[英]Try Except not catching IndexError
I'm trying to create a linked list for practice.我正在尝试创建一个用于练习的链接列表。
Here is my code so far:到目前为止,这是我的代码:
class Node:
def __init__(self, data, _next):
self.data = data
self._next = _next
test_list = [1,4,9]
def Linked(nodes):
try:
linked = []
for i in range(len(nodes)-1):
j = Node(nodes[i], nodes[i+1])
linked.append(j)
return linked
except IndexError:
print('Tail Found')
test = Linked(test_list)
test[2].data
I'm trying to catch the Index Error with the try/except but I'm still getting the following error:我正在尝试使用 try/except 捕获索引错误,但我仍然收到以下错误:
IndexError Traceback (most recent call last)
<ipython-input-193-5fa4e0d033a1> in <module>
20
21
---> 22 test[2].data
IndexError: list index out of range
Why isn't it printing 'Tail Found'?为什么不打印“发现尾巴”?
EDIT:编辑:
I changed the code to this:我将代码更改为:
class Node:
def __init__(self, data, _next):
self.data = data
self._next = _next
test_list = [1,4,9]
def Linked(nodes):
linked = []
for i in range(len(nodes)-1):
j = Node(nodes[i], nodes[i+1])
linked.append(j)
return linked
test = Linked(test_list)
try:
print(test[2].data)
except IndexError:
print('Tail Found: '+test[1]._next)
And it works now.它现在有效。
If test[2] doesn't exist then I know that test_list[2] is the last one in the list (ie the tail).如果 test[2] 不存在,那么我知道 test_list[2] 是列表中的最后一个(即尾部)。
Your error are in line 22. There is no indexerror occur in your try catch block您的错误在第 22 行。您的 try catch 块中没有 indexerror 发生
Why "Tail Found"?为什么“发现尾巴”? If you pass a list with length greater than 2 to function Linked, there won't be any exception, so no "Tail Found" !
如果将长度大于 2 的列表传递给 function Linked,则不会有任何异常,因此不会出现“Tail Found”!
You gave a list with length 3, then you got a test with length 3-1.你给出了一个长度为 3 的列表,然后你得到了一个长度为 3-1 的测试。 Index text allowed only for 0 or 1, test[2] certainly get a result
索引文本只允许 0 或 1,test[2] 肯定会得到结果
IndexError: list index out of range
IndexError:列表索引超出范围
If you print(Linked(test_list))
, you will get:如果你
print(Linked(test_list))
,你会得到:
[<__main__.Node object at 0x02A62220>, <__main__.Node object at 0x02AC1220>]
See?看? Only 2 elements in the list, so you can't call index
2
(third element) .列表中只有 2 个元素,因此不能调用 index
2
(third element) 。
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