递归函数以生成/打印斐波那契数列

问题描述

我正在尝试创建一个递归函数调用方法，该方法将打印斐波那契直到特定位置:

I am trying to create a recursive function call method that would print the Fibonacci until a specific location:

1 function f = fibonacci(n)
2 fprintf('The value is %d\n', n)
3 if (n==1)
4     f(1) = 1;
5     return;
6 elseif (n == 2)
7     f(2) = 2;
8 else
9     f(n) = fibonacci(n-1) + fibonacci(n-2);   
10 end
11 end

根据我的理解，将递归调用fibonacci函数，直到传递给它的参数n的值为1.然后函数栈将相应地回滚.因此，当我从命令中调用此函数时:

As per my understanding the fibonacci function would be called recursively until value of argument n passed to it is 1. Then the function stack would rollback accordingly. So when I call this function from command:

>> fibonacci(4)

n的值为4，因此第9行的执行方式如下:

The value of n is 4, so line 9 would execute like:

9 f(4) = fibonacci(3) + fibonacci(2);

现在，我相信第一个fibonacci(3)将被调用-因此，再次调用fibonacci(3)

Now I believe that that first fibonacci(3) would be called - hence again for fibonacci(3)

9 if(3) = fibonacci(2) + fibonacci(1);

第3行和第6行的ifs很重要.

The ifs in line number 3 and 6 would take care.

但是现在fibonacci(2)+ fibonacci(1)语句将变为:

But now how fibonacci(2) + fibonacci(1) statement would change to:

 if(3) = 2 + 1;

我收到以下错误，无法进一步调试以解决该问题:

I am receiving the below error and unable to debug further to resolve it:

>> fibonacci(4)
The value is 4
The value is 3
The value is 2
The value is 1
In an assignment  A(I) = B, the number of elements in B and I must be the same.

Error in fibonacci (line 9)
    f(n) = fibonacci(n-1) + fibonacci(n-2);

Error in fibonacci (line 9)
    f(n) = fibonacci(n-1) + fibonacci(n-2);

请提供一些解决方案的见解，并首先在第9行递归调用哪个参数斐波那契函数.

Please provide some insight for the solution and with which parameter would fibonacci function be recursively called at line number 9 first and consequently.

例如n = 4

f(n) = fibonacci(3) + fibonacci(2);

那么MATLAB将首先调用fibonacci(3)还是fibonacci(2)?

So will MATLAB call fibonacci(3) or fibonacci(2) first?

代码不应该像下面这样:

Shouldn't the code be some thing like below:

1 function f = fibonacci(n)
2 fprintf('The valus is %d\n', n)
3 if (n==1)
4     f(1) = 1;
5     return f(1);
6 elseif (n == 2)
7     f(2) = 2;
8    return f(2);
9 else
10   f(n) = fibonacci(n-1) + fibonacci(n-2);   
11 end
12 end

为什么函数中的返回表达式会导致错误?

Why return expression in a function is resulting in an error?

推荐答案

尝试一下:

 function f = fibonacci(n)
 if (n==1)
     f= 1;
 elseif (n == 2)
     f = 2;
 else
     f = fibonacci(n-1) + fibonacci(n-2);   
 end

请注意，这也是一个递归(仅对每个n计算一次):

Note that this is also a recursion (that only evaluates each n once):

function f=fibonacci(n)
  f=additive(n,1,2);

function a=additive(n,x0,x1)
  if(n==1)
    a=x0;
  else 
    if(n==2)
      a=x1;
    else 
      a=additive(n-1,x1,x0+x1);
    end
  end

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