如何在 C++ 中使用复数“i”
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How to use complex number "i" in C++
提问by Cancan
I am coding a simple DFT algorithm now and I want to use the complex number i in complex exponential. I saw somebody use #include<complex>and #include<cmath>, and then they used the overloaded symbol Isuch as exp(2*I). But it seems it doesn't work in my visual studio compiler. So, can anyone give a simple example of using complex exponential? Thanks!
我现在正在编写一个简单的 DFT 算法，我想在复指数中使用复数 i。我看到有人使用#include<complex>and #include<cmath>，然后他们使用了重载符号，I例如exp(2*I)。但它似乎在我的 Visual Studio 编译器中不起作用。那么，谁能举一个使用复指数的简单例子？谢谢！
采纳答案by Floris
Here is a short complete example:
这是一个简短的完整示例：
#include <iostream>
#include <complex>
#include <cmath>
using namespace std;
typedef complex<double> dcomp;
main() {
dcomp i;
dcomp a;
double pi;
pi = 2 * asin(1);
i = -1;
i = sqrt(i);
a = exp(2*pi*i);
cout << "i is " << i << "and Euler was right: e(i pi) = " << a << endl;
}
Tested with g++
用 g++ 测试
回答by Mike22LFC
I get this question recently as well and find a easy way for future reader:
我最近也收到了这个问题，并为未来的读者找到了一个简单的方法：
Just use <complex>library like the following
只需使用<complex>如下所示的库
#include <iostream>
#include <complex>
using namespace std ;
int main(int argc, char* argv[])
{
const complex<double> i(0.0,1.0);
cout << i << endl ;
return(0) ;
}
回答by ogni42
回答by Ayxan
Another way is to use std::literals::complex_literals::operator""iafter C++14:
另一种方式是std::literals::complex_literals::operator""i在C++14之后使用：
#include <iostream>
#include <complex>
int main() {
using namespace std::complex_literals;
auto c = 1.0 + 3.0i;
std::cout << "c = " << c << '\n';
}
Output:
输出：
c = (1,3)
回答by Paul Mackenzie
The following code in C++ shows a macro for implementing the imaginary number j. It is well known that in programming the terms i and j are commonly used as counter variables. I instead use the capital letter J to represent the imaginary number to avoid any confusion.
以下 C++ 代码显示了一个用于实现虚数 j 的宏。众所周知，在编程中，术语 i 和 j 通常用作计数器变量。我改为使用大写字母 J 来表示虚数以避免任何混淆。
/ * dcomplex.h
/* dcomplex.h
#ifndef DCOMPLEX_H_
#define DCOMPLEX_H_
#define J dcomplex(0.0,1.0)
typedef std::complex<double> dcomplex;
#endif /* DCOMPLEX_H_ */
Using this macro, the imaginary number J [together with the complex library] can be used in the main code. An example of its use is shown below:
使用这个宏，可以在主代码中使用虚数 J [与复数库一起]。其使用示例如下所示：
....
....
#include <complex>
#include "dcomplex.h"
....
....
tmp = tmp + t[n]*exp( (2.0*PI*(double)n*(double)l/(double)tab_size)*J );
....
....
....
where tmp, t[n] are variables of a complex type, and J is the imaginary number. The variables n, l, and tab_size are of an integer type. The constant PI is the well known constant 3.14... The function exp() is overloaded to handled complex numbers. [n.b. this code sample is part of a simple DFT]
其中 tmp、t[n] 是复数类型的变量，J 是虚数。变量 n、l 和 tab_size 是整数类型。常数 PI 是众所周知的常数 3.14... 函数 exp() 被重载以处理复数。[nb 此代码示例是简单 DFT 的一部分]
Using this macro, the code is more readable..
使用这个宏，代码更易读..
回答by user3405743
pi, being an irrational number, cannot be exactly represented by a double. cosof an inexact approximation of pi is likely to yield a result which is close to but perhaps not exactly 1. Likewise sinof an inexact approximation of an inexact approximation of piis like to result in a number is has a very small magnitude that is perhaps not exactly 0. Why not just define I to be std::complex(0.0, 1.0)and avoid the gratuitous inexactness.
pi是一个无理数，不能用double精确表示。 pi 的不精确近似值的cos可能会产生接近但可能不完全是 1 的结果。同样，pi的不精确近似值的不精确近似值的sin就像导致一个数字的大小非常小，即也许不完全是 0。为什么不将 I 定义为std::complex(0.0, 1.0)并避免无端的不精确。
