通过将类作为该函数的参数传递来访问“另一个python文件的函数”中的“类的方法”

Question

I have a class file. Let's call it "C1.py". The sample code looks like below.

class C1(object):

    def __init__(self):
        self.greeting = "Hello, world!"

    def M1(ans):
        if ans == 1 or ans == 2:
            return True
        else:
            return False

Now, I have another python file in the same folder, which will access the class file shown above.

from trial import C1

def getAns(class1):
    while True:
        ans = input("Answer: ")
        if class1.M1(ans):
            return ans
            break

sample = C1()
print sample.greeting
ans = getAns(sample)
print ans

When I run those files, sample.greeting prints fine. Everything is fine until when the execution reaches the line "ans = getAns(C1)", which gives the error "M1() takes exactly 1 argument (2 given)".

So, where in the code should I change so that I can call that method successfully?

Note here that the above code is only the abstraction of my whole program to highlight my problem. It sounds stupid with just the code above alone. So, please, please bear with that for me.

Answer 1

M1 is currently defined as a method of C1, as such it needs to have an additional self argument which would be the instance of the class. Ie

class C1(object):

    def __init__(self):
        self.greeting = "Hello, world!"

    def M1(self, ans):
        if ans == 1 or ans == 2:
            return True
        else:
            return False

In other languages such as C++ or Java the presence of self (or this) is implicit, but in python it's explicit.

alternatively if you don't want or need M1 to access C1's state you could make M1 static via @staticmethod ie

class C1(object):

    def __init__(self):
        self.greeting = "Hello, world!"

    @staticmethod
    def M1(ans):
        if ans == 1 or ans == 2:
            return True
        else:
            return False