遍历 for 循环时生成器中断的问题

Question

分配使用生成器从头开始设计链表类来遍历列表，但我在调试此问题时遇到了问题：

class LinkedList:
def __init__(self, data = None):
self.node = Node(data)
self.first = self.node 
self.last = self.node
self.n = 1
def append(self, data = None):
new = Node(data)
self.last.next = new
self.last = new
self.n += 1
def __iter__(self):
self.index = self.first
return self.generator()
def generator(self):
for i in range(0, self.n):
yield self.index
self.index = self.index.next

电流发生器通过的测试之一是：

for n in a:
print n
# which should and does output
0
1
2

但是，在这种情况下，它无法输出正确的值

for n in a:
if n == 2:
break
else:
print n
# Outputs 0 1 2, instead of:
0
1

我认为我对发电机缺乏了解，并希望你们提供任何帮助！

Answer 1

• 您不应该使用n == 2作为中断条件。

试试这个：

# Assume your defined class `Node.py` like this:
class Node:
def __init__(self, data=None):
self.val = data
self.next = None
# I didn't modify your `LinkedList.py`
class LinkedList:
def __init__(self, data=None):
self.node = Node(data)
self.first = self.node
self.last = self.node
self.n = 1
def append(self, data=None):
new = Node(data)
self.last.next = new
self.last = new
self.n += 1
def __iter__(self):
self.index = self.first
return self.generator()
def generator(self):
for i in range(0, self.n):
yield self.index
self.index = self.index.next

if __name__ == '__main__':
ass = LinkedList(0)
ass.append(1)
ass.append(2)
ass.append(3)
for n in ass:
if n.val == 2:  # Notice here
break
else:
print(n.val)  # And here