Python函数默认参数的一个小陷阱

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