Question

当我使用while循环时，该函数返回一个正确的值，但是当我使函数递归时，它返回一个nan。出于调试目的，我在返回它之前输入了值（x）并给出了正确的答案，但是在将值返回给调用函数之后，它是一个nan。另一件事，程序对x的系数不取0。任何尝试都会导致nan。下面是我的代码（所有这些只是为了确保我没有提供足够的信息）：

// This is my first useful program
// to calculate the root (Solution) of exponential
// functions up to the fourth degree using
// Newton-Raphson method and applying a recursive function
#include <iostream>
#include <cstdlib>
#include <string>


using namespace std;

// Below is the function fGetRoot's prototype declaration
// c4, c3, etc are the coefficients of x in the 4th power, 3rd power
// etc, while c is the constant

float fGetRoot (float c4, float c3, float c2, float c1, float c);
float fPowerFunc(float fPowered, int iPower);
float x;    // declaring this as global variables so they can be initialized in the calling function
int i=10;  //  and be used in the recursive function without being reinitialized during each recursion


int main()
{
    float c4, c3, c2, c1, c;
    float fRoot;

    cout << "Hello, I am a genie for calculating the root of your problems\
up to the fourth power" << endl;
    cout << "Please enter the values of the coefficients of x to the power 4\
, 3, 2, 1 and the constant respectively" << endl;

    cout << "x^4:" << endl;
    cin >> c4;
    cout << "\n x^3:" << endl;
    cin >> c3;
      cout << "x^2:" << endl;
    cin >> c2;
    cout << "\n x:" << endl;
    cin >> c1;
      cout << "Constant, C:" << endl;
    cin >> c;
    cout << "\nEnter the initial iteration. Any figure is fine. The closer to the answer, the better" << endl;
    cin >> x;

    i=10; // gets the number of times to iterate. the larger, the more accurate the answer
    fRoot = fGetRoot(c4, c3, c3, c1, c);
    cout <<"\nAnd the root is: " << fRoot << "\n";

    return 0;
}


// The fGetRoot function

float fGetRoot(float c4, float c3, float c2, float c1, float c)
{
    float fTemp1, fTemp2, fTemp3, fTemp4;

    // the series of lines below are equivalent to the one line below but clearer
    fTemp1 = c4*fPowerFunc(x,4);
    cout << "This is the value of c4*x^4: "<< fTemp1 << endl; // for debugging purposes
    fTemp2 = c3*fPowerFunc(x,3);
    fTemp3 = fTemp2 + fTemp1;
    fTemp1 = c2*fPowerFunc(x,2);
    cout << "This is the value of c2*x^2: "<< fTemp1 << endl; //for debugging purposes
    fTemp2 = c1*fPowerFunc(x,1);
    fTemp4 = fTemp1 + fTemp2 + c;
    fTemp1 = fTemp3 + fTemp4;
    fTemp2 = 4*c4*fPowerFunc(x,3);
    fTemp3 = 3*c3*fPowerFunc(x,2);
    fTemp4 = fTemp2 + fTemp3 + 2*c2*x;
    fTemp2 = fTemp4;
    fTemp3 = fTemp1/fTemp2;
    fTemp1 = fTemp3;
    fTemp2 = x - fTemp1;
    x = fTemp2;
    i--;

    // The line below is equivalent to the "fTemp" series of lines... just to be sure

//x=x-(c4*fPowerFunc(x,4)+c3*fPowerFunc(x,3)+c2*fPowerFunc(x,2)+c1*fPowerFunc(x,1)+c)/(4*c4*fPowerFunc(x,3)+3*c3*fPowerFunc(x,2)+2*c2*x);

    cout << "\nThis is x: " << x << endl;
    i--;
    if(i==0)
        return x;
    x=fGetRoot(c4, c3, c2, c1, c); // Let the recursion begin...

}

// A simpler approach to the fPowerFunc(). This gets two numbers and powers the left one by the right one. It works right
float fPowerFunc(float fPowered, int iPower)
{
    float fConstant=fPowered;
    while(iPower>0)
    {
        fPowered *=fConstant;
        iPower--;
    }
    return fPowered;
}

Answer 1

你在功能结束时忘了return x。编译器可以在寄存器中保存临时值，使其保持返回值并使其看起来有效。

您应该编译并启用更多警告，以便编译器可以告诉您类似的事情。

Answer 2

此声明中还有另一个问题：if(i==0)，应由if(i<=0)替换。

在浮动领域，严格来说没有平等的概念。范围中的数字是更正确的抽象。例如，要测试f是否大约为零，您可以这样做： if(f >= -epsilon && f <= epsilon) ... 其中epsilon是标准FLT_EPSILON或其他特定于应用的精度常量。

递归函数返回nan（c ++）

2 个答案: