斐波那契和非斐波那契编码
我很抱歉，因为这可能是一个常见问题，但我认为我正在寻找一个在其他主题中找不到的非常具体的答案。基本上，我对添加数字的流程感到很困惑。这是两个类似的代码，它们以不同的方式计算数字。对此有什么简单的解释吗？
>>> a = 0
>>> b = 1
>>> while b <1000:
print b
a, b = b, a+b
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987
>>> a =0
>>> b=1
>>> while b<1000:
print b
a = b
b = a+b
1,2,4,8,16,32,64,128,256,512
I apologize since this might be a common question, but I think I am looking for a quite specific answer that wouldn't be found in other topics. Basically, I am quite confused about the flow of adding numbers. Here are two similar codes that compute numbers differently. Is there any simple explanation for this?
>>> a = 0
>>> b = 1
>>> while b <1000:
print b
a, b = b, a+b
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987
>>> a =0
>>> b=1
>>> while b<1000:
print b
a = b
b = a+b
1,2,4,8,16,32,64,128,256,512
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评论（3）
区别在于交换时的值是什么
a, b = b, a+b将 a 设置为 b 并将 a 设置为 a+b 但交换是相对同时完成的，因此不是按顺序，即 b 中的更改不考虑 a 首先更改。
在第二个示例中，
值已更改，第二条语句尊重第一个语句的更改
The difference lies in what the values are WHEN swapped
a, b = b, a+bsets a to b and sets a to a+b but the swaps are done relatively at the same time so it's not in order, ie the change in b doesn't respect that a was changed first.
In the second example
the values are changed and the 2nd statement respects the change of the first
交换存在优先级差异。
在第一个示例中，您分配：
一个= 1
b = 1
在第二个示例中，您要分配：
一个= 1
b = 2
为了实现与第一个示例相同的操作顺序，您必须使用临时变量。
There is a precedence difference in the swap.
In the first example you are assigning:
a = 1
b = 1
In the second example you are assigning:
a = 1
b = 2
In order to achieve the same order of operations as the first example you'll have to use a temp var.
第一个代码示例中的分配同时发生，第二个代码示例中的分配连续发生，从而导致不同的答案。
The assignments happen at the same time in the 1st code example, and serially in the 2nd, leading to a different answer.