如何存储指向在 Fsm 中接受参数的函数的指针？

Question

我想在 FSM 中实现一个带有参数的函数。当我尝试这个时，出现这个错误

error: initializer element is not constant
         {(&DriveCenter)(86),50,{stop,right,left,stop}},
 note: (near initialization for 'fsm[0].fun')
 error: initializer element is not constant
{(&DriveRight)(45),50,{stop,right,left,stop}},

代码如下：

void DriveCenter(unsigned long out){
    printf("\ncenter = %d",out);
}

typedef struct  {
    void (*fun)(unsigned long out);
    unsigned long delay;
    unsigned long Next_State[4];
} state ;


state fsm[4] ={
        {(&DriveCenter)(86),50,{stop,right,left,stop}},
        {(&DriveRight)(45),50,{stop,right,left,stop}},
        {(&DriveLeft)(787),50,{stop,right,left,stop}},
        {(&DriveStop)(33),50,{stop,right,left,stop}}
};

Answer 1

初始化函数指针时不能设置参数，参数应声明为struct

的另一个成员

typedef struct  {
    void (*fun)(unsigned long);
    unsigned long out;
    unsigned long delay;
    unsigned long Next_State[4];
} state ;

state fsm[4] = {
    {DriveCenter,86,50,{stop,right,left,stop}},
    {DriveRight,45,50,{stop,right,left,stop}},
    {DriveLeft,787,50,{stop,right,left,stop}},
    {DriveStop,33,50,{stop,right,left,stop}}
};

在 C11 下，您可以使用匿名 structs 来说明这两个变量一起工作：

typedef struct  {
    struct {
        void (*fun)(unsigned long);
        unsigned long out;
    };
    unsigned long delay;
    unsigned long Next_State[4];
} state ;

state fsm[4] = {
    {{DriveCenter,86},50,{stop,right,left,stop}},
    {{DriveRight,45},50,{stop,right,left,stop}},
    {{DriveLeft,787},50,{stop,right,left,stop}},
    {{DriveStop,33},50,{stop,right,left,stop}}
};

如何存储指向在 Fsm 中接受参数的函数的指针？

how to store pointer to function that takes arguments in Fsm?

c

fsm