Can ParamSpec be used to type individual arguments?-CodePudding

I'm trying to type the following wrapper function. It takes another function and the function's arguments and runs it with some side effect.

from typing import Callable, ParamSpec, TypeVar

P = ParamSpec("P")
R = TypeVar("R")


def wrapper(func: Callable[P, R], *args: P.args, **kwargs: P.kwargs) -> R:
    # Run some side effects
    return func(*args, **kwargs)


def f(a: int, b: int) -> int:
    return a   b

# Prints 3 with the side effect. 
print(wrapper(f, 1, 2))

The above snippet works and mypy is happy. However, I don't like the black boxes that *args and **kwargs are and would like to make wrapper less generic. How do I type something similar to this:

from typing import Callable, ParamSpec, TypeVar

P = ParamSpec("P")
R = TypeVar("R")

# How do I express the fact that a and b are the input params of func?
def wrapper(func: Callable[P, R], a: ?, b: ?) -> R:
    # Run some side effects
    return func(a, b)


def f(a: int, b: int) -> int:
    return a   b

# Mypy should be happy here!
print(wrapper(f, 1, 2))

Is this possible?

CodePudding user response：

When you know the arguments that will be passed to func, you don't need ParamSpec; ordinary TypeVars will do.

AType = TypeVar('AType')
BType = TypeVar('BType')


def wrapper(func: Callable[[AType, BType], R], a: AType, b: BType) -> R:
    return func(a, b)

If there were additional arguments, then you would use Concatenate to combine the known arguments with ParamSpec.

AType = TypeVar('AType')
BType = TypeVar('BType')
P = ParamSpec('P')


def wrapper(func: Callable[Concatenate[AType, BType, P], R],
            a: AType,
            b: BType,
            *args: P.args,
            **kwargs: P.kwargs) -> R:
    return func(a, b, *args, **kwargs)