How do I add default parameters to functions when using type hinting?

Question

If I have a function like this:

def foo(name, opts={}):
  pass

And I want to add type hints to the parameters, how do I do it? The way I assumed gives me a syntax error:

def foo(name: str, opts={}: dict) -> str:
  pass

The following doesn't throw a syntax error but it doesn't seem like the intuitive way to handle this case:

def foo(name: str, opts: dict={}) -> str:
  pass

I can't find anything in the typing documentation or on a Google search.

Edit: I didn't know how default arguments worked in Python, but for the sake of this question, I will keep the examples above. In general it's much better to do the following:

def foo(name: str, opts: dict=None) -> str:
  if not opts:
    opts={}
  pass

Answer 1

Your second way is correct.

def foo(opts: dict = {}):
    pass

print(foo.__annotations__)

this outputs

{'opts': <class 'dict'>}

Although it is not explicitly mentioned in PEP 484, type hints are a specific use of function annotations, as outlined in PEP 3107. The syntax section clearly demonstrates that keyword arguments can be annotated in this manner.

I strongly advise against using mutable keyword arguments. More information here.

Answer 2

If you're using typing (introduced in Python 3.5) you can use typing.Optional, where Optional[X] is equivalent to Union[X, None]. It is used to signal that the explicit value of None is allowed . From typing.Optional:

def foo(arg: Optional[int] = None) -> None:
    ...

With Python 3.10 and above, as mentioned in joel's comment, this can equivalently be written as:

def foo(arg: int | None = None) -> None:
    ...

Answer 3

I recently saw this one-liner:

def foo(name: str, opts: dict=None) -> str:
    opts = {} if not opts else opts
    pass

Answer 4

Correct method

def foo(name: str, opts: dict = {}) -> str:
  pass

Type hinting syntax

<name>: <type> = <default_value>