Python输入：给它实例方法声明可调用类型

Question

请考虑以下代码：

import typing

def a(x: int, y: int) -> int:
    return x + y

class F(object):
    def b(self, x: int, y: int) -> int:
        return x + y

def call(operation: typing.Callable[[int, int], int]) -> int:
    return operation(2, 2)

call(a)

f = F()
call(f.b)

我的PyCharm IDE指示最后一行的输入错误：

这是打字/类型声明错误吗？ PyCharm类型检查器失败了吗？ 如果是输入错误，应该是什么？

Answer 1

这是一个PyCharm类型检测程序错误。 mypy typechecker接受您的示例，没有警告或错误：

$ bin/mypy --verbose so_41869174.py
LOG:  Mypy version 0.470
LOG:  Parsing so_41869174.py (so_41869174)
LOG:  Parsing lib/mypy/typeshed/stdlib/3/typing.pyi (typing)
LOG:  Parsing lib/mypy/typeshed/stdlib/3/builtins.pyi (builtins)
LOG:  Parsing lib/mypy/typeshed/stdlib/3/sys.pyi (sys)
LOG:  Parsing lib/mypy/typeshed/stdlib/3/abc.pyi (abc)
LOG:  Parsing lib/mypy/typeshed/stdlib/3/types.pyi (types)
LOG:  Parsing lib/mypy/typeshed/third_party/2and3/mypy_extensions.pyi (mypy_extensions)
LOG:  Parsing lib/mypy/typeshed/stdlib/3/_importlib_modulespec.pyi (_importlib_modulespec)
LOG:  Loaded graph with 8 nodes
LOG:  Found 2 SCCs; largest has 7 nodes
LOG:  Processing SCC of size 7 (_importlib_modulespec mypy_extensions types abc typing sys builtins) as inherently stale
LOG:  Processing SCC singleton (so_41869174) as inherently stale
LOG:  No fresh SCCs left in queue
LOG:  Build finished in 0.482 seconds with 8 modules, 1708 types, and 0 errors

因为F().b是一个绑定方法 ，所以它在没有self参数的情况下继承了底层函数的签名（因为它是绑定方法在绑定实例中传递的工作）。

例如，应用于绑定方法的typing.get_type_hints()函数正确地省略了self ：

>>> typing.get_type_hints(f.b)
{'x': <class 'int'>, 'y': <class 'int'>, 'return': <class 'int'>}