如何使用assert语句来确认python3s的类型

Question

如何确保此函数的参数必须是set类型？你知道吗

我尝试使用assert语句中的isinstance方法来验证剩余的参数和使用的参数只是set类型，但是失败了。你知道吗

def listAllSubset(remaining = set(), used = set()):
    assert(isinstance(remaining, set))
    assert(isinstance(used, set))

    if (len(remaining) == 0):
        print(used)
    else:
        element = remaining.pop()
        listAllSubset(remaining, used)
        listAllSubset(remaining, used.add(element))

我想看看如何确保只设置remaining和used参数。你知道吗

Answer 1

不是一个有效的方法，而是一个棘手的方法：

def listAllSubset(remaining = set(), used = set()):
    remaining.discard
    used.discard
    ...

Answer 2

isinstance是权威的，重点在代码的最后一行

这是密码

def listAllSubset(remaining = set(), used = set()):
    assert(isinstance(remaining, set))
    assert(isinstance(used, set))

    if (len(remaining) == 0):
        print(used)
    else:
        element = remaining.pop()    # OK: element will be None
        listAllSubset(remaining, used)   # OK: remaining == used == set() 
        listAllSubset(remaining, used.add(element))    # Error: used.add(element) will return None, and the None will give to keyword argument "used", then "assert" will raise a AssertionError.

if __name__ == "__main__":
    listAllSubset()

好吧， 最后一行代码listAllSubset(remaining, used.add(element))等于listAllSubset(set(), None）， 然后assert(isinstance(used, set))将引发断言错误

---

看看这个密码

used = set()
flag = used.add('_')
print(flag)    # print value of flag

None

---

已更改

我改了密码，我的，它能满足你的需要

def listAllSubset(remaining = set(), used = set()):
    assert(isinstance(remaining, set))
    assert(isinstance(used, set))

    if (len(remaining) == 0):
        print(used)
    else:
        element = remaining.pop()
        listAllSubset(remaining, used)
        used.add(element)
        listAllSubset(remaining, used)

请原谅我英语不好，英语太难了！！你知道吗