Decorator execution order

Question

def make_bold(fn):
    return lambda : "<b>" + fn() + "</b>"

def make_italic(fn):
    return lambda : "<i>" + fn() + "</i>"

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

helloHTML = hello()

Output: "<b><i>hello world</i></b>"

I roughly understand about decorators and how it works with one of it in most examples.

In this example, there are 2 of it. From the output, it seems that @make_italic executes first, then @make_bold.

Does this mean that for decorated functions, it will first run the function first then moving towards to the top for other decorators? Like @make_italic first then @make_bold, instead of the opposite.

So this means that it is different from the norm of top-down approach in most programming lang? Just for this case of decorator? Or am I wrong?

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 : "<b>" + fn() + "</b>" #  where fn() ->
    lambda : "<i>" + fn() + "</i>" # where fn() -> 
        return "hello world"

so the output becomes:

"<b>" + ("<i>" + ("hello world") + "</i>") + "</b>"