这项计划中的"返回斐波纳契......"是怎么回事？

Question

我有问题了解return fibonacci( number-1 ) + fibonacci( number-2 )以下程序中的功能:

import sys
def fibonacci( number ):
    if( number <= 2  ):
        return 1
    else:
        return fibonacci( number-1 ) + fibonacci( number-2 )

问题是我无法想象这条线是如何工作的:

return fibonacci( number-1 ) + fibonacci( number-2 )

"fibonacci(数字-1)"和"斐波那契(数字-2)"是否同时处理？或者"斐波纳契(数字-1)"是第一个被处理,然后是第二个？

我只看到处理它们最终都会返回'1'所以最后的结果我希望看到它是'1 + 1'='2'

如果有人能够详细解释其计算过程,我将非常感激.

我认为这是一个非常新的问题,但我无法真正了解其过程.

Answer 1

你为什么不这样做:

>>> def fibonacci(number):
...     if number < 2:
...         return number
...     print "Number is currently %d, getting fibonacci(%d)" % (number, number - 1)
...     minus_one = fibonacci(number-1)
...     print "Number is currently %d, just got fibonacci(%d), now getting fibonacci(%d)" % (number, number - 1, number - 2)
...     minus_two = fibonacci(number-2)
...     print "Number is currently %d, returning %d + %d" % (number, minus_one, minus_two)
...     return minus_one + minus_two

所以当你打电话给fibonacci你的时候会得到这样的东西:

>>> fibonacci(4)
Number is currently 4, getting fibonacci(3)
Number is currently 3, getting fibonacci(2)
Number is currently 2, getting fibonacci(1)
Number is currently 2, just got fibonacci(1), now getting fibonacci(0)
Number is currently 2, returning 1 + 0
Number is currently 3, just got fibonacci(2), now getting fibonacci(1)
Number is currently 3, returning 1 + 1
Number is currently 4, just got fibonacci(3), now getting fibonacci(2)
Number is currently 2, getting fibonacci(1)
Number is currently 2, just got fibonacci(1), now getting fibonacci(0)
Number is currently 2, returning 1 + 0
Number is currently 4, returning 2 + 1
3

它仍然很复杂,但至少现在你可以看到函数正在做什么来计算你的数字.