I got this from the documentation. the Fibonacci series:

1 def fib(n):

2 a, b = 0, 1

3 while a < n:

4 print(a, end=' ')

5 # a, b = b, a+b # This is the original line

6 a = b # Tried replacing with these lines

7 b = a+b # This too

8 print()

9 fib(100)

My question is, why it produces wrong result (with line 6 and 7)

How (a,b = b, a+b ) is different from (a=b and b=a+b) ???

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Otto Von Chesterfield
· August 20, 2015
a, b = b, a+b happens at the same time. In order to do the same thing, you have to do:
In this line of code:
A hasn't changed yet when also calculating b. Also, use the code block. It's for the better. |

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Are Riff
· August 19, 2015
It is taken from the python documentation.
The to working code is a, b = b, a+b So, I thought that is more difficult to read. So, I tried to break the lines into two: a = b b = a + b logically (for me) it is the same. So I comment out the original code to test the new one. But it gave different result, means that I was wrong. But why? |

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Otto Von Chesterfield
· August 19, 2015
First, make sure to cover your code with the code tag.
It makes the code much easier to read. The code button is the piece of paper with the <> to the left of the dialogue box. Next. By defining a to be b before you say b = a+b, by algebra, you're saying that b = b+b, or b = 2*b. Why a, b = b, a+b works is that a in b = a+b does not change - it's still a. It sounds confusing, but in summary, by doing both operations at the same time, each computation works as if none of the equations happened yet. |

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