为什么 append() 在 Python 中总是返回 None?
问题描述
list = [1, 2, 3]
print(list.append(4)) ## WRONG, print does not work, append() returns None
## RIGHT:
list.append(4)
print(list) ## [1, 2, 3, 4]
我正在学习 Python，我不确定这个问题是否特定于语言以及 append 在 Python 中是如何实现的.
I'm learning Python and I'm not sure if this problem is specific to the language and how append is implemented in Python.
解决方案
append 是一个变异(破坏性)操作(它修改列表而不是返回一个新列表).进行 append 的非破坏性等效的惯用方法是
append is a mutating (destructive) operation (it modifies the list in place instead of of returning a new list). The idiomatic way to do the non-destructive equivalent of append would be
l = [1,2,3]
print l + [4] # [1,2,3,4]
print l # [1,2,3]
回答你的问题，我的猜测是如果append返回新修改的列表，用户可能会认为它是非破坏性的，即他们可能会编写类似的代码
to answer your question, my guess is that if append returned the newly modified list, users might think that it was non-destructive, ie they might write code like
m = l.append("a")
n = l.append("b")
并期望 n 为 [1,2,3,"b"]
相关文章