C++ 浮点精度

2022-01-09 00:00:00 floating-point precision c floating-accuracy c++

可能重复:
浮点不准确示例 

double a = 0.3;
std::cout.precision(20);
std::cout << a << std::endl;

结果:0.2999999999999999889

result: 0.2999999999999999889

double a, b;
a = 0.3;
b = 0;
for (char i = 1; i <= 50; i++) {
  b = b + a;
};
std::cout.precision(20);
std::cout << b << std::endl;

结果:1??5.000000000000014211

result: 15.000000000000014211

所以.. 'a' 比它应该的要小.但是如果我们取 'a' 50 次 - 结果会比它应该的要大.

So.. 'a' is smaller than it should be. But if we take 'a' 50 times - result will be bigger than it should be.

这是为什么?在这种情况下如何得到正确的结果?

Why is this? And how to get correct result in this case?

推荐答案

要获得正确的结果，请不要设置大于此数值类型可用的精度:

To get the correct results, don't set precision greater than available for this numeric type:

#include <iostream>
#include <limits>
int main()
{
        double a = 0.3;
        std::cout.precision(std::numeric_limits<double>::digits10);
        std::cout << a << std::endl;
        double b = 0;
        for (char i = 1; i <= 50; i++) {
                  b = b + a;
        };
        std::cout.precision(std::numeric_limits<double>::digits10);
        std::cout << b << std::endl;
}

虽然如果该循环运行 5000 次而不是 50 次迭代，即使使用这种方法也会显示累积的错误 - 这就是浮点数的工作原理.

Although if that loop runs for 5000 iterations instead of 50, the accumulated error will show up even with this approach -- it's just how floating-point numbers work.

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