自制pow（）c ++ [英] self made pow() c++

问题描述

我正在阅读我如何写一篇功能函数自己？和dan04给出的答案吸引了我的注意，主要是因为我不知道fortran给出的答案，但我采取了并实现这个：

  #include< iostream> 
 using namespace std; 
 float pow（float base，float ex）{
 // power of 0 
 if（ex == 0）{
 return 1; 
 // negative exponenet 
} else if（ex< 0）{
 return 1 / pow（base， 
 // even exponenet 
} else if（（int）ex％2 == 0）{
 float half_pow = pow（base，ex / 2） 
 return half_pow * half_pow; 
 //整数exponenet 
} else {
 return base * pow（base，ex  -  1）; 
} 
} 
 int main（）{
 for（int ii = 0; ii <10; ii ++）{\ 
 cout< pow（< ii<<.5）=< pow（ii，.5）<< endl; 
 cout<< pow（< ii<<，2）=< pow（ii，2）<< endl; 
 cout<< pow（< ii<<，3）=< pow（ii，3）< endl; 
} 
} 
  

虽然我不确定是否翻译所有调用给出.5作为指数返回0.在答案中，它表示它可能需要一个log2（x）基于 a ^ b = 2 ^（b * log2（a）），但是我不确定把它放在哪里，或者如果我甚至考虑这个问题。

注意：我知道这可能在数学库中定义，但是我不需要为一些函数增加整个数学库的所有费用。

编辑：有没有人知道分数指数的浮点实现？ （我已经看到一个双重实现，但是这是使用一个技巧与寄存器，我需要浮点，并添加一个库只是为了做一个戏法我会更好，只是包括数学库）

$ b

我认为更简单的解决方案会缺乏准确性在结果中或不处理InF和NaN参数。

http://opensource.apple.com/source/Libm/Libm-2026/Source/Intel/expf_logf_powf.c

http://opensource.apple.com/source/Libm/Libm- 315 / Source / ARM / powf.c

I was reading through How can I write a power function myself? and the answer given by dan04 caught my attention mainly because I am not sure about the answer given by fortran, but I took that and implemented this:

#include <iostream>
using namespace std;
float pow(float base, float ex){
    // power of 0
    if (ex == 0){
        return 1;
    // negative exponenet
    }else if( ex < 0){
        return 1 / pow(base, -ex);
    // even exponenet
    }else if ((int)ex % 2 == 0){
        float half_pow = pow(base, ex/2);
        return half_pow * half_pow;
    //integer exponenet
    }else{
        return base * pow(base, ex - 1);
    }
}
int main(){
    for (int ii = 0; ii< 10; ii++){\
        cout << "pow(" << ii << ".5) = " << pow(ii, .5) << endl;
        cout << "pow(" << ii << ",2) = " << pow(ii,  2) << endl;
        cout << "pow(" << ii << ",3) = " << pow(ii,  3) << endl;
    }
}

though I am not sure if I translated this right because all of the calls giving .5 as the exponent return 0. In the answer it states that it might need a log2(x) based on a^b = 2^(b * log2(a)), but I am unsure about putting that in as I am unsure where to put it, or if I am even thinking about this right.

NOTE: I know that this might be defined in a math library, but I don't need all the added expense of an entire math library for a few functions.

EDIT: does anyone know a floating-point implementation for fractional exponents? (I have seen a double implementation, but that was using a trick with registers, and I need floating-point, and adding a library just to do a trick I would be better off just including the math library)

解决方案

Below are links to real implementations of powf.

I expect simpler solutions would lack accuracy in the result or not handle InF and NaN parameters.

http://opensource.apple.com/source/Libm/Libm-2026/Source/Intel/expf_logf_powf.c

http://opensource.apple.com/source/Libm/Libm-315/Source/ARM/powf.c

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