我正在创建一个链表实现,我无法修复这个错误,即必须双击node.val.val
来打印数据,而不是内存地址.
以下是我的实现:
class LinkedNode:
def __init__(self, val, nxt=None):
self.val = val
self.nxt = nxt
class LinkedList:
def __init__(self, head=None):
self.head = head
def append(self, new_val):
node = LinkedNode(new_val, None)
if self.head:
curr = self.head
while curr.nxt:
curr = curr.nxt
curr.nxt = node
else:
self.head = node
def print(self):
curr = self.head
while curr:
**print(curr.val)**
curr = curr.nxt
l1 = LinkedList()
l1.append(LinkedNode(2))
l1.append(LinkedNode(3))
l1.append(LinkedNode(4))
l1.print()
当打印函数中的line为"打印(当前值)"时,该函数打印内存地址.当line为"print(curr.val.val)"时,函数打印2,3,4.
有人有可能的解决方案吗?