c++ - 这个循环有什么问题吗？

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如果我删除 while并在 if, else if, else 中选择一个条件，我可以得到一个结果。但是当我添加 while循环到函数solve中，就会发生无限循环。请问问题出在哪里？

#include<stdio.h>
#include<math.h>
int a, b, c, d;
float fun(float x);
float solve(void);

int main()
{
    printf("Put in the coefficient of the equation:\n");
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    scanf("%d", &d);

    float ans = solve();

    printf("A solve for the equation：x=%.3f", ans);

    return 0;
}

float fun(float x)
{
    float value = a * x * x * x + b * x * x + c * x + d;

    return value;
}

float solve(void)
{
    float x1 = -100, x2 = 100, x3;
    float diff = fabs(fun(x1) - fun(x2));
    while (diff > 0.001)
    {
        x3 = (x1 * fun(x2) - x2 * fun(x1)) / (fun(x2) - fun(x1));

        if (fun(x3) == 0)
        {
            x1 = x3;
            x2 = x3;
        }
        else if ((fun(x3) * fun(x1)) < 0)
        {
            x2 = x3;
        }
        else
        {
            x1 = x3;
        }

        diff = fabs(fun(x1) - fun(x2));
    }

    return diff;
}

Answer 1

首先那是c而不是c++!如果你返回 diff 你每次都会得到一个小于 0.001 的值作为解决方案。我认为你应该返回x3。如果你问 fun(x3) == 0 fun(x3) 必须完全等于 0。在大多数情况下，使用 float 你永远无法达到这一点。最好这样做:

        funx3=fun(x3);

        if (funx3 == 0.0 || funx3>0.0 &&funx3<0.001 || funx3<0.0 && funx3>-0.001 )
        {
            x1 = x3;
            x2 = x3;
        }

这也不是在所有情况下都有效。要解决这个问题，请在整个程序中使用 double 。您还将获得更多的小数位。我不完全确定所有小数位是否正确。也许你不应该削减一个时代。

#include<stdio.h>
#include<math.h>
int a, b, c, d;
double fun(float x);
double solve(void);

int main()
{
    printf("Put in the coefficient of the equation:\n");
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    scanf("%d", &d);

    float ans = solve();

    printf("A solve for the equation：x=%.8f", ans);

    return 0;
}

double fun(float x)
{
    double value = a * x * x * x + b * x * x + c * x + d;

    return value;
}

double solve(void)
{
    double x1 = -100, x2 =100, x3;
    double diff = fabs(fun(x1) - fun(x2));
    double funx3;
    while (diff > 0.0000001)
    {
        x3 = (x1 * fun(x2) - x2 * fun(x1)) / (fun(x2) - fun(x1));

        funx3=fun(x3);

        if (funx3 == 0.0 || funx3>0.0 &&funx3<0.0000001 || funx3<0.0 && funx3>-0.0000001 )
        {
            x1 = x3;
            x2 = x3;
        }
        else if ((fun(x3) * fun(x1)) < 0.0)
        {
            x2 = x3;
        }
        else
        {
            x1 = x3;
        }

        diff = fabs(fun(x1) - fun(x2));
    }

    return x3;
}