Python 3.4+inspectfunction

inspect.unwrap()

signatures

Peels off decorator layers: follows the __wrapped__ chain to its last object and returns the original function.

Syntax

inspect.unwrap(func, *, stop=None)

The chain exists only if the decorator uses functools.wraps: without it there is no __wrapped__ and unwrap returns the wrapper as is. The stop parameter halts the descent at the layer you want; the predicate receives every object. A cycle (f.__wrapped__ = f) raises ValueError instead of hanging. signature() and getsource() already unwrap on their own; you need it explicitly to reach the original function, bypassing a cache or a logger.

Examples

import functools, inspect
def cached(fn):
    @functools.wraps(fn)
    def wrapper(*args):
        return fn(*args)
    return wrapper
@cached
@cached
def area(w, h):
    return w * h
base = inspect.unwrap(area)
print(base is area, base.__name__)  # → False area
print(hasattr(base, '__wrapped__'))  # → False — это исходная функция
first = inspect.unwrap(area, stop=lambda f: not hasattr(f.__wrapped__, '__wrapped__'))
print(first is area.__wrapped__)  # → True — снят один слой

import inspect
def f():
    pass
f.__wrapped__ = f
try:
    inspect.unwrap(f)
except ValueError as e:
    print(type(e).__name__)  # → ValueError

See also

Common errors

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