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将分数转换为浮点数

[英]Converting fractions to floating point

I'm trying to convert a fraction to floating point and use it for comparison.我正在尝试将分数转换为浮点数并将其用于比较。 but the values are too small and it returns true for the results of the Boolean variables.但这些值太小,对于布尔变量的结果,它返回 true。 is my converision correct ?我的转换正确吗? or should I do it in another way which I don't know ?或者我应该以另一种我不知道的方式来做吗?

A test case:一个测试用例:

  // result is -0.0074
  float coilh0re = fr32_to_float(GO_coil_H[0].re)*0.8f;
  // result is -0.0092
  float coilrefundamental = fr32_to_float(CoilEepromData.coilboardhspule.reFundamental);
  // result is -0.01123
  float coilh0re2 = fr32_to_float(GO_coil_H[0].re)*1.2f;
  -0.0074>-0.0092> -0.01123

here is a snipped of the code这是代码的片段

       bool resultA  = fr32_to_float(GO_coil_H[0].re)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.reFundamental)  ? 1 : 0;
       bool resultB  = fr32_to_float(CoilEepromData.coilboardhspule.reFundamental)  <= fr32_to_float(GO_coil_H[0].re)*1.2f ? 1 : 0;

       bool resultAB = !(resultA & resultB); // always true

       bool resultC  =  fr32_to_float(GO_coil_H[1].re)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.reHarmonic) ? 1:0;
       bool resultD  = fr32_to_float(CoilEepromData.coilboardhspule.reHarmonic)  <= fr32_to_float(GO_coil_H[1].re)*1.2f ? 1:0;

       bool resultCD = !(resultC & resultD); // always true

       bool resultE  =  fr32_to_float(GO_coil_H[0].im)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.imFundamental)? 1 : 0;
       bool resultF  =  fr32_to_float(CoilEepromData.coilboardhspule.imFundamental) <= fr32_to_float(GO_coil_H[0].im)*1.2f ? 1 : 0;

       bool resultEF = !(resultE & resultF);// always true

       bool resultG  =  fr32_to_float(GO_coil_H[1].im)*0.8f < CoilEepromData.coilboardhspule.imHarmonic ? 1 : 0;
       bool resultH  =  fr32_to_float(CoilEepromData.coilboardhspule.imHarmonic) <= fr32_to_float(GO_coil_H[1].im)*1.2f ? 1 : 0;

       bool resultGH = !(resultG & resultH);// always true

        if(! ((fr32_to_float(GO_coil_H[0].re)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.reFundamental)) && (fr32_to_float(CoilEepromData.coilboardhspule.reFundamental) <= fr32_to_float(GO_coil_H[0].re)*1.2f) ) 
        || ! ((fr32_to_float(GO_coil_H[1].re)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.reHarmonic))    && (fr32_to_float(CoilEepromData.coilboardhspule.reHarmonic) <= fr32_to_float(GO_coil_H[1].re)*1.2f)    )
        || ! ((fr32_to_float(GO_coil_H[0].im)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.imFundamental)) && (fr32_to_float(CoilEepromData.coilboardhspule.imFundamental) <= fr32_to_float(GO_coil_H[0].im)*1.2f) )
        || ! ((fr32_to_float(GO_coil_H[1].im)*0.8f < fr32_to_float(CoilEepromData.coilboardhspule.imHarmonic))    && (fr32_to_float(CoilEepromData.coilboardhspule.imHarmonic) <= fr32_to_float(GO_coil_H[1].im)*1.2f)    ) )

        {
            eUserCode           = E_USER_SOIL_FAILED;
            eProcessState       = E_ERROR_HANDLING;  
        }

}

If appears OP wants to test if a value reFundamental is in range +/-20% of re .如果出现 OP 想要测试值reFundamental是否在re +/-20% 范围内。 This is not a float precision issue, but a math one.这不是float精度问题,而是数学问题。

// Simplified problem
float re = -0.01123f/1.2f;
float reFundamental = -0.0092f;
bool resultA  = re*0.8f < reFundamental;
bool resultB  = reFundamental <= re*1.2f;
bool resultAB = !(resultA & resultB); // always true

But the values are negative and so the < and <= should be reversed .但是值是负的,所以<<=应该颠倒

Various alternatives.各种替代品。 Example: (Adjust to taste)示例:(根据口味调整)

bool in_range(float x, float limit, float factor) {
  float limitp = limit*(1.0f + factor);
  float limitm = limit*(1.0f - factor);

  if (x > limitm) return x <= limitp;
  if (x < limitm) return x >= limitp;
  return x == limitp;
}

bool resultAB = !in_range(fr32_to_float(CoilEepromData.coilboardhspule.reFundamental),   
    fr32_to_float(GO_coil_H[0].re), 0.20);

If you want to compare fractions - do not use floating-point at all.如果您想比较分数 - 根本不要使用浮点数。 Convert them to the same denominator and compare numerators.将它们转换为相同的分母并比较分子。

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