使用参数print_popcorn_time()
定义函数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
后跟"秒"。以换行符结束。 bag_ounces = 7
的示例输出:
42秒。
对于这个程序,我很容易出错:
"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)
答案 0 :(得分:1)
你的最后一个条件应该是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))
答案 1 :(得分:0)
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)
您也可以通过此操作,因为您没有使用时间功能。
答案 2 :(得分:0)
"is"