Python - Get the source code of the line that called me

Question

Using the python inspect module, in a function, I would like to get the source code of the line that called that function.

So in the following situation:

def fct1():
    # Retrieve the line that called me and extract 'a'
    return an object containing name='a'

a = fct1()

I would like to retrieve the string "a = fct1()" in fct1

All I can do so far is to retrieve the code of the whole module with :

code = inspect.getsource(sys._getframe().f_back)

Please note that fct1() can be called many times in the main module.

Eventually, what I want is to retrieve the variable name "a" which is easy if I can get s = "a = fct1()" in fct1() :

a_name = s.split("=")[0].strip()

So you want the code in a variable, not actually find the code on the editor, right? — NicoCaldo, Jun 30 '22 at 11:58
Does [this answer](https://stackoverflow.com/a/427504/19330634) help you ? — l -_- l, Jun 30 '22 at 12:01
l -_- l, no, I want the reverse. I want the caller in the called. — u2gilles, Jun 30 '22 at 13:34
NicoCaldo, I want s = "a = fct1()" in the fct1() then I will s.split[0] myself to get "a". I updated my question. Thanks — u2gilles, Jun 30 '22 at 13:40
@u2gilles Does https://stackoverflow.com/questions/3056048/filename-and-line-number-of-python-script help? — jtlz2, Jun 30 '22 at 13:56

IamFr0ssT · Accepted Answer · 2022-06-30T15:30:56.017

A really dumb solution would be to capture a stack trace and take the 2nd line:

import traceback

def fct1():
    stack = traceback.extract_stack(limit=2)
    print(traceback.format_list(stack)[0].split('\n')[1].strip())  # prints "a = fct1()"
    return None

a = fct1()

@jtlz2 asked for it in a decorator

import traceback

def add_caller(func):
    def wrapper(*args, **kwargs):
        stack = traceback.extract_stack(limit=2)
        func(*args, caller=traceback.format_list(stack)[0].split('\n')[1].strip(), **kwargs)
    return wrapper

@add_caller
def fct1(caller):
    print(caller)

fct1()

And it does work

Sure it works. Fantastic. Now I can easilly retrieve the variable name 'a'. Thanks a million. — u2gilles, Jun 30 '22 at 16:05

Python - Get the source code of the line that called me

1 Answers1