【C语言库函数源代码】

【本程序在Dev C++ 4.9.9.2 下编译通过】

/*

hypot函数对于给定的直角三角形的两个直角边，

求其斜边的长度。

*/

//一般的常规算法：

doublemy_hypot01(double x, double y)

{

double hypotenuse;

x = fabs(x);

y = fabs(y);

if (x < y)

{

double temp = x;

x = y;

y = temp;

}

if (x == 0.)

return 0.;

else

{

hypotenuse = y/x;

return x*sqrt(1.+hypotenuse*hypotenuse);

}

}

#define__SQRT_DBL_MAX 1.3e+154

#define__SQRT_DBL_MIN 2.3e-162

doublemy_hypot02(double x, double y)

{

double ratio;

double r, t, s, p, q;

x = fabs(x), y = fabs(y);

if (x < y)

{

double temp = x;

x = y;

y = temp;

}//保持x 是两个直角边中较长的边，y是较短的边。

if (y == 0.)

return x;

/*

主要考虑的是当x很大而y很小，那么它们的平方和将会造成

丢失小数的现象。首先要判断y是否是太小，x是不是太大。

如果出现这种情况则用，第一个公式来处理。其他的则用

这样可以让求出的斜边更加精确一些。

*/

if ((ratio = y / x) > __SQRT_DBL_MIN && x < __SQRT_DBL_MAX)

return x * sqrt(1. + ratio*ratio);

else

{//使用3次迭代是增加精确度。

r = ratio*ratio, p = x, q = y;

do

{

t = 4.+ r;

if (t == 4.)

break;

s = r / t;

p += 2. * s * p;

q *= s;

r = (q / p) * (q / p);

} while (1);

return p;

}

}

structcomplex

{

double x;

double y;

}

doublecabs(struct complex x)

{

return hypot(z.x,z.y);

}//hypot 函数的封装(这里不再举调用的例子了。)

# defineDBL_MAX1.79769313486231e+308

# defineDBL_MIN2.22507385850721e-308

intmain(void)

{

printf("hypot(3, 4)=%25.17e/n",hypot(3., 4.));

printf("hypot(3*10^150,4*10^150)=%25.17g/n",hypot(3.e+150, 4.e+150));

printf("hypot(3*10^306,4*10^306)=%25.17g/n",hypot(3.e+306, 4.e+306));

printf("hypot(3*10^-320,4*10^-320)=%25.17g/n",hypot(3.e-320, 4.e-320));

printf("hypot(0.7*DBL_MAX,0.7*DBL_MAX) =%25.17g/n",hypot(0.7*DBL_MAX,0.7*DBL_MAX));

printf("hypot(DBL_MAX, 1.0)=%25.17g/n",hypot(DBL_MAX, 1.0));

printf("hypot(1.0, DBL_MAX)=%25.17g/n",hypot(1.0, DBL_MAX));

printf("hypot(0.0, DBL_MAX)=%25.17g/n",hypot(0.0, DBL_MAX));

printf("/n************************************************************/n");

printf("hypot(3, 4)=%25.17e/n",my_hypot01(3., 4.));

printf("hypot(3*10^150,4*10^150)=%25.17g/n",my_hypot01(3.e+150, 4.e+150));

printf("hypot(3*10^306,4*10^306)=%25.17g/n",my_hypot01(3.e+306, 4.e+306));

printf("hypot(3*10^-320,4*10^-320)=%25.17g/n",my_hypot01(3.e-320, 4.e-320));

printf("hypot(0.7*DBL_MAX,0.7*DBL_MAX) =%25.17g/n",my_hypot01(0.7*DBL_MAX,0.7*DBL_MAX));

printf("hypot(DBL_MAX, 1.0)=%25.17g/n",my_hypot01(DBL_MAX, 1.0));

printf("hypot(1.0, DBL_MAX)=%25.17g/n",my_hypot01(1.0, DBL_MAX));

printf("hypot(0.0, DBL_MAX)=%25.17g/n",my_hypot01(0.0, DBL_MAX));

printf("/n************************************************************/n");

printf("hypot(3, 4)=%25.17e/n",my_hypot02(3., 4.));

printf("hypot(3*10^150,4*10^150) =%25.17g/n",my_hypot02(3.e+150, 4.e+150));

printf("hypot(3*10^306,4*10^306)=%25.17g/n",my_hypot02(3.e+306, 4.e+306));

printf("hypot(3*10^-320,4*10^-320)=%25.17g/n",my_hypot02(3.e-320, 4.e-320));

printf("hypot(0.7*DBL_MAX,0.7*DBL_MAX) =%25.17g/n",my_hypot02(0.7*DBL_MAX,0.7*DBL_MAX));

printf("hypot(DBL_MAX, 1.0)=%25.17g/n",my_hypot02(DBL_MAX, 1.0));

printf("hypot(1.0, DBL_MAX)=%25.17g/n",my_hypot02(1.0, DBL_MAX));

printf("hypot(0.0, DBL_MAX)=%25.17g/n",my_hypot02(0.0, DBL_MAX));

system("pause");

return 0;

}