def foo(a1, args = []):
    print "args before = %s" % (args)
    args.insert(0, 10)
    args.insert(0, 99999)
    print "args = %s " % (args)

def main():

if __name__ == "__main__":


args before = []
args = [99999, 10] 
args before = [99999, 10]
args = [99999, 10, 99999, 10] 


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

  •, jobs)