在函数中使用函数

Question

下面是我尝试运行的一段代码，其中我想要运行 (dN) 在我的主函数中 returns a complex<double>.

类型的值

#include <iostream>
#include <complex>
#include <cmath>
using namespace std;

const complex<double> Im1(0.0,1.0); //imaginary number definition

class Functions{
public:
    complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1){
        complex<double> OUT = Im1*(N[k][i]+kN)/(T1);
        return OUT;
    }; 

};

int main(int argc, const char * argv[]) {

    //...more code here

    complex<int> **NM = new complex<int>*[1000]; //1000x500 array
    //run loop to initialize
    for (int i = 0; i < 1000; ++i)
    {
        NM[i] = new complex<int>[500];
    }

    complex<double> dN_OUT = Functions::dN(**NM,1,20,0.,20.);

    return 0;
};

虽然这段代码没有运行和returns错误：Call to non-static member function without an object argument

根据我的理解，C++ 不允许 nested functions 并且我上面的代码无法运行，因为我在主函数中调用了一个单独的函数。尽管（基于 link）它似乎确实可以通过在结构中定义一个函数来实现 "local classes"，该函数必须在主函数中。虽然当我尝试这样做时：

int main(int argc, const char * argv[]) {

    complex<int> **NM = new complex<int>*[1000]; //1000x500 array
    //run loop to initialize
    for (int i = 0; i < 1000; ++i)
    {
        NM[i] = new complex<int>[500];
    }

    struct Functions{
        complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1){
            complex<double> OUT = Im1*(N[k][i]+kN)/(T1);
            return OUT;
        }; 

    };

    complex<double> dN_OUT = Functions::dN(**NM,1,20,0.,20.);

    return 0;
};

错误仍然存在。最终，我只是想在我的主函数中使用函数 dN，即 returns 类型 complex<double> 的输出，但我不确定 best/operational 的实现方式。

Answer 1

我相信你误解了什么是嵌套函数。嵌套函数如下所示：

int main()
{
    void nested() {} // not allowed in C++
}

您的问题的解决方案在您的编译器提供的错误消息中：

Call to non-static member function without an object argument

看看下面的内容：

// Example 1
struct Functions {
   void func() {}
};

int main() 
{
   // to call Functions::func() you would need to have an object
   // of type Functions because Functions::func() is not a static function
   Functions f;
   f.func();
}

// Example 2
// by making func() static you can call it without an object:
struct Functions {
   static void func() {}
};

int main() 
{
    Functions::func(); // OK
}

Answer 2

Ultimately, I am simply wanting to use the function dN that returns output of type complex within my main function, but am unsure of the best/operational way to implement this.

使用自由函数，就像 main 一样，除非 dN 有特定原因成为 class 的一部分：

complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1)
{
...
}

int main(int argc, const char * argv[]) {
    ...
    //like this 
    complex<double> dN_OUT = dN(NM,1,20,0.,20.);
    //not like this
    //complex<double> dN_OUT = dN(**NM,1,20,0.,20.);
}

Answer 3

选项 1：您可以使用 class 来完成此操作，如下所示

#include <iostream>
#include <complex>
#include <cmath>

using namespace std;

const complex<double> Im1 (0.0, 1.0); //imaginary number definition

class Functions {
public:
    complex<double> dN (complex<double> **N, int k, int i, complex<double> kN, double T1) 

    {
        complex<double> OUT = Im1*(N[k][i] + kN) / (T1);

        return OUT;
    };

};

int main (int argc, const char * argv[]) {

    //...more code here

    complex<double> **NM = new complex<double>*[1000]; //1000x500 array
                                                 //run loop to initialize
    for (int i = 0; i < 1000; ++i)
    {
        NM[i] = new complex<double>[500];
    }

    Functions fun; //create class instance 

     //call the function NOTE the changes here i.e not correct passing **NM
    complex<double> dN_OUT = fun.dN (NM, 1, 20, 0., 20.);

    return 0;
};

选项 2（其他人直接调用更改而不是 **NM 你应该使用 NM。

complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1){
    ...
}

int main(int argc, const char * argv[]) {
    ...
    complex<double> dN_OUT = dN(NM,1,20,0.,20.);
}

在函数中使用函数

Using functions within functions

c++

function

nested

complex-numbers