Question

我正在尝试将main中声明的变量放入我的类的私有变量中，而不将其作为构造函数的参数传递。我需要将中断控制器链接到多个硬件中断，而无需重新初始化中断实例，从而覆盖它。

XScuGic InterruptInstance;

int main()
{
    // Initialize interrupt by looking up config and initializing with that config
    ConfigPtr = XScuGic_LookupConfig(INTERRUPT_DEVICE_ID);
    XScuGic_CfgInitialize(&InterruptInstance, ConfigPtr, ConfigPtr->BaseAddr);

    Foo foo;
    foo.initializeSpi(deviceID,slaveMask);

    return 0;
}

和Foo类的实现：

class Foo
{
     // This should be linked to the one in the main
     XScuGic InterruptInstance;
public:
     // In here, the interrupt is linked to the SPI device
     void initializeSpi(uint16_t deviceID, uint32_t slaveMask); 
};

deviceID和slaveMask在包含的标头中定义。

有没有办法实现这个目标？

Answer 1

您可以使用使用全局变量的构造函数初始化私有类引用成员，因此无需在构造函数中传递它：

XScuGic InterruptInstance_g; // global variable

class Foo {
private:
    const XScuGic& InterruptInstance; // This should be linked to the one in the main
public:
    Foo() : InterruptInstance{InterruptInstance_g} {}; // private variable is initialized from global variable
    void initializeSpi(uint16_t deviceID,uint32_t slaveMask); // In here, the interrupt is linked to the SPI device
};

int main()
{
    // Initialize interrupt by looking up config and initializing with that config
    ConfigPtr = XScuGic_LookupConfig(INTERRUPT_DEVICE_ID);
    XScuGic_CfgInitialize(&InterruptInstance,ConfigPtr,ConfigPtr->BaseAddr);

    Foo foo{}; // reference is not required as it will come from the global variable to initialize the private reference member
    foo.initializeSpi(deviceID,slaveMask);

    return 0;
}

C ++将外部变量导入私有类变量

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