Question

In a comment on this answer to another question, someone said that they weren’t sure what functools.wraps was doing. So, I’m asking this question so that there will be a record of it on StackOverflow for future reference: what does functools.wraps do, exactly?

Answer 1

When you use a decorator, you’re replacing one function with another. In other words, if you have a decorator

def logged(func):     def with_logging(*args, **kwargs):         print(func.__name__ + " was called")         return func(*args, **kwargs)     return with_logging 

then when you say

@logged def f(x):    """does some math"""    return x + x * x 

it’s exactly the same as saying

def f(x):     """does some math"""     return x + x * x f = logged(f) 

and your function f is replaced with the function with_logging. Unfortunately, this means that if you then say

print(f.__name__) 

it will print with_logging because that’s the name of your new function. In fact, if you look at the docstring for f, it will be blank because with_logging has no docstring, and so the docstring you wrote won’t be there anymore. Also, if you look at the pydoc result for that function, it won’t be listed as taking one argument x; instead it’ll be listed as taking *args and **kwargs because that’s what with_logging takes.

If using a decorator always meant losing this information about a function, it would be a serious problem. That’s why we have functools.wraps. This takes a function used in a decorator and adds the functionality of copying over the function name, docstring, arguments list, etc. And since wraps is itself a decorator, the following code does the correct thing:

from functools import wraps def logged(func):     @wraps(func)     def with_logging(*args, **kwargs):         print(func.__name__ + " was called")         return func(*args, **kwargs)     return with_logging  @logged def f(x):     """does some math"""     return x + x * x  print(f.__name__)  # prints 'f' print(f.__doc__)   # prints 'does some math'