • 3
    How is that answering the poster's question? Can you expand this to include an example how the function name is known from within the function?
    – parvus
    Commented Mar 18, 2019 at 12:37
  • @parvus: my answer as is is an example that demonstrates an answer to OP's question Commented Mar 22, 2019 at 4:01
  • Ok, my_function is the random user's function of the OP. Blame this to my lack of understanding of decorators. Where the @? How will this work for functions whose arguments you don't want to adapt? How I understand your solution: when I want to know the function name, I have to append it with @get_function_name, and add the name argument, hoping it is not already there for another purpose. I'm likely missing something, sorry for that.
    – parvus
    Commented Mar 29, 2019 at 6:41
  • 3
    This just looks like a convoluted way to get to the __name__ attribute of a function. The usage requires knowing the thing you are trying to get, which doesn't seem very useful to me in simple cases where functions aren't defined on the fly.
    – Avi
    Commented May 14, 2019 at 17:55
  • 1
    This is rather a non-answer. You just hint the very basic idea. The real usable execution of the idea is in the answer by Jeff Laughlin Commented Jun 3, 2022 at 14:49
6

It seems from all the answer above that use the inspect library, all are writing something like:

import inspect

inspect.stack()[0][3]

But, since the return of inspect.stack()[0] is a NamedTuple of the form:

FrameInfo(frame=<frame at 0x103578810, file '<stdin>', line 1, code <module>>, filename='<stdin>', lineno=1, function='<module>', code_context=None, index=None)

One can simply call by the name, i.e. inspect.stack()[0].function

A small dummy example can be seen here:

    def test_train_UGRIZY_noZ(self, architecture, dataset, hyperrun, wloss):
        log.warning(f"{inspect.stack()[0].function} -- Not Implemented Yet")
        pass

Which when run prints:

WARNING - test_train_UGRIZY_noZ -- Not Implemented Yet
3
  • I like being able to use .function to reference the function name. Unfortunately the timings for this option are still in the microsecond range, as opposed to the nanosecond range for inspect.currentframe().f_code.co_name and sys._getframe().f_code.co_name as in Alex Granovsky's answer. Also in the microsecond range: inspect.getframeinfo(inspect.currentframe()).function, but about 2x as fast as inspect.stack().
    – hlongmore
    Commented Feb 26, 2023 at 9:25
  • It was already mentioned in a 2016 answer. stackoverflow.com/a/34665538/674039
    – wim
    Commented Feb 21, 2024 at 23:28
  • @wim True, but it failed to elaborate on anything or provide examples like this answer. Maybe that answer should've been updated with this information instead, but I can understand someone overlooking it since it's a one-line answer. Those usually aren't helpful.
    – Maybe
    Commented Sep 9, 2024 at 15:23
5

I do my own approach used for calling super with safety inside multiple inheritance scenario (I put all the code)

def safe_super(_class, _inst):
    """safe super call"""
    try:
        return getattr(super(_class, _inst), _inst.__fname__)
    except:
        return (lambda *x,**kx: None)


def with_name(function):
    def wrap(self, *args, **kwargs):
        self.__fname__ = function.__name__
        return function(self, *args, **kwargs)
return wrap

sample usage:

class A(object):

    def __init__():
        super(A, self).__init__()

    @with_name
    def test(self):
        print 'called from A\n'
        safe_super(A, self)()

class B(object):

    def __init__():
        super(B, self).__init__()

    @with_name
    def test(self):
        print 'called from B\n'
        safe_super(B, self)()

class C(A, B):

    def __init__():
        super(C, self).__init__()

    @with_name
    def test(self):
        print 'called from C\n'
        safe_super(C, self)()

testing it :

a = C()
a.test()

output:

called from C
called from A
called from B

Inside each @with_name decorated method you have access to self.__fname__ as the current function name.

5

I suggest not to rely on stack elements. If someone use your code within different contexts (python interpreter for instance) your stack will change and break your index ([0][3]).

I suggest you something like that:

class MyClass:

    def __init__(self):
        self.function_name = None

    def _Handler(self, **kwargs):
        print('Calling function {} with parameters {}'.format(self.function_name, kwargs))
        self.function_name = None

    def __getattr__(self, attr):
        self.function_name = attr
        return self._Handler


mc = MyClass()
mc.test(FirstParam='my', SecondParam='test')
mc.foobar(OtherParam='foobar')
1
  • 1
    Oh Lord, what a creative misuse of __getattr__() and passing of an argument as an object attribute! Please do not use this in normal (non-experimental code). Commented Jun 3, 2022 at 21:40
5

I recently tried to use the above answers to access the docstring of a function from the context of that function but as the above questions were only returning the name string it did not work.

Fortunately I found a simple solution. If like me, you want to refer to the function rather than simply get the string representing the name you can apply eval() to the string of the function name.

import sys
def foo():
    """foo docstring"""
    print(globals()[sys._getframe().f_code.co_name].__doc__)
2
  • Thanks for the feedback. I've updated the answer. Let me know if you have any other thoughts. Commented Mar 10, 2024 at 8:30
  • 2
    Yes, sys._getframe is undocumented/implementation detail and should not be used directly.
    – wim
    Commented Mar 11, 2024 at 14:58
4

Sincesys._getframe().f_back.f_code.co_name does not work at all in python 3.9, following could be used from now:

from inspect import currentframe


def testNameFunction() -> str:
    return currentframe().f_back.f_code.co_name


print(f'function name is {testNameFunction()}(...)')

Result:

function name is testNameFunction(...)
3
  • 1
    Except that the question is not about line numbers.
    – MEMark
    Commented Aug 20, 2021 at 7:35
  • @MEMark i updated my answer accordently, you should have noticed the textual error, but i have also investigate the issue, and i came to the conclusion, that the call sys._getframe().f_back.f_code.co_name works allthough the IDE PyCharm does not recognize it Cannot find reference '_getframe' in 'sys.pyi | sys.pyi'. That,s why i wrote that answer before. Commented Aug 20, 2021 at 10:30
  • @MEMark here is the post, i wrote on this issue : https://stackoverflow.com/q/68772415/5667103 Commented Aug 20, 2021 at 10:42
3

I like the idea of using a decorator but I'd prefer to avoid touching the function arguments. Hence, I'm providing yet another alternative:

import functools

def withname(f):
    @functools.wraps(f)
    def wrapper(*args, **kwargs):
        global __name
        __saved_name = globals().get("__name")
        __name = f.__name__
        ret = f(*args, **kwargs)
        __name = __saved_name
        return ret
    return wrapper

@withname
def f():
    print(f"in f: __name=={__name}")
    g()
    print(f"back in f: __name=={__name}")

@withname
def g():
    print(f"in g: __name=={__name}")

We need to save and restore __name when calling the function as consequence of it being a global variable. Calling f() above produces:

in f: __name==f
in g: __name==g
back in f: __name==f

Unfortunately, there is no alternative to the global variable if we don't change the function arguments. Referencing a variable, that is not created in the context of the function, will generate code that would look for a global variable:

>>> def f(): print(__function__)
>>> from dis import dis
>>> dis(f)
  1           0 LOAD_GLOBAL              0 (print)
              2 LOAD_GLOBAL              1 (__function__)
              4 CALL_FUNCTION            1
              6 POP_TOP
              8 LOAD_CONST               0 (None)
             10 RETURN_VALUE
3

@jeff-laughlin's answer is beautiful. I have modified it slightly to achieve what I think is the intent: to trace out the execution of functions, and also to capture the list of arguments as well as the keyword arguments. Thank you @jeff-laughlin!

from functools import wraps                                                                                                                                                                                                     
import time                                                                                                                                                                                                                     
                                                                                                                                                                                                                                
def named(func):                                                                                                                                                                                                                
    @wraps(func)                                                                                                                                                                                                                
    def _(*args, **kwargs):                                                                                                                                                                                                     
        print(f"From wrapper function: Executing function named: {func.__name__}, with arguments: {args}, and keyword arguments: {kwargs}.")                                                                                    
        print(f"From wrapper function: {func}")                                                                                                                                                                                 
        start_time = time.time()                                                                                                                                                                                                
        return_value = func(*args, **kwargs)                                                                                                                                                                                    
        end_time = time.time()                                                                                                                                                                                                  
        elapsed_time = end_time - start_time                                                                                                                                                                                    
        print(f"From wrapper function: Execution of {func.__name__} took {elapsed_time} seconds.")                                                                                                                              
        return return_value                                                                                                                                                                                                     
    return _                                                                                                                                                                                                                    
                                                                                                                                                                                                                                
@named                                                                                                                                                                                                                          
def thanks(message, concepts, username='@jeff-laughlin'):                                                                                                                                                                       
    print(f"From inner function: {message} {username} for teaching me about the {concepts} concepts of closures and decorators!")                                                                                               
                                                                                                                                                                                                                                
thanks('Thank you', 'two', username='@jeff-laughlin')                                                                                                                                                                           
print('-'*80)                                                                                                                                                                                                                   
thanks('Thank you', 'two', username='stackoverflow')
print(thanks) 

From wrapper function: Executing function named: thanks, with arguments: ('Thank you', 'two'), and keyword arguments: {'username': '@jeff-laughlin'}.

From wrapper function: <function thanks at 0x7f13e6ceaa60>

From inner function: Thank you @jeff-laughlin for teaching me about the two concepts of closures and decorators!

From wrapper function: Execution of thanks took 2.193450927734375e-05 seconds.

--------------------------------------------------------------------------------

From wrapper function: Executing function named: thanks, with arguments: ('Thank you', 'two'), and keyword arguments: {'username': 'stackoverflow'}.

From wrapper function: <function thanks at 0x7f13e6ceaa60>

From inner function: Thank you stackoverflow for teaching me about the two concepts of closures and decorators!

From wrapper function: Execution of thanks took 7.152557373046875e-06 seconds.
<function thanks at 0x7f13e6ceaca0>

What is most surprising to me is that there is a way to intercept functions at runtime, inspect them, and take some actions based on this. The other surprising thing is the memory address of the inner function was the same both times. Does anyone know why this is? I have a ways to go before I can understand this decorator/closure magic.

2
  • Decorator works in roughly the same way as replacing all your thanks() calls to named(thanks)(). So the two thanks() calls are targeting the same function. The wrapper function (underscore (_)) is dynamically instantiated (and soon destroied) everytime.
    – Cychih
    Commented Oct 13, 2022 at 13:00
  • By the way, if you put some print(_) calls before return _, you will see the output upon every @named call. print() can be added between def for better understanding. In Python, almost all things happen in run time, including function "declarations", which creates a function object and registers it to current namespace.
    – Cychih
    Commented Oct 13, 2022 at 13:27
1

implementing your own decorator

#foo.py

from mydecorators import *

@print_my_name
def bar():
    #do something else
    pass
#in terminal: my name is: bar

#mydecorators.py

def resolve_function(func):
    #in case annotated func is an staticmethod
    if isinstance(func,staticmethod):
        return func.__func__
    return func

def print_my_name(func):
    def function_caller(*args,**kwargs):
        _func = resolve_function(func)
        print("my name is: %s" %_func.__name__)
        return _func(*args,**kwargs)
    return function_caller
0
str(str(inspect.currentframe())).split(' ')[-1][:-1]
0
import inspect


def method_name():
    return inspect.stack()[1][3]


def method_name_caller():
    return inspect.stack()[2][3]


def asdf():
    print(method_name_caller())
    print(method_name())


def asdf2():
    print(method_name_caller())
    print(method_name())
    asdf()

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