如何避免在C ++中浮点运算的舍入问题? [英] How to get around rounding issues in floating point arithmetic in C++?
问题描述
我遇到一些问题,浮点运算不准确。我试图基于一个加权公式计算得分,其中每个输入变量的权重大约是下一个重要的20倍。然而,输入是实数,因此我最后使用double来存储结果。下面的代码有丢失E1和E2之间的区别的问题。
Im running into some issues with floating point arithmetic not being accurate. I'm trying to calculate a score based on a weighted formula where every input variable weighs about as much as 20 times the next significant one. The inputs however are real numbers, so I ended up using a double to store the result. The code below has the problem of losing the difference between E1 and E2.
这段代码是性能敏感的,所以我需要找到一个有效的答案这个问题。我想到将我的输入乘以一百,然后使用一个int(因为那将是精确到足够我认为),但我怀疑这是最好的解决方案,因此的问题。
This code is performance sensitive, so I need to find an efficient answer to this problem. I thought of multiplying my inputs by a hundred and then using an int (since that would be precise enough I think), but I doubt that is the best solution, hence the question.
#include <iostream>
int main()
{
double score1, score2;
float a = 2.75 ;
float b = 5.25 ;
float c = 5.25 ;
float d = 2.75 ;
float E1 = 3 ;
float E2 = 6 ;
score1 = 20 * b - 1 * a + 0.05 * d /* - 0.0025 * c*/ + 0.0001 * E1 ;
score2 = 20 * b - 1 * a + 0.05 * d /* - 0.0025 * c*/ + 0.0001 * E2 ;
std::cout << score1 << std::endl;
std::cout << score2 << std::endl;
std::cin.get();
return 0;
}
//ouputs:
//102.388
//102.388
推荐答案
- 您不是输出整个值,使用
cout< setprecision(number_of_digits)< score1<< endl;
- 您的分数计算需要多少有效数字?
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