问题
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?
回答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>"
来源:https://stackoverflow.com/questions/27342149/decorator-execution-order