无幂函数的浮点数求幂 [英] Floating point exponentiation without power-function

问题描述

当前，我必须在电源操作员被窃听的环境中工作.谁能想到一种方法可以暂时解决此错误，并在没有幂函数或运算符的情况下计算a ^ b(两个浮点数)?

Currently I have to work in an environment where the power-operator is bugged. Can anyone think of a method temporarily work around this bug and compute a^b (both floating point) without a power function or operator?

推荐答案

如果有sqrt()可用:

if you have sqrt() available:

double sqr( double x ) { return x * x; }
// meaning of 'precision': the returned answer should be base^x, where
//                         x is in [power-precision/2,power+precision/2]
double mypow( double base, double power, double precision )
{   
   if ( power < 0 ) return 1 / mypow( base, -power, precision );
   if ( power >= 10 ) return sqr( mypow( base, power/2, precision/2 ) );
   if ( power >= 1 ) return base * mypow( base, power-1, precision );
   if ( precision >= 1 ) return sqrt( base );
   return sqrt( mypow( base, power*2, precision*2 ) );
}
double mypow( double base, double power ) { return mypow( base, power, .000001 ); }

测试代码:

void main()
{
   cout.precision( 12 );
   cout << mypow( 2.7, 1.23456 ) << endl;
   cout << pow  ( 2.7, 1.23456 ) << endl;
   cout << mypow( 1.001, 1000.7 ) << endl;
   cout << pow  ( 1.001, 1000.7 ) << endl;
   cout << mypow( .3, -10.7 ) << endl;
   cout << pow  ( .3, -10.7 ) << endl;
   cout << mypow( 100000, .00001 ) << endl;
   cout << pow  ( 100000, .00001 ) << endl;
   cout << mypow( 100000, .0000001 ) << endl;
   cout << pow  ( 100000, .0000001 ) << endl;
}

输出:

3.40835049344
3.40835206431
2.71882549461
2.71882549383
393371.348073
393371.212573
1.00011529225
1.00011513588
1.00000548981
1.00000115129

这篇关于无幂函数的浮点数求幂的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！