将成员函数传递给静态函数

问题描述

Math.h

Math.h

class Math
{
    public:
    static double integrate(double (*f)(double x), double a, double b);
};

Math.cpp

Math.cpp

#include "Math.h"

double Math::integrate(double (*f)(double x), double a, double b)
{
    int m = 1000;
    double dt = (b-a)/m;
    double c = 0;

    for(int i = 0; i<m;>    {
        c += (*f)(a+i*dt);
    }

    return c/m;
}

型号.cpp

Model.cpp

...
double Model::i_L_charging(double t)
{
    return V_in/(R_L+RDSON) - V_in/(R_L+RDSON)*exp(-t/tau_charging());
}

double Model::I_d_avg()
{
    return math.integrate(&i_L_charging, 0.0, T_charging()); //Error is here
}
...

缩短的错误是(GNU GCC编译器):

||在成员函数`double Model :: I_d_avg()'':|

| 45 |错误:ISO C ++禁止使用不合格或带括号的非静态成员函数的地址来形成指向成员函数的指针.说`& Model :: i_L_charging''|

| 45 |错误:没有匹配函数可调用`Math :: integrate(double(Model :: *)(double)，double，double)''|

| 5 |注意:候选对象是:静态double Math :: integrate(double(*)(double)，double，double)|

|| ===构建完成:2个错误，0个警告=== |


就像测试一样，我将"sqrt"放入函数中，而不是"i_L_charging"，并且效果很好.这里发生了什么?

感谢您的帮助.

问候，
Johnny

The shortend error is (GNU GCC Compiler):

||In member function `double Model::I_d_avg()'':|

|45|error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say `&Model::i_L_charging''|

|45|error: no matching function for call to `Math::integrate(double (Model::*)(double), double, double)''|

|5|note: candidates are: static double Math::integrate(double (*)(double), double, double)|

||=== Build finished: 2 errors, 0 warnings ===|


Just as a test I put "sqrt" in as the function instead of "i_L_charging" and it works perfectly. What is going on here?

Your help is appreciated.

Regards,
Johnny

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