返回 C 中斐波那契数列的特定数字

Question

我正在编写一个 C 程序来计算斐波那契序列中的特定数字，尽管我无法将序列作为数组返回......

我做错了什么？

int fibonacci(int ceiling)
{
  int counter;
  int num1 = 1, num2 = 1;
  static int fibArray[1000];
  for (counter = 1; counter < ceiling; counter+=2)
    {
      fibArray[counter] = num1;
      fibArray[counter+1] = num2;
      num2 += num1;
      num1 += num2;
    }
  return &(fibArray);
}

我也收到错误：

fibonacci.c:28: warning: return makes integer from pointer without a cast

？

Answer 1

由于您要返回指向数组的指针，因此返回类型应为 int*。这是示例代码：

int* fibonacci(int ceiling) //Modified return type
{
  int counter;
  int num1 = 1, num2 = 1;
  static int fibArray[1000];
  for (counter = 1; counter < ceiling; counter+=2)
    {
      fibArray[counter] = num1;
      fibArray[counter+1] = num2;
      num2 += num1;
      num1 += num2;
    }
  return (fibArray); //Return the address of the array's starting position
}

Answer 2

您想从 fibArray 返回一个元素，对吗？在这种情况下，请使用 return fibArray[...];，其中 ... 是元素索引。

Answer 3

你不需要数组。您只需要两个整数来保存当前值和先前值。然后你做：

newValue = previousValue + newValue;
previousValue = newValue - previousValue;`

你应该被设置。当达到限制时，只需返回 newValue 即可。

如果您确实想返回整个数组（您说的是“计算斐波那契序列中的特定数字”，这不是整个序列）。将函数的返回类型设置为 int* 并使用数组。

Answer 4

//you dont need all the "includes" I have in here. These are just a basic format I work with. This should get you going if you arent going already. cheers. :) 

#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<stdio.h>
#include<cctype>
#include<list>
#include<string>

using namespace std;
int main()
//-------------------------------declaration of variables----------------------------------
{
   int num1, num2;
   int initial_value, final_value;
//-----------------------------------------inputs------------------------------------------
   cout << "What is your first number? :";
   cin  >> num1;
   initial_value = num1;
   cout << "What is your second number? :";
   cin >> num2;
   final_value = num2;
//-----------------------------------------dasloop----------------------------------------
   do
   {
       final_value = initial_value + final_value;
       initial_value = final_value - initial_value;
       cout  <<  final_value  <<  endl;
   }
   while(final_value <= 1000);

//---------------------------exits perfectly when greater than 1000------------------------
   cout << endl << endl;
   system("pause");
   return 0;
}
//I have it exit at 1000 because its a nice round number that allows you enough room
//to see the code, and sequence are both correct.