python之python 构造函数中缺少 1 个必需的位置参数

我正在尝试学习 python 中的继承概念。我有一个 Employee 类和派生类 Executive。

class Employee: 
    'Class defined for employee' 
 
    def __init__(self, name, dept, salary): 
        self.name = name 
        self.dept = dept 
        self.salary = salary 

子类

class Executive(Employee): 
 
    def __init__(self, name, dept, salary, hascar): 
        Employee.__init__(name, dept, salary) 
        self.hascar = hascar 

has car 是传递给构造函数的 bool 值，但是这会给我一个错误:

File "I:\Python_practicals\com\python\oop\Executive.py", line 7, in init Employee.init(name, dept, salary) TypeError: init() missing 1 required positional argument: 'salary'

当我尝试实例化 Executive 的对象时。
emp4 = Executive("Nirmal", "Accounting", 150000, True)

请您参考如下方法:

虽然 __init__ 是实例方法，但您是在类 而不是实例上调用它。此调用称为未绑定(bind)，因为它未绑定(bind)到实例。因此，您需要显式传递 self:

class Executive(Employee): 
    def __init__(self, name, dept, salary, hascar): 
        Employee.__init__(self, name, dept, salary) 
#                         ^^^^ 
        self.hascar = hascar 

然而，推荐的方法是使用 super :

Return a proxy object that delegates method calls to a parent or sibling class of type. This is useful for accessing inherited methods that have been overridden in a class.

使用 super 你的代码看起来像这样:

class Executive(Employee): 
 
    def __init__(self, name, dept, salary, hascar): 
        super(Executive, self).__init__(name, dept, salary) 
#       ^^^^^^^^^^^^^^^^^^^^^^ 
        self.hascar = hascar 

Python 3 添加了一些语法糖来简化这个公共(public)父类调用:

class Executive(Employee): 
 
    def __init__(self, name, dept, salary, hascar): 
        super().__init__(name, dept, salary)  # Py 3 
#       ^^^^^^^ 
        self.hascar = hascar