C++ 将指针强制转换为静态方法

Question

我好久没写C++代码了；但是现在我必须在 texas instruments F28335 DSP 上工作，并且我正在尝试从 C 迁移到 C++。我有以下代码试图用类的静态方法初始化中断服务例程:

//type definition for the interrupt service routine
typedef interrupt void (*PINT)(void);
//EPWMManager.h
class EPWMManager
{
public:
    EPWMManager();      
    static interrupt void Epwm1InterruptHandler(void);  
};
//EPWMManager.cpp
interrupt void EPWMManager::Epwm1InterruptHandler(void)
{
 //some code to be called on interruption
}   
//main.cpp
int main(void)
{
    PINT p;
    p = &(EPWMManager::Epwm1InterruptHandler);
    return 0;
 }

编译时我得到以下信息:

error: a value of type "void (*)()" cannot be assigned to an entity of type "PINT"

我想我错过了一些类型转换。

Answer 1

我认为最根本的问题是 & 符号在你赋值给 p 的 RHS 前加上前缀。此外，“PINT”在其他操作系统中是“指向整数的指针”。因此，让我们避免任何潜在的名称冲突。但我认为这对你有用:

// you may have to move "interrupt" keyword to the left of the "void" declaration.  Or just remove it.
typedef void (interrupt *FN_INTERRUPT_HANDLER)(void);

interrupt void EPWMManager::Epwm1InterruptHandler(void)
{
 //some code to be called on interruption
}  

int main(void)
{
    FN_INTERRUPT_HANDLER p;
    p = EPWMManager::Epwm1InterruptHandler; // no ampersand

    // and if for whatever reason you wanted to invoke your function, you could just do this:

   p(); // this will invoke your function.

    return 0;
}