main()函数不会输出coin_to_bag的值

时间:2015-12-04 17:46:08

标签: python dictionary main

我收到错误:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File ".\file.py", line 113, in <module>
    main()
  File ".\file.py", line 111, in main
    cs_in_the_b = assign_c2b(bs, cs, c2b)
NameError: global name 'c2b' is not defined

当我运行我的代码时

cs = ['quarter','dime','nickel']
bs = ['sm','med','lg']

def assign_c2b(bs, cs):
    '''
    assign_c2b() assigns cs to bs
    specifically c2b{} stores this association between the c and b
    '''
    c2b = {}#'test'}
    print('Here are your bs:\n')
    print(bs)
    print('\n')
    print('Here are your c types:\n')
    print (cs)

    for b in bs:
        c_type = input('Number of this type?    e.g. type 1 for Quarter "["Quarter", "Nickel"]" ')  #e.g. 0.25 * 2
        c_amount = input('Number of this type?  e.g. 15')  #e.g. 0.25 * 2
        for c in cs:
            c2b[b] = [c_type, c_amount]

    print(c2b)

    return (c2b)


def main():
    bs = gather_b()
    cs = gather_c()
    cs_in_the_b = assign_c2b(bs, cs, c2b)

main()

我只是试图让我的var c2bmain()中调用时显示一些内容。也许我完全忽略了原因,但还是没能弄清楚这一点。有人可以帮忙吗?谢谢!!

1 个答案:

答案 0 :(得分:1)

def main():
    bags = gather_bag()
    coins = gather_coin()
    coins_in_the_bag = assign_coin_to_bag(bags, coins, coin_to_bag)
在调用coin_to_bag时,永远不会定义

assign_coin_to_bag

即使已定义,assign_coin_to_bag也需要2个参数(def assign_coin_to_bag(bags, coins),但你试图用3来调用它。