如何计算 C++ 中的欧拉常数或欧拉？
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How to calculate Euler constant or Euler powered in C++?
提问by Tony
I am trying to find the more 'natural' way to use the number ein C/C++. I am focused on calculating the function e^n.
我试图找到在 C/C++ 中使用数字e的更“自然”的方式。我专注于计算函数e^n。
I think that 'cmath', by default, does not provide support for both (function and constant). However, it can be enabled to include constants defined by the compiler, in this case, M_E. This can be done by including the statement #define _USE_MATH_DEFINES.
我认为“cmath”默认情况下不支持（函数和常量）。但是，可以启用它以包含编译器定义的常量，在本例中为M_E. 这可以通过包含语句来完成#define _USE_MATH_DEFINES。
On the other hand, ecan be defined as a constant:
另一方面，e可以定义为常数：
#define E 2.71828182845904523536;
or
或者
const double EULER = 2.71828182845904523536;
Said this. Which one is the most 'standard' way to approach to this mathematical constant? Is it any other library?
说了这个。哪个是接近这个数学常数的最“标准”的方法？是其他图书馆吗？
回答by IInspectable
If you can avoid using a preprocessor symbol you should. It will cause you trouble when you least expect it. Eis likely going to be a variable.
如果您可以避免使用预处理器符号，则应该这样做。它会在您最不期望的时候给您带来麻烦。E很可能会成为一个变量。
Proposed solution:
建议的解决方案：
#include <cmath>
const double EulerConstant = std::exp(1.0);
The advantage of calculating the constant instead of assigning a floating point literal is that it will produce a result with precision that matches the precision of the doubledata type for your particular C++ implementation. And it removes the possibility of introducing an error by accidentally skipping a digit.
计算常量而不是分配浮点文字的优点是它会产生一个精度与double特定 C++ 实现的数据类型精度相匹配的结果。它消除了因意外跳过数字而引入错误的可能性。
As illustrated above, <cmath>does declare std::exp, so there is no need for you to roll your own.
如上所示，<cmath>确实声明了std::exp，所以你不需要自己滚动。
