typed python: using the classes' own type insi

问题:

This question already has answers here:

Self-reference or forward-reference of type annotations in Python [duplicate] (1 answer)

How do I specify that the return type of a method is the same as the class itself? (4 answers)

Closed 2 years ago.

The following code does not work as expected. Apparently, I cannot use the classes' own type inside class definition:

class Foo:
    def __init__(self, key :str) -> None:
        self.key = key

    def __eq__(self, other :Foo) -> bool:
        return self.key == other.key

print('should be true: ', Foo('abc') == Foo('abc'))
print('should be false: ', Foo('abc') == Foo('def'))

The result of running it is:

Traceback (most recent call last):
  File "class_own_type.py", line 1, in <module>
    class Foo:
  File "class_own_type.py", line 5, in Foo
    def __eq__(self, other :Foo) -> bool:
NameError: name 'Foo' is not defined

Also, checking the code with mypy returns:

class_own_type.py:5: error: Argument 1 of "__eq__" incompatible with supertype "object"

How can I correct this code to be valid, both for Python and for mypy?

回答1:

The name Foo isn't bound yet, because the class itself has not yet been defined at the time that you try to use the name (remember: function arguments are evaluated at function definition time, not at function call time).

You can use string literals, to delay evaluation of the type:

class Foo:
    def __init__(self, key :str) -> None:
        self.key = key

    def __eq__(self, other: 'Foo') -> bool:
        return self.key == other.key

print('should be true: ', Foo('abc') == Foo('abc'))
print('should be false: ', Foo('abc') == Foo('def'))