CODEWITHSUNDEEP
python-3.xpython-decorators
showkey
showkey

Reputation: 290

function's returned value lost after decorated by other function

def myfunc():
    print(" myfunc() called.")
    return 'ok'

The 'ok' was function's returned value.

>>> myfunc()
 myfunc() called.
'ok'

Now to decorate it with other function. The decorate function.

def deco(func):
    def _deco():
        print("before myfunc() called.")
        func()
        print("  after myfunc() called.")
    return _deco

To decorate myfunc with deco function.

@deco
def myfunc():
    print(" myfunc() called.")
    return 'ok'

>>> myfunc()
before myfunc() called.
 myfunc() called.
  after myfunc() called.

Why the result is not as following?

>>> myfunc()
before myfunc() called.
 myfunc() called.
'ok'
  after myfunc() called.

Upvotes: 1

Views: 62

Answers (1)

skrx
skrx

Reputation: 20448

If you call the undecorated myfunc function in the shell, it prints the returned value automatically. After the decoration, myfunc is set to the _deco function which only returns None implicitly and doesn't print the returned value of myfunc, so 'ok' doesn't appear in the shell anymore.

If you want to print 'ok', you have to do it in the _deco function:

def deco(func):
    def _deco():
        print("before myfunc() called.")
        returned_value = func()
        print(returned_value)
        print("  after myfunc() called.")
    return _deco

And if you want to return the value you have to return it from _deco:

def deco(func):
    def _deco():
        print("before myfunc() called.")
        returned_value = func()
        print("  after myfunc() called.")
        return returned_value
    return _deco

Upvotes: 1

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