Question

我是C的新手，我正在尝试寻找机器epsilon（1.0 + macheps＆gt; 1.0），eta（eta＆gt; 0.0）和MAX（MAX＆lt; infinity），但我的代码不起作用如预期。首先，使用80位精度计算macheps。如何强制它单，双和长双？其次，代码根本没有完成双精度计算结果。

编辑：修正了格式错误。

/* macheps eta max */

#include <stdio.h>
#include <math.h>
#include <float.h>
#define TYPE long double

int main(void)
{
    TYPE macheps = (TYPE) 1.0;
    TYPE eta = (TYPE) 1.0;
    TYPE maksymilian = (TYPE) 2.0;
    TYPE real_macheps;
    TYPE real_eta;
    TYPE real_maksymilian;

    TYPE something = (TYPE) 1.0 + (TYPE) macheps;

    while ((TYPE) something > (TYPE) 1.0)
    {
        real_macheps = (TYPE) macheps;
        printf("%e ", (TYPE) real_macheps);
        macheps = (TYPE) macheps/(TYPE) 2.0;
        something = (TYPE) 1.0 + (TYPE) macheps;
    }
    printf("%e\n", (TYPE) real_macheps);

    while ((TYPE) eta > (TYPE) 0.0)
    {
        real_eta = (TYPE) eta;
        eta = (TYPE) eta/(TYPE) 2.0;
    }
    printf("%e\n", (TYPE) real_eta);

    while ((TYPE) maksymilian != INFINITY)
    {
        (real_maksymilian) = (TYPE) maksymilian;
        maksymilian = (TYPE) maksymilian*(TYPE) 2.0;
    }
    real_maksymilian = (TYPE) real_maksymilian * (TYPE) (2.0-(TYPE) real_macheps);
    printf("%e\n", (TYPE) real_maksymilian);
}

EDIT2：上面的代码不应该强制精确吗？我错过了什么？

EDIT3：仍然没有为长双打给出正确的目标。

Answer 1

long是long signed int的简写，因此是整数类型。请改用long double。

根据无法呈现的数字的舍入方式，(1.0 + machepsfloat) > 1.0可能始终为真。检查<float.h>中的FLT_ROUNDS以确定实施的舍入行为。

Answer 2

您的程序将调用未定义的行为。在C中，long表示long int而非long double。

Answer 3

忘记发布更正的代码。我不得不做一个解决方案，但它有效：

/* macheps eta max */

#include <stdio.h>
#include <math.h>
#include <float.h>
#define TYPE long double /* printf 'e' for float and double, 'Le' for long double */
/* used_type: float 0, double 1, long double 2 */
int main(void)
{
    int used_type = 2;
    TYPE macheps = (TYPE) 1.0;
    TYPE eta = (TYPE) 1.0;
    TYPE maksymilian = (TYPE) 1.0;
    TYPE real_macheps = 0.0;
    TYPE real_eta = 0.0;
    TYPE real_maksymilian = 0.0;
    TYPE something = (TYPE) 1.0 + (TYPE) macheps;

    while ((TYPE) something > (TYPE) 1.0)
    {
        real_macheps = (TYPE) macheps;
        macheps = (TYPE) macheps/(TYPE) 2.0;
        something = (TYPE) 1.0 + (TYPE) macheps;
    }

    while ((TYPE) eta > (TYPE) 0.0)
    {
        real_eta = (TYPE) eta;
        eta = (TYPE) eta/(TYPE) 2.0;
    }

    while ((TYPE) maksymilian != INFINITY)
    {
        (real_maksymilian) = (TYPE) maksymilian;
        maksymilian = (TYPE) maksymilian*(TYPE) 2.0;
    }
    real_maksymilian = (TYPE) real_maksymilian * (TYPE) (2.0-(TYPE) real_macheps);

    if (used_type == 2)
    {
        printf("%Le\n", (TYPE) real_macheps);
        printf("%Le\n", (TYPE) real_eta);
        printf("%Le\n", (TYPE) real_maksymilian);
    }
    else
    {
        printf("%e\n", (TYPE) real_macheps);
        printf("%e\n", (TYPE) real_eta);
        printf("%e\n", (TYPE) real_maksymilian);
    }
}

C中的机器epsilon估计

3 个答案: