Decorator execution order

Question

def make_bold(fn):
    return lambda : \"\" + fn() + \"\"

def make_italic(fn):
    return lambda : \"\" + fn() + \"\"

@make_b         


        

Answer 1

Decorators wrap the function they are decorating. So make_bold decorated the result of the make_italic decorator, which decorated the hello function.

The @decorator syntax is really just syntactic sugar; the following:

@decorator
def decorated_function():
    # ...

is really executed as:

def decorated_function():
    # ...
decorated_function = decorator(decorated_function)

replacing the original decorated_function object with whatever decorator() returned.

Stacking decorators repeats that process outward.

So your sample:

@make_bold
@make_italic
def hello():
  return "hello world"

can be expanded to:

def hello():
  return "hello world"
hello = make_bold(make_italic(hello))

When you call hello() now, you are calling the object returned by make_bold(), really. make_bold() returned a lambda that calls the function make_bold wrapped, which is the return value of make_italic(), which is also a lambda that calls the original hello(). Expanding all these calls you get:

hello() = lambda : "" + fn() + "" #  where fn() ->
    lambda : "" + fn() + "" # where fn() -> 
        return "hello world"

so the output becomes:

"" + ("" + ("hello world") + "") + ""