How does function call another function in Python

Question

I want call function dynamically in Python, and code like as:

class A:
    def show1(self, x) :
        print x

    def show2(self, x, y) :
        print x, y

    def callfunc(self, f, args) :
        #TODO: call function f with args
        pass

c = A()
c.callfunc(c.show1, [1])
c.callfunc(c.show2, [1,2])

But I do not know how to call "show1" or "show2" in callfunc. Because "show1" and "show2" has different number of args, and "args" is a list.

Ignacio Vazquez-Abrams · Accepted Answer · 2013-08-30 05:06:32Z

5

Same as always.

def callfunc(self, f, args):
  f(*args)

answered Aug 30, 2013 at 5:06

Ignacio Vazquez-Abrams

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6 Comments

Adam Rosenfield Over a year ago

And the language spec for that is here.

Hyperboreus Over a year ago

Now this answer in python2.7 (inferring from using print as a statement) raises: TypeError: show1() takes exactly 2 arguments (3 given).

Ignacio Vazquez-Abrams Over a year ago

@Hyperboreus: Hardly my fault.

Hyperboreus Over a year ago

Yes, but in which respect does your post answer OP's question, taking his code snippet into consideration?

Ignacio Vazquez-Abrams Over a year ago

@Hyperboreus: His typo is still not my fault.

|

thefourtheye · Accepted Answer · 2013-08-30 05:43:15Z

1

If you can pass function reference as the parameter, you can instead call the function directly. Here is a more flexible way to do this

class A:
    def show1(self, x) :
        print x

    def show2(self, x, y) :
        print x, y

    def callfunc(self, f, args) :
        return getattr(self, f)(*args)

c = A()
c.callfunc("show1", [1])
c.callfunc("show2", [1,2])

In this case, the function to be called can be determined and invoked dynamically.

answered Aug 30, 2013 at 5:43

thefourtheye

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