[issue21794] stack frame contains name of wrapper method, not that of wrapped method

[Zachary Ware]

Zachary Ware added the comment:

I agree with Josh; I don't think there's any bug here.  To confirm, I expanded 
on your example, reproduced below.  Have a look at and play around with this, 
and if you still believe something is wrong, ask on python-list and the folks 
there should be able to help.  If you determine that there really is a bug, 
please reopen the issue.


import functools
import inspect

def lineno():
return inspect.getlineno(inspect.stack()[1][0])

try: # just to make it easy to see code and output together...
with open(__file__) as file: data = file.read()
print(data[:data.rfind(# output:) + 9])
except Exception:
pass

print('#line {}: defining decorator'.format(lineno()))

def decorator(f):
print('#  line {}: decorator called'.format(lineno()))
print('#  line {}: defining inner'.format(lineno()))
@functools.wraps(f)
def inner(arg1, arg2):
print('#line {}: inner called, arg1 {} arg2 {}'.format(lineno(), 
arg1, arg2))
for i, frame in enumerate(inspect.stack()):
print(#  line {}: printed in inner, frame {}, name 
{}.format(lineno(), i, frame[3]))
f(arg1 + 1 , arg2 + 1)

print('#  line {}: printed in decorator, inner.__name__ == 
{}'.format(lineno(),inner.__name__))
print('#  line {}: printed in decorator, f.__name__ == {}'.format(lineno(), 
f.__name__))
return inner

print('#line {}: defining wrapped'.format(lineno()))

@decorator
def wrapped(warg1, warg2):
print('#  line {}: wrapped called, warg1 {} warg2 {}'.format(lineno(), 
warg1, warg2))
print(#  line {}: printed in wrapped, wrapped.__name__ == 
{}.format(lineno(), wrapped.__name__))
stack = inspect.stack()
for i, frame in enumerate(stack):
print(#line {}: printed in wrapped, frame {}, name 
{}.format(lineno(), i, frame[3]))

print('#line {}: Calling wrapped...'.format(lineno()))
wrapped(1,2)

print(#line {}: done.format(lineno()))

# output:

# expected output:
#line 13: defining decorator
#line 29: defining wrapped
#  line 16: decorator called
#  line 17: defining inner
#  line 25: printed in decorator, inner.__name__ == wrapped
#  line 26: printed in decorator, f.__name__ == wrapped
#line 39: Calling wrapped...
#line 20: inner called, arg1 1 arg2 2
#  line 22: printed in inner, frame 0, name inner
#  line 22: printed in inner, frame 1, name module
#  line 33: wrapped called, warg1 2 warg2 3
#  line 34: printed in wrapped, wrapped.__name__ == wrapped
#line 37: printed in wrapped, frame 0, name wrapped
#line 37: printed in wrapped, frame 1, name inner
#line 37: printed in wrapped, frame 2, name module
#line 42: done


--
nosy: +zach.ware
resolution:  - not a bug
stage:  - resolved
status: open - closed

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[issue21794] stack frame contains name of wrapper method, not that of wrapped method

[the mulhern]

New submission from the mulhern:

 def decorator(f):
... @functools.wraps(f)
... def new_func(self, junk):
... stack = inspect.stack()
... for i, frame in enumerate(stack):
... print(%s % frame[3])
... f(self, junk)
... return new_func
... 
 @decorator
... def junk(self, p):
... print(%s % junk.__name__)
... 
 junk(a, b)
new_func
module
junk
 junk.__name__
'junk'


Note that the wrapper function itself inspects the stack, printing out the 
names of methods on the stack.
Note that junk, the name of the wrapped function does not appear on the 
stack, it is only printed out by the junk method itself.

I think that the name of the function at the top of the stack should be the 
name of the wrapped function, not of its wrapper. The name of the wrapper 
function should not appear at all.

--
components: Interpreter Core
messages: 220863
nosy: the.mulhern
priority: normal
severity: normal
status: open
title: stack frame contains name of wrapper method, not that of wrapped method
type: behavior
versions: Python 2.7

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[issue21794] stack frame contains name of wrapper method, not that of wrapped method

[Josh Rosenberg]

Josh Rosenberg added the comment:

I don't think you understand how wrapping works. At the time you're enumerating 
the stack, the wrapped function has not actually been called. Only the wrapper 
has been called. If the stack included junk when junk had not yet been 
executed, the stack would be a lie.

From the interpreter's point of view, it doesn't even know that wrapping is in 
play, aside from the chain of __wrapped__ values attached to the wrapper by 
functools.wraps (as a convenience). Until you actually call the wrapped 
function, it's possible you could change your mind and call some other 
function instead; Python won't stop you, and Python can't tell the difference 
before the call has been made.

--
nosy: +josh.rosenberg

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