如何存储指向Fsm中接受参数的函数的指针?

问题描述 投票:1回答:1

我想在FSM中使用参数实现一个函数。当我尝试这个时,会出现此错误

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}},

这是代码:

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}}
};
c fsm
1个回答
1
投票

初始化指向函数的指针时无法设置参数,该参数应声明为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来阐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}}
};
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