Question

我正在尝试使用牛顿法计算x的平方根。一切正常，a等于x的平方根，直到它返回给我一个完全不同的（总是恒定的）更大的数字时才返回它。

int main()
{
    float newtonA;
    float newtonX = 35735;
    float epsilon = 0.001;

    newtonA = newtonX / 2;

    printf("\n********RECURSIVE NEWTON********\n");
    printf("The square root of %0.1f is: %0.2f\n", newtonX, newtonRec(newtonX, newtonA, epsilon));

    return 0;
}


float newtonRec (float x, float a, float eps)
{
    if (abs(a * a - x) <= eps )
    {
        printf("\n****%0.2f****\n", a);*/
        return a;
    }
    else
    {
        printf("\n***a: %.1f x: %.1f***\n", a, x);
        a = (a + x / a) / 2;
        newtonRec(x, a, eps);
    }

    return a;
}

Answer 1

请更改

a = (a + x / a) / 2;
newtonRec(x, a, eps);

到

a = (a + x / a) / 2;
return newtonRec(x, a, eps);

> $./a.out
> ****189.04**** The square root of 35735.0 is: 189.04

递归语句中返回的值与计算得出的值不同

1 个答案: