装饰器的Python 3类型提示 [英] Python 3 type hinting for decorator

问题描述

请考虑以下代码：

from typing import Callable, Any

TFunc = Callable[..., Any]

def get_authenticated_user(): return "John"

def require_auth() -> Callable[TFunc, TFunc]:
    def decorator(func: TFunc) -> TFunc:
        def wrapper(*args, **kwargs) -> Any:
            user = get_authenticated_user()
            if user is None:
                raise Exception("Don't!")
            return func(*args, **kwargs)
        return wrapper
    return decorator

@require_auth()
def foo(a: int) -> bool:
    return bool(a % 2)

foo(2)      # Type check OK
foo("no!")  # Type check failing as intended

这段代码按预期工作。现在想象一下，我想扩展它，而不是仅仅执行 func（* args，** kwargs），我想将用户名注入参数中。因此，我修改了函数签名。

This piece of code is working as intended. Now imagine I want to extend this, and instead of just executing func(*args, **kwargs) I want to inject the username in the arguments. Therefore, I modify the function signature.

from typing import Callable, Any

TFunc = Callable[..., Any]

def get_authenticated_user(): return "John"

def inject_user() -> Callable[TFunc, TFunc]:
    def decorator(func: TFunc) -> TFunc:
        def wrapper(*args, **kwargs) -> Any:
            user = get_authenticated_user()
            if user is None:
                raise Exception("Don't!")
            return func(*args, user, **kwargs)  # <- call signature modified

        return wrapper

    return decorator


@inject_user()
def foo(a: int, username: str) -> bool:
    print(username)
    return bool(a % 2)


foo(2)      # Type check OK
foo("no!")  # Type check OK <---- UNEXPECTED

我不知道正确的方法键入方式。我知道在此示例中，修饰函数和返回函数在技术上应具有相同的签名（但即使未被检测到）。

I can't figure out a correct way to type this. I know that on this example, decorated function and returned function should technically have the same signature (but even that is not detected).

推荐答案

您不能使用 Callable 谈论其他参数。它们不是通用的。唯一的选择是说您的装饰器采用 Callable ，并返回不同的 Callable 。

You can't use Callable to say anything about additional arguments; they are not generic. Your only option is to say that your decorator takes a Callable and that a different Callable is returned.

在您的情况下，您可以可以使用类型变量确定返回类型：

In your case you can nail down the return type with a typevar:

RT = TypeVar('RT')  # return type

def inject_user() -> Callable[[Callable[..., RT]], Callable[..., RT]]:
    def decorator(func: Callable[..., RT]) -> Callable[..., RT]:
        def wrapper(*args, **kwargs) -> RT:
            # ...

即使是装饰后的 foo（）函数的键入签名为 def（* Any，** Any）->当您使用 reveal_type（）时，builtins.bool * 。

Even then the resulting decorated foo() function has a typing signature of def (*Any, **Any) -> builtins.bool* when you use reveal_type().

当前有各种建议正在讨论使 Callable 更加灵活，但这些措施尚未实现。参见

Various proposals are currently being discussed to make Callable more flexible but those have not yet come to fruition. See

• 允许可变参数的泛型


• 建议：概括 Callable 能够指定参数名称和种类


• TypeVar表示可调用对象的参数


• 出色地支持函数装饰器

• Allow variadic generics
• Proposal: Generalize Callable to be able to specify argument names and kinds
• TypeVar to represent a Callable's arguments
• Support function decorators excellently

例如。该列表中的最后一个是一张总括票，其中包括您的特定用例，修饰器，该修饰器更改了可调用的签名：

for some examples. The last one in that list is an umbrella ticket that includes your specific usecase, the decorator that alters the callable signature:

Mess with return类型或带有参数

对于任意函数，您根本无法做到这一点-甚至没有语法。这是我为它编写的一些语法。

Mess with the return type or with arguments

For an arbitrary function you can't do this at all yet -- there isn't even a syntax. Here's me making up some syntax for it.

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