X86斐波那契程序 [英] X86 Fibonacci program

问题描述

我的任务是写一个程序，计算斐波那契数列的前七个值.给出的公式是:

My assignment is to write a program that calculates first seven values of fibonacci number sequence. the formula given is:

Fib(1) = 1, Fib(2) = 1, Fib(n) = Fib(n-1) + Fib(n-2)

我相信这是一个函数，但是我不明白如何将其合并到代码中.我需要将值放在EAX寄存器中.我正在使用MASM，但这没有任何区别.有提示吗?

I believe that is a function but I do not understand how to incorporate it into code. I need to place the values in EAX register. I am using MASM not that makes any difference. Any hints?

推荐答案

我怀疑这是一项学术任务，所以我只打算部分回答这个问题.

I suspect that this is an academic assignment so I'm only going to partially answer the question.

斐波那契数列对非负整数的正式定义如下:

The fibonacci sequence is formally defined for non-negative integers as follows:

F(n) = n                   | n < 2
     = F(n - 1) + F(n - 2) | n >= 2

这给出了:

  n | F(n)
  0 |   0
  1 |   1
  2 |   1
  3 |   2
  4 |   3
  5 |   5
  6 |   8
  7 |  13
etc etc...

您仅需几个寄存器即可完成操作，让我们对其进行识别:

You can do it with just a few registers, let's identify them:

• R _n (请求的斐波那契数的编号)
• R _f1 (用于计算斐波那契数)
• R _f2 (也用于计算斐波那契数)
• R _x (用于保存返回值的寄存器.可以与任何其他寄存器重叠)

• R_n (the number of the requested fibonacci number)
• R_f1 (used to calculate fibonacci numbers)
• R_f2 (also used to calculate fibonacci numbers)
• R_x (the register to hold the return value. can overlap with any other register)

R _n 作为参数传递给函数.R _f1 应该从0开始，R _f2 应该从1开始.

R_n is passed as the argument to the function. R_f1 shall start at 0, and R_f2 shall start at 1.

这是我们得到答案的方法，按例程划分:

Here's what we do to get the answer, split up by routines:

开始

1. 将R _f1 初始化为0.
2. 将R _f2 初始化为1.
3. 继续循环.

1. Initialize R_f1 to 0.
2. Initialize R_f2 to 1.
3. Continue to Loop.

循环

1. 从R _n 中减去2.
2. 如果R _n 小于0，请跳至完成".
3. 将R _f2 添加到R _f1 ，并将结果存储在R _f1 中.
4. 将R _f1 添加到R _f2 ，并将结果存储在R _f2 中.
5. 跳到循环.

1. Subtract 2 from R_n.
2. If R_n is less than 0, jump to Finish.
3. Add R_f2 to R_f1, storing the result in R_f1.
4. Add R_f1 to R_f2, storing the result in R_f2.
5. Jump to Loop.

完成

1. 如果 R_n AND 1 为假(暗示 R_n 是偶数)跳转到 FinishEven.
2. 将R _f1 存储为返回值.
3. 返回.

1. If R_n AND 1 is false (implying that R_n is even) jump to FinishEven.
2. Store R_f1 as the return value.
3. Return.

FinishEven

1. 将R _f2 存储为返回值.
2. 返回.

通过R _n = 5进行跟踪:

Tracing through for R_n = 5:

1. R _f1 = 0
2. R _f2 = 1
3. R _n = R _n -2//R _n = 3
4. 测试R _n <0//错误
5. R _f1 = R _f1 + R _f2 //R _f1 = 0 +1 = 1
6. R _f2 = R _f1 + R _f2 //R _f2 = 1 + 1 = 2
7. 无条件跳转到循环
8. R _n = R _n -2//R _n = 1
9. 测试R _n <0//错误
10. R _f1 = R _f1 + R _f2 //R _f1 = 1 + 2 = 3
11. R _f2 = R _f1 + R _f2 //R _f2 = 3 + 2 = 5
12. 无条件跳转到循环
13. R _n = R _n -2//R _n = -1
14. 测试R _n <0//是
15. 跳到结束
16. 测试R _n &1//是
17. R _x = R _f2 //5

1. R_f1 = 0
2. R_f2 = 1
3. R_n = R_n - 2 // R_n = 3
4. Test R_n < 0 // false
5. R_f1 = R_f1 + R_f2 // R_f1 = 0 + 1 = 1
6. R_f2 = R_f1 + R_f2 // R_f2 = 1 + 1 = 2
7. Unconditional Jump to Loop
8. R_n = R_n - 2 // R_n = 1
9. Test R_n < 0 // false
10. R_f1 = R_f1 + R_f2 // R_f1 = 1 + 2 = 3
11. R_f2 = R_f1 + R_f2 // R_f2 = 3 + 2 = 5
12. Unconditional Jump to Loop
13. R_n = R_n - 2 // R_n = -1
14. Test R_n < 0 // true
15. Jump to Finish
16. Test R_n & 1 // true
17. R_x = R_f2 // 5

我们的表显示F(5)= 5，所以这是正确的.

Our table shows that F(5) = 5, so this is correct.

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