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Question

定义功能print_popcorn_time()与参数bag_ounces。如果bag_ounces小于3，则打印“太小”。如果大于10，则打印“太大”。否则，计算并打印6 * bag_ounces，然后加上“秒”。以换行符结束。例如输出bag_ounces = 7：

42 seconds.

对于这个程序我得到吸吮的错误：

"File "main.py", line 6 
elif bag_ounces = bag_ounces * 6" 
       ^

def print_popcorn_time(bag_ounces): 
    if bag_ounces < 3: 
     print("Too small") 
    elif bag_ounces > 10: 
     print("Tool large") 
    elif bag_ounces = bag_ounces * 6 
    else: 
     print(bag_ounces) 

print_popcorn_time(7) 

Answer 1

你的最后一个条件应该仅仅是else

def print_popcorn_time(bag_ounces): 
    if bag_ounces < 3: 
     print("Too small") 
    elif bag_ounces > 10: 
     print("Tool large") 
    else: 
     print('{} seconds'.format(bag_ounces * 6)) 

Answer 2

def print_popcorn_time(bag_ounces): 
    if bag_ounces < 3: 
     print'Too small') 
    elif bag_ounces > 10: 
     print('Too large') 
    else: 
     print(bag_ounces,':', (bag_ounces * 6), 'seconds') 

print_popcorn_time(7) 

你也可以去这是因为你没有使用时间函数。