Optional parameters in Python functions and their default values
Possible Duplicate:
“Least Astonishment” in Python: The Mutable Default Argument
我对可选参数在Python函数/方法中的工作方式有点困惑。
我有以下代码块：
1 2 3 4 5 6 7 8 9 | >>> def F(a, b=[]): ... b.append(a) ... return b ... >>> F(0) [0] >>> F(1) [0, 1] >>> |
为什么
我是说，里面发生了什么？
几年前Pycon的Good Doc-解释了默认参数值。但基本上，由于列表是可变对象，关键字参数在函数定义时进行计算，所以每次调用函数时，都会得到相同的默认值。
正确的方法是：
1 2 3 4 5 | def F(a, b=None): if b is None: b = [] b.append(a) return b |
从直观上看，默认参数有点像函数对象上的成员变量。
Default parameter values are evaluated when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that same"pre-computed" value is used for each call. This is especially important to understand when a default parameter is a mutable object, such as a list or a dictionary: if the function modifies the object (e.g. by appending an item to a list), the default value is in effect modified.
http://docs.python.org/reference/compound-stmts.html函数
Lists are a mutable objects; you can change their contents. The correct way to get a default list (or dictionary, or set) is to create it at run time instead, inside the function:
1 2 3 4 5 | def good_append(new_item, a_list=None): if a_list is None: a_list = [] a_list.append(new_item) return a_list |