CODEWITHSUNDEEP
pythonpycharmarguments
Hapson
Hapson

Reputation: 13

PyCharm. Unexpected argument(s) Possible callees

I write in PyCharm and get a bug (in line "func(dict_data)"): Unexpected argument(s) Possible callees: A.foo(dict_data: dict) A.bar(dict_data: dict).

Is it a PyCharm bug or am I doing something wrong?

PyCharm 2020.3

    class A:
    
        def __init__(self):
            self.functions = {
                "foo": self.foo,
                "bar": self.bar
            }
    
        def typing(self, dict_data: dict):
            for key, value in dict_data.items():
                if key in self.functions:
                    func = self.functions[value["type"]]
                    func(dict_data)
    
        def foo(self, dict_data: dict):
            print(dict_data)
    
        def bar(self, dict_data: dict):
            print(dict_data)
    
    
    class B:
    
        def __init__(self):
            self.data = {
                "foo": {"type": "foo"},
                "bar": {"type": "bar"}
            }
    
        def get(self) -> dict:
            return self.data
    
    
    if __name__ == "__main__":
    
        b = B()
        data = b.get()
        a = A()
        a.typing(data)

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Upvotes: 0

Views: 5059

Answers (4)

grey
grey

Reputation: 1

you can get the warning away by adding lambdas:

self.functions = {
                "foo": lambda x: self.foo(x),
                "bar": lambda x: self.bar(x)
            }

Upvotes: 0

Cody A. Ray
Cody A. Ray

Reputation: 6017

A better approach is to add a type hint to func like

func: Callable[[dict], None] = self.functions[value["type"]]

(Don't forget to add from typing import Callable)

This will get rid of the IDE warning by giving it additional information to correctly determine the types. Basically, it will no longer see func as a Union type but directly know that it's a Callable a known method signature.

Upvotes: 2

simon lombard
simon lombard

Reputation: 33

you can get the warning away by adding lambdas:

self.functions = {
                "foo": lambda x: self.foo,
                "bar": lambda x: self.bar
            }

Upvotes: 0

Ananth
Ananth

Reputation: 831

func is actually a variable in your code, but you are calling it as a function which does not exist there, so PyCharm is guessing what function you might want to use instead which can accommodate dict_data as it's argument.

1

This 'def function_name(arguments : datatype)' in the pic is how function is identified but your ss has the keyword 'Union' and therefore func's datatype is Union.

Upvotes: 1

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