使用返回函数的变量作为另一个函数的参数

Question

I have a function that has 4 parameters defined as:

call(a,b,c,d):

I have another function called

return_args() #that returns 4 variables. 

When I do the following:

call(return_args()) #it errors out saying, that's 1 arguments, not 4 arguments 

I know I can store the 4 variables in another line previous to the call(), but it it possible to return the variables all within call() line.

Answer 1

You can use * argument unpacking

call(*return_args())

This is a shorthand form of the following:

a, b, c, d = return_args()
call(a, b, c, d)

or even

tup = return_args()
call(*tup)

Python's * operator used in this way essentially says "unpack this tuple and pass the unpacked values to this function as arguments".

The related operator ** does a similar trick for keyword arguments:

def call(arg1=None, arg2=None, **kwargs):
    pass

kwargs = {'arg1': 'test', 'arg3': 'whatever'}
call(**kwargs)

* and ** can be used together:

def call(*args, **kwargs):
    pass
call(*return_args(), **kwargs)

Answer 2

您可以使用*运算符将列表作为单独的参数发送：

call(*return_args())

Answer 3

Use the 'variable unpacking' syntax: call( *return_args() )

Alternatively you can use an obsolete syntax: apply( call, return_args() )

Answer 4

return_args() #that returns 4 variables. 

This is a misunderstanding, it doesn't return 4 variables. A function like:

def return_args():
    return 1, 2, 3, 4

Is actually doing this:

def return_args():
    mytuple = (1, 2, 3, 4)
    return mytuple

and returning a single thing.

Another side of this is "destructuring assignment" in Python, which is the ability to do this:

a, b = 1, 2

It's a way to assign/bind two variables at once, but it's actually creating then unpacking the sequence (1,2) . You can write:

a, b, c, d = return_args()

and it looks like you returned four things and bound them to four variable names and that's a clean, useful abstraction, but that's not what happened - actually one sequence was created (with 4 things in it), then it was unpacked to match a sequence of variable names.

The two abstractions leak, and you find out that return_args() is returning a single thing when you try to do this:

call(return_args()) #it errors out saying, that's 1 arguments, not 4 arguments

The other answers are rightly suggesting call(*return_args()) as one solution, it's documented here under "Unpacking argument lists": http://docs.python.org/2/tutorial/controlflow.html#unpacking-argument-lists

(The other side to this is a function created to accept variable numbers of arguments discussed here: https://stackoverflow.com/a/11550319/478656 )