因为函数的参数指定为* a，这就像说您的参数是未定义维的元组

because the argument of your function is specified as *a, which is like saying your argument is a tuple of undefined dimension

当您尝试对带有嵌套列表的元组进行排序时，该值不会更改

when you try to sort a tuple with a nested list, the value will not change

事实上，您获得了列表列表(您获得了[[3，2，1]]而不是[3，2，1])

infact as result you got a list of list (you got [[3, 2, 1]] not [3, 2, 1])

如果您尝试这样做，它将起作用

if you try this, it will work

def foo(*x): 
    y=sorted(x[0]) 
    print(y)