Python: pass dictionary variable as function argument-CodePudding

I am reusing a piece of code from another developer (it is a socket listener) which has the following function:

def start_my_client(self, *, config: dict):

I am new to Python, so I am assuming it requires a dictionary variable to be passed as an argument, so I created this variable:

config_dic = {'name': 'my_socket', 'hostname': 'test', 'description': 'TEST'}

However, I tried many different ways to pass the variable when calling the function, but none of them worked for me, for example:

def handle_messages(conn, addr):
    conn.start_my_client(config=config_dic)

I receive nothing from the function. I tried this too:

def handle_messages(conn, addr):
    conn.start_my_client(**config_dic)

It throws this error:

TypeError: start_my_client() got an unexpected keyword argument 'name'

Any hint on how to pass the variable?

CodePudding user response：

The first call looks reasonable

def handle_messages(conn, addr):
    conn.start_my_client(config=config_dic)

What is expected output for this?

If you pass config_dic like that

def handle_messages(conn, addr):
    conn.start_my_client(**config_dic)

you perform unpacking arguments here. This means that the method start_my_client get arguments the same way as if you would call method like that

conn.start_my_client(name='my_socket', hostname='test', description='TEST')

This is the reason you got error

TypeError: start_my_client() got an unexpected keyword argument 'name'

start_my_client doesnt expect argument called name.

CodePudding user response：

As mentioned in:

the function you are using can receive several different arguments:

def start_my_client(self, *, config: dict):

So, it could take a number of arguments when called, something like this:

conn.start_my_client(a, b, c)

As for the :, this

Just indicate that each of the parameters passed should be dicts.

So, you could just call the function as:

def handle_messages(conn, addr):
    conn.start_my_client(config_dic)