函数参数的类型注释 [duplicate]-解网

问：

2个月前关闭。

请看下面的例子：

from typing import TypeVar
from collections.abc import Callable

T0 = TypeVar("T0")
T1 = TypeVar("T1")
T2 = TypeVar("T2")

def my_fun(a: str, b: int) -> None:
    print(int(a) + b)
    return None

def fun_caller(f: Callable[[T0, T1], T2], *args: tuple[T0, T1]) -> T2:
    return f(*args)

fun_caller(my_fun, "1", 2)

我想将一个函数后跟该函数的参数传递给另一个函数。上面的示例起作用，因为类型检查器知道中使用了正确数量的参数和正确的参数类型。但是，此示例仅适用于具有 2 个输入参数的函数，我想将其推广到任意数量的未知类型的参数。fun_caller(my_fun, "1", 2)

另请注意以下示例：

from typing import TypeVar, Any
from collections.abc import Callable

T = TypeVar("T")

def my_fun(a: str, b: int) -> None:
    print(int(a) + b)
    return None

def fun_caller(f: Callable[..., T], *args: tuple[Any, ...]) -> T:
    return f(*args)

fun_caller(my_fun, 0, 2, "a") # I want the type checker to complain, e.g.,
                              # first arg should be str
                              # there should only be 2 args, not 3

这是行不通的，因为现在类型检查器不会抱怨无效的参数数量和无效的参数类型。此代码将按预期引发，但我想在类型检查器中捕获此问题。TypeError: my_fun() takes 2 positional arguments but 3 were given

python 类型

# pyright: strict

from typing import ParamSpec, TypeVar
from collections.abc import Callable

P = ParamSpec("P")
T = TypeVar("T")

def my_fun1(a: str, b: int) -> None:
    print(int(a) + b)
    return None

def my_fun2(a: str, b: int, c: float) -> None:
    print(int(a) + b*c)
    return None

def fun_caller(f: Callable[P, T], *args: P.args, **kwargs: P.kwargs) -> T:
    return f(*args, **kwargs)

fun_caller(my_fun1, "1", 2)             # no issues reported
fun_caller(my_fun2, "1", 2, 3.14)       # no issues reported
fun_caller(my_fun2, c=3.14, a="1", b=2) # no issues reported
fun_caller(my_fun1, 0, 2, "a")          # Argument of type "Literal[0]" cannot be assigned to parameter of type "str"
                                        # Expected 2 positional arguments

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函数参数的类型注释 [duplicate]

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