test_func()我期望这会有一些智能错误，但事实并非如此。但是，这确实显示出一些错误，有人知道为什么吗？
ndimage.generic_filter()答案 0 :(得分：2)
通过设计，将对象定义为文字时，TS仅在多余属性上出错。 https://www.typescriptlang.org/docs/handbook/interfaces.html#excess-property-checks
答案 1 :(得分：2)
除了@kingdaro关于对象文字的多余属性检查的正确观点之外，
class Char:
name = "undefined"
def __init__(self, race, str, int, dex, con, spd, mp_bonus):
self.race = race
self.exp = 0
self.lvl = 1
self.str = str
self.int = int
self.dex = dex
self.con = con
self.spd = spd
self.hp = (con + str) / 2
self.current_hp = self.hp
self.mp_bonus = mp_bonus
self.mp = (int * mp_bonus)
self.current_mp = self.mp
def save(self):
with open("save.pk1", "wb") as fp:
pickle.dump(self.__dict__, fp, protocol=pickle.HIGHEST_PROTOCOL)
def load():
with open('save.pk1', 'rb') as fp:
Char.__init__ = pickle.load(fp) # no idea what to put here
# or if it should be in the Char class or not
def options(dude):
cls()
print("OPTIONS")
print("_____________________")
print("s. Save Game")
print("l. Load Game")
print("x. Quit Game")
print("_____________________")
select = input("please type in the corresponding letter or number: ")
if select == "s":
Char.save(player)
cls()
print("Save Complete")
wait()
main(dude)
elif select == "l":
cls()
print("Load Complete")
wait()
main(dude)
elif select == "x":
exit_screen(dude)
else:
print("you chose the wrong key")
wait()
main(dude)
def main(dude):
#menu as written in options above
select = input("please type in the corresponding letter or number: ")
if select == "s":
stats(dude)
elif select == "i":
inventory(dude)
elif select == "1":
rand_enemy()
elif select == "o":
options(dude)
else:
print("you chose the wrong key")
wait()
main(dude)
def start_screen(char):
#menu as written in options above
select = input("Please type in the corresponding number: ")
if select == "1":
get_char(char)
elif select == "2":
load()
main(char)
elif select == "3":
exit()
else:
print("you chose the wrong key")
wait()
start_screen(char)
start_screen(Char)
的定义只是扩大了Omit<T,K>的类型，因此它在T处不包含 known 属性（“我不知道或关心K是否是一个属性；我将忽略它。”），这与在K上禁止一个属性不同。也就是说，K是T的子类型，并且类型Omit<T, K>的任何值也可以分配给类型T。如果您的意图实际上是在Omit<T, K>上禁止一个属性，则可以（通过某种方式）通过指定K属性为可选属性并将其值类型设置为{ {1}}（或K，请参见下文）。我们称之为never：
undefined让我们看看它是否有效：
ForcefullyOmit<T, K>看起来不错。
这只是禁止密钥的“一种”，因为它仍将允许类型为type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;
type ForcefullyOmit<T, K extends keyof T> = Omit<T, K> & Partial<Record<K, never>>;
（或interface ObjInterface {
first: string;
second: string;
}
const obj = {
first: "first",
second: "second"
};
const out: ForcefullyOmit<ObjInterface, "first"> = obj; // error
// types of property "first" are incompatible
// string is not assignable to undefined
）的那个密钥的属性，但这有点像unresolved issue TypeScript ...语言并不总是能区分缺少的属性和undefined的属性。
由您决定用例是否确实需要更严格的行为never。无论如何，希望能有所帮助；祝你好运！