英文:
Why does my Python function enter the else block when I use recursion to generate a Fibonacci list with 'else' in the if statement?
问题
递归。该函数返回一个大致长度的简单斐波那契列表。
我们有一个函数，可以在没有 else 块的情况下正常工作。
但令人困惑的是，它也可以在有 else 块的情况下正确工作。
def sp_fibo_generation(length):
if length <= 0:
return [1]
if length == 1:
return [1, 1]
else:
sp = sp_fibo_generation(length - 1)
n = (sp[-1] + sp[-2])
sp.append(n)
return sp
是的，这段代码在没有 else 的情况下也可以正常工作。但为什么它带有 else 也能工作呢？
我尝试理解深度递归的原理。
英文:
Recursion. The function returns a simple Fibonacci list in an approximate length.
We have a function that can work properly without the else block.
But confusingly, it also works correctly with an else block.
def sp_fibo_generation(length):
if length <= 0:
return [1]
if length == 1:
return [1, 1]
else:
sp = sp_fibo_generation(length - 1)
n = (sp[-1] + sp[-2])
sp.append(n)
return sp
Yes, the code will work correctly without else. But why it does it work with else?
I tried to understand the principle of deep recursion.
答案1
得分: 3
因为return退出函数。所以如果你跳过了第二个if并且仍然在函数内部，那么else:无论如何都是隐含的，除非你明确指定。
这两种风格都相当常见，所以我不会推荐其中一种优于另一种；但你可能想要尽量保持一致：要么将第二个if改为elif，要么删除最后的else:并将该块的缩进去掉。
英文:
Because return exits the function. So if you fall off the second if and are still within the function, the else: is implicit anyway if you don't make it explicit.
Both styles are reasonably common, so I won't recommend one above the other; but you probably want to try to be consistent: Either change the second if to elif, or remove the final else: and remove the indentation from that block.
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