Python: typed method arguments

Question

In Python3 it's possible to add the type to the arguments of a function:

def foo(bar: str = "default") -> str:
    """
    @param bar: a textual value
    """
    return "test"

Now I got two questions. First, how can you do that for a callback function? Meaning, how to define the signature of that callback in the function header?

def foo(callback) -> str:
    """
    @param callback: function(value: str) -> str
    """
    # calculate some intermediate stuff
    my_var = ...    
    return callback(my_var)

Second, how to do that for tuples. This would include the defining that the value is of type tuple and should have two values (no triple etc.).

def foo(value) -> str:
    """
    @param value: tuple of strings
    """

    v1, v2 = value
    return v1 + v2

Thanks for your comments and answers.

but a callback must satisfy a signature in which is needs to be invoked. See example above. — Matthias
– Matthias, Commented Dec 1, 2017 at 12:30
In your example callback annotation would be typing.Callable[[str], str] — Michael Butscher
– Michael Butscher, Commented Dec 1, 2017 at 12:32
One more problem: if you define a static method inside a class whereas that method has arguments of the same type/class, then that type is unknown ... — Matthias
– Matthias, Commented Dec 1, 2017 at 12:36

Laurent LAPORTE · Accepted Answer · 2017-12-01 12:46:15Z

2

To specify the signature of your callback function, you can use Callable, like this:

from typing import Callable


def foo(callback: Callable[[str], str]) -> str:
    # calculate some intermediate stuff
    my_var = '...'
    return callback(my_var)

For a tuple of two strings, you can use:

from typing import Tuple


def foo(value: Tuple[str, str]) -> str:
    v1, v2 = value
    return v1 + v2

edited Dec 1, 2017 at 12:46

answered Dec 1, 2017 at 12:41

Laurent LAPORTE

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