出于某种原因,要捕获变量,比如 i 这里需要定义一个函数,然后调用它 不 捕获 我 并且,在调用时,使用3,即用于 我 .

是否有更好的方法来捕获变量?(没有像那些函数定义/调用那样多余的语法噪音)

class Node(object):
    def __init__(self, value, next=None):
        self.value = value
        self.next = next
    
    def __str__(self):
        return str(self.value) + ',' + str(self.next)
    
    def list2LinkedListFoldrImpPb(nums):
      ret = {0:lambda x:x}
      i = 0  
      for num in nums:
        ret[i+1] = lambda rs: ret[i](Node(num, rs))#---- i NOT captured !!
        i = i+1
      return ret[i](None)
    
    def list2LinkedListFoldrImp(nums):
      ret = {0:lambda x:x}
      i = 0
      def setf(ret, i, num):
          ret[i+1] = lambda rs: ret[i](Node(num, rs))
      for num in nums:
        setf(ret, i, num) #---- i captured !!
        i = i+1
      return ret[i](None)
    
    
    print(list2LinkedListFoldrImpPb([5,4,1])) # maximum recursion depth exceeded  !!!
    print(list2LinkedListFoldrImp([5,4,1])) # works 

解决方案

作为参考,正如重复链接中提到的那样,解决方案是确保列出所有要作为本地参数捕获的变量。

有 不 捕获体内,创建范围/环境(闭包的一部分) 只有 调用函数时(和默认值 论据 -我想这是通话的一部分- 是 捕获)

class Node(object):
  def __init__(self, value, next=None):
    self.value = value
    self.next = next

  def __str__(self):
    return str(self.value) + ',' + str(self.next)

# Creating a function and calling it works
def list2LinkedListFoldrOK(nums):
  ret = {0:lambda x:x}
  i = 0
  def setf(ret, i, num):
      ret[i+1] = lambda rs: ret[i](Node(num, rs))
  for num in nums:
    setf(ret, i, num)
    i = i+1
  return ret[i](None)

#Pb : The i in each lambdas refers to the *last* value that i had in the scope it came from, i.e., 3
def list2LinkedListFoldrImpKO(nums):
  ret = {0:lambda x:x}
  i = 0  
  for num in nums:
    ret[i+1] = lambda rs: ret[i](Node(num, rs))
    i = i+1
  return ret[i](None)

#Solution : List all captured variables as locals via default
def list2LinkedListFoldrImpOK2(nums):
  ret = {0:lambda x:x}
  i = 0  
  for num in nums:
    ret[i+1] = lambda rs, i=i, num=num: ret[i](Node(num, rs))
    i = i+1
  return ret[i](None)

#Solution : or make an actual call 
def list2LinkedListFoldrImpOK3(nums):
  ret = {0:lambda x:x}
  i = 0  
  for num in nums:
    ret[i+1] = (lambda i,num: lambda rs: ret[i](Node(num, rs)))(i,num)
    i = i+1
  return ret[i](None)



print(list2LinkedListFoldrImpOK2([5,4,1]))
print(list2LinkedListFoldrImpOK3([5,4,1]))
print(list2LinkedListFoldrImp([5,4,1]))