def foo(a1, args = []): print "args before = %s" % (args) args.insert(0, 10) args.insert(0, 99999) print "args = %s " % (args) def main(): foo('a') foo('b') if __name__ == "__main__": main()
以上小程序会有如下输出:
args before = [] args = [99999, 10] args before = [99999, 10] args = [99999, 10, 99999, 10]
按照通常的理解,第二次调用的args应该为默认值[],但为什么会变成上一次的结果呢?
查阅Python manual有如下的说法:
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 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. This is generally not what was intended. A way around this is to use None as the default, and explicitly test for it in the body of the function, e.g.:
def whats_on_the_telly(penguin=None): if penguin is None: penguin = [] penguin.append("property of the zoo") return penguin
至此,原因已经很清楚了:函数中的参数默认值是一个可变的list, 函数体内修改了原来的默认值,而python会将修改后的值一直保留,并作为下次函数调用时的参数默认值