Question

i = 0
numbers = ["1","1","1","1","1"]
while 0 == 0:
   print(numbers[i])
   i+=1

一旦完成，就会说超出指数/范围。我怎么能说它用完了数字而不是错误？

我认为它被称为错误处理。

Answer 1

而不是while，我会使用for。

numbers = ["1","1","1","1","1"]
for n in numbers:
  print(n)
print("ran out of numbers")

Answer 2

堆栈跟踪告诉您要捕获的异常

>>> i = 0
>>> numbers = ["1","1","1","1","1"]
>>> while 0 == 0:
...    print(numbers[i])
...    i+=1
... 
1
1
1
1
1
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
IndexError: list index out of range

所以添加一个try / except块

>>> i = 0
>>> numbers = ["1","1","1","1","1"]
>>> try:
...     while 0 == 0:
...         print(numbers[i])
...         i += 1
... except IndexError:
...     print("Ran out of numbers!")
... 
1
1
1
1
1
Ran out of numbers!

但请通过一些优秀的python教程。这将是一种学习语言的痛苦方法！

Answer 3

您可以使用try和except

处理错误

try:
   print(numbers[i])
except Exception:
   print("ran out of numbers")

用您receiving

替换Exception

Answer 4

在不修改代码的情况下，您可以raise a custom exception message然后break来避免无限循环：

i = 0
numbers = ["1","1","1","1","1"]
while 0 == 0:
   try:
      print(numbers[i])
      i+=1
   except IndexError:
      print('ran out of numbers')
      break

Answer 5

您的循环条件始终为true，因此它会打印每个元素0-4，然后尝试打印不存在的第6个元素 - 因此超出索引/范围错误。请尝试使用“for”循环。 Psuedocode：对于x的数字打印（数字[x]）

Answer 6

使用(546,1024,1024,2)。您可以使用try..except：

而不是0 == 0

True

如何在python中创建自定义错误处理？

6 个答案: